physics secondary stage 2
DESCRIPTION
scienceTRANSCRIPT
™HÉ£e
Book Sector
2011 - 2012
Arab republic of egyptMinistry of Education
Book Sector
Foreword:Unit 1: Waves:
Chapter 1 : Wave Motion Chapter 2 : Sound Chapter 3 : Light
Unit 2: Fluid Mechanics: Chapter 4 : HydrostaticsChapter 5 : Hydrodynamics
Unit 3 : Heat:Chapter 6 : Gas LawsChapter 7 : Kinetic Theory of Gases Chapter 8 : Cryogenics (Low Temperature Physics)
Unit 4 : Dynamic Electricity and Electromagnetism:Chapter 9 : Electrical Current and Ohm's Law Chapter 10 : Magnetic Effects of Electric Current
and Measuring Instruments.Chapter 11 : Electromagnetic Induction.
Unit 5 : Introduction to Modern Physics:Chapter 12 : Wave Particle Duality Chapter 13 : Atomic Spectra Chapter 14 : Lasers Chapter 15 : Modern Electronics
General Revision : Appendixes :
Appendix 1 : Symbols and Units of Physical Quantities Appendix 2 : Fundamental Physical ConstantsAppendix 3 : Standard Prefixes Appendix 4 : Greek Alphabet Appendix 5 : Gallery of ScientistsAppendix 6 : Selected Physics Sites on the Internet
Table of Contents
1 : 7922347
81 : 13282117
134 : 182135159173
184 : 271185202
231273 : 382
274306324350
383 : 404407:418
406409411412413418
intensively. It is targeted to use genetics, atoms and lasers in the computer of the
future. It is a limitless world, enriched by imagination, where sky is the limit.
The scientific progress is a cumulative effort. This collective endeavor has led to
where we are today. A scholar of physics must be acquainted with such accumulated
knowledge in a short time, so that he could add to it within the limited span of his
life. In studying what others have found, we must skip details and trials, and extract
the end results and build on them.A global view is, therefore, more important at this
stage than being drowned in minute details that could be postponed to a later stage of
study.
This book is divided into 5 units. Unit 1 deals with waves, which are the basis of
communication in the universe. (Chapter 1) deals with wave motion, (chapter 2) with
sound, and (chapter 3) with light. Unit 2 deals with fluid mechanics, : hydrostatics
(chapter 4) and hydrodynamics (chapter 5). Unit 3 deals with heat, where (chapter 6)
deals with gas laws, (chapter 7) with the kinetic theory of gases and (chapter 8) deals
with low temperature physics. Unit 4 treats electricity, where (chapter 9) covers the
electric current and Ohm’s law, (chapter 10) covers the magnetic effects of electric
current and measuring instruments, while (chapter 11) covers electromagnetic
induction. Unit 5 gives an introduction to modern physics, where (Chapter 12) deals
Physics is the cornerstone of basic sciences. It deals with the understanding of
nature and what goes around us, big and small in this universe. It is the root of all
sciences. Interwined with it is chemistry which focuses on reactions between
materials, biology which deals with living creatures, geology which is involved with
the layers of the Earth, and astronomy which treats celestial objects. But in the end,
physics remains the mother of all sciences and the basis for the tremendous present
scientific and technological progress. Understanding physics means understanding
the laws governing this universe. Such understanding has led to the current industrial
development spearheaded by the West. The Arabs and Moslems were once the
pioneers of civilization in the world when they realized the importance of
understanding the laws of this universe. We owe them the discovery of most laws of
physics centuries before the West. The foundations of medicine, physics, chemistry,
astronomy, mathematics and music were all laid by Arab and Moslem scientists.
In fact, understanding physics and its applications converts a poor, and
underdeveloped society into an affluent and developed one. This has taken place in
Europe, US, Japan and South East Asia. Computers, satellites, cellular (mobile)
phones, and TV are all byproducts of physics. Genetics is currently being looked into
Foreword
with wave particle duality, (Chapter 13) deals with atomic spectra, and (chapter 14)
deals with lasers and their applications, while (chapter 15) covers modern electronics.
Suzanne Mubarak Science Exploration Center has carried out the preparation, and
the typing of manuscript as well as the design of the artwork.
In the end, we want the student to take liking to physics. For this is the way to the
future. We want the teacher to teach the subject of physics in an innovative way, to
arouse the interest of the students by constantly referring to the use and applications
of physics in the daily life. We hope that one day we will have great inventors and
industrialists among today's students.
Committee for the preparation of this new version
of the textbook.
Prof. Mustafa Kamal Mohammad Yussef,Ph.D.Prof. Mohammad Sameh Said ,Ph.D.Mustafa Mohammad El-Sayed ,Ph.D.Tarik Mohammed Tala'at Salama,Ph.D.Karima Abdel-Alim Sayed Ahmad
Un
it 1: W
ave
s C
ha
pte
r 1: W
ave
Mo
tion
electromagnetic waves spreading in space and the surrounding medium . When received bythe mobile antenna at the receiver, electromagnetic waves are transformed back intoelectrical signals and then to sound or even to an image.
We can see water waves but we cannot see the radio, TV or mobile waves. However,we can detect them. Water waves are mechanical waves, so are sound waves and waves invibrating strings. But radio, TV, and mobile waves are electromagnetic waves. Amongthese electromagnetic (em) waves, there are, for example, light waves and X-rays whichare used in radiology. Mechanical waves require a medium to propagate through, while(em) waves do not require a medium. They can propagate in space.
Mechanical Waves Mechanical waves require the following :-
1 ) a vibrating source.2 ) a disturbance transmitted from the source to the medium.3 ) a medium that carries a vibration.
There are many forms of vibrating sources :1 ) a simple vibrating pendulum (Fig 1 - 2).2 ) a tuning fork (Fig 1 - 3).3 ) a vibrating stretched wire (or string) (Fig 1- 4).4 ) a plumb (bob) attached to a vibrating spring (Yoyo) (Fig 1 - 5).
3
Un
it 1
:
W
ave
s
Ch
ap
ter
1:
W
ave
Mo
tio
n
Overview :
Many of us enjoy watching waves on the surface of water pushing a fishing float or aboat up and down, or even making waves by throwing a pebble in a pond or still water.Each pebble becomes a source of disturbance in the water, spreading waves as concentriccircles (Fig 1-1). Hence, waves are disturbances that spread and carry along energy.
Waves are not only water waves. There are, for example, radio waves. We often hear theannouncer say: "This is Radio Cairo on the medium wave 366.7 m". Also, TV stationstransmit both sound and image in the form of waves which are received by the aerial(antenna) . Such waves are transformed into electrical signals in the receiver, where theyare eventually converted back to sound (audio) and image (video). Also, the mobile phoneruns on waves. Sound signals are transformed into electrical signals then into
2
Fig (1 -1)Waves spreading from a
point source
Chapter 1 Wave Motion
Un
it 1: W
ave
s C
ha
pte
r 1: W
ave
Mo
tion
5
Amplitude (A) (meter): is the maximum displacement of the vibrating object or the
distance between two points along the path of the object, where the velocity at one point is
maximum and zero at the other.
Complete Oscillation: is the motion of a vibrating body in the interval between theinstants of passing by one point along the path of its motion twice successively withmotion in the same direction and same displacement, i.e., at the same phase, relative to thestarting point of motion.
Frequency (ν) (Hertz or Hz): is the number of complete oscillations made by
a vibrating body in one second.
Periodic Time (T) (seconds): is the time taken by a vibrating body to make onecomplete oscillation, or the time taken by the vibrating body to pass by the same pointalong the path of motion twice successively with motion in the same direction and thesame displacement.
Simple Harmonic Motion: A vibrational motion in its simplest form is
called a simple harmonic motion, e.g., a
swing (Fig 1-6) or a simple pendulum (Fig
1-7). The vibration starts from point "a" then
increases to a positive maximum at "b" then
to zero at "a" then to negative maximum at
"c" then to zero at "a". and the cycle is
repeated continually (Fig 1-7 a ).
ν (1-1)1T=
Fig (1-6)A swing as an example ofa simple harmonic motion
Un
it 1
:
W
ave
s
Ch
ap
ter
1:
W
ave
Mo
tio
n
Fig (1 -5)
Yoyo
rest position
amplitude
4
To study vibrations, we need to define some relevant physical quantities such as:displacement, amplitude, complete oscillation, periodic time and frequency as follows:
Displacement (meter): is the distance of a vibrating body at any instant from its restposition or its equilibrium origin. It is a vector quantity.
Fig (1 -2)A pendulum
Fig (1 -3)A tuning fork
Fig (1 -4)A vibrating string
Un
it 1
:
W
ave
s
Ch
ap
ter
1:
W
ave
Mo
tio
n
Fig (1 -5)
Yoyo
rest position
amplitude
4
To study vibrations, we need to define some relevant physical quantities such as:displacement, amplitude, complete oscillation, periodic time and frequency as follows:
Displacement (meter): is the distance of a vibrating body at any instant from its restposition or its equilibrium origin. It is a vector quantity.
Fig (1 -2)A pendulum
Fig (1 -3)A tuning fork
Fig (1 -4)A vibrating string
Un
it 1
:
W
ave
s
Ch
ap
ter
1:
W
ave
Mo
tio
n
Fig (1 -5)
Yoyo
rest position
amplitude
4
To study vibrations, we need to define some relevant physical quantities such as:displacement, amplitude, complete oscillation, periodic time and frequency as follows:
Displacement (meter): is the distance of a vibrating body at any instant from its restposition or its equilibrium origin. It is a vector quantity.
Fig (1 -2)A pendulum
Fig (1 -3)A tuning fork
Fig (1 -4)A vibrating string
Un
it 1
:
W
ave
s
Ch
ap
ter 1
:
Wa
ve
Mo
tio
n
Fig (1 -5)
Yoyo
rest position
amplitude
4
To study vibrations, we need to define some relevant physical quantities such as:displacement, amplitude, complete oscillation, periodic time and frequency as follows:
Displacement (meter): is the distance of a vibrating body at any instant from its restposition or its equilibrium origin. It is a vector quantity.
Fig (1 -2)A pendulum
Fig (1 -3)A tuning fork
Fig (1 -4)A vibrating string
Un
it 1: W
ave
s C
ha
pte
r 1: W
ave
Mo
tion
7
Learn at Leisure
ResonanceAt a certain frequency, the amplitude of the
mechanical vibration may get out of hand, e.g., thecrushing of a glass cup due to nearby sound waves (Fig1-8), and the collapse of Tacoma bridge (USA) due tostrong winds in November 1940 (Fig 1-9). This conditionis called resonance. It is the cause of the collapse ofmany buildings. It occurs when a simple harmonicmotion is set at the natural (or resonant) frequency of thebuilding. Similar to mechanical resonance, there is alsoelectrical resonance which is the basis of tuning a radioor a TV receiver to a certain station, where one out ofmany electrical signals picked up by the aerial isamplified and made to coincide with the resonant frequency ofthe amplifier in the receiver when tuned to that particularstation.
Fig (1-9)Collapse of Tacoma bridge
(USA) due to the wind causingthe vibration of the bridge at thenatural (resonant) frequency of
the bridge
Fig (1-8)A glass cup crushed dueto nearby sound waves
Un
it 1
:
W
ave
s
Ch
ap
ter
1:
W
ave
Mo
tio
n
Fig (1-7 a)Displacement of a
pendulum bob as timegoes by
Fig (1-7 b)Displacement of a pendulum bob at different phases A, D are of the
same phase (same displacement and direction)B,C are not of the same phase (the same direction but not the
same displacement)
a bc
rest position
6
Un
it 1: W
ave
s C
ha
pte
r 1: W
ave
Mo
tion
5
Amplitude (A) (meter): is the maximum displacement of the vibrating object or the
distance between two points along the path of the object, where the velocity at one point is
maximum and zero at the other.
Complete Oscillation: is the motion of a vibrating body in the interval between theinstants of passing by one point along the path of its motion twice successively withmotion in the same direction and same displacement, i.e., at the same phase, relative to thestarting point of motion.
Frequency (ν) (Hertz or Hz): is the number of complete oscillations made by
a vibrating body in one second.
Periodic Time (T) (seconds): is the time taken by a vibrating body to make onecomplete oscillation, or the time taken by the vibrating body to pass by the same pointalong the path of motion twice successively with motion in the same direction and thesame displacement.
Simple Harmonic Motion: A vibrational motion in its simplest form is
called a simple harmonic motion, e.g., a
swing (Fig 1-6) or a simple pendulum (Fig
1-7). The vibration starts from point "a" then
increases to a positive maximum at "b" then
to zero at "a" then to negative maximum at
"c" then to zero at "a". and the cycle is
repeated continually (Fig 1-7 a ).
ν (1-1)1T=
Fig (1-6)A swing as an example ofa simple harmonic motion
Un
it 1
:
W
ave
s
Ch
ap
ter
1:
W
ave
Mo
tio
n
Longitudinal Waves:Imagine a mass “m” on a smooth horizontal surface attached to one end of a spring
whose other end is attached to a vertical wall. If we pull the mass in the direction of thespring and let it go, the mass moves around its rest position in an oscillatory motion towardthe spring and away (Fig 1-10). This is a simple harmonic motion. If we draw the curvethat the center of gravity of the mass makes with respect to its rest position, we will obtaina sine wave (Fig 1-11). This is what distinguishes a simple harmonic motion from anyother type of motion.
Fig (1-10)A vibrating spring
Fig (1-11)A sine wave resulting from a
simple harmonic motion
rest position
pulling the spring
releasing the spring
8
Un
it 1: W
ave
s C
ha
pte
r 1: W
ave
Mo
tion
Fig (1-13 b)Vertical displacement as a sine wave
Fig (1-14)A pulse resulting from part of a simple harmonic motion
spreading along a stretched rope
You can do this experiment yourself by using a long stretched rope. The far end isattached to a vertical wall while the near end is in your hand. When you move your hand upand down in the form of a pulse, you note that the wave spreads in a pulse form along therope. This is known as a traveling wave (Fig 1-14).
v
v
distance
vert
ical
dis
plac
emen
t
11
Un
it 1
:
W
ave
s
Ch
ap
ter
1:
W
ave
Mo
tio
n
Fig (1-13 a)Vertical displacement in a simple harmonic motion
Thus, a vibrating source making a simple harmonic motion may generate a wavepropagating at velocity “v”. Each particle of the medium performs, in turn, a simpleharmonic motion about its equilibrium position. An example of this motion is thelongitudinal waves of sound in air.
Transverse Waves:
Imagine a mass “m” attached to a vertical spring. A long horizontal taut (stretched) rope
is also attached to this mass at the near end, while the other (far) end of the rope is attachedto a vertical wall.
When the mass “m” performs a simple harmonic motion in the vertical direction, thenthe near end of the rope performs the same motion. Consequently, the following parts ofthe rope do the same thing successively. Then the motion transfers horizontally along therope in the form of a wave at velocity “v”, while the other parts of the rope oscillatevertically in a simple harmonic motion about their rest positions. This wave is called atransverse wave (Fig 1-13 ).
10
Un
it 1: W
ave
s C
ha
pte
r 1: W
ave
Mo
tion
In conclusion, we may classify mechanical waves into two types:
1 ) Transverse waves
2 ) Longitudinal waves
In transverse waves, the particles of the medium oscillate about their equilibrium
positions in a direction perpendicular to the direction of the propagation of the wave.
In longitudinal waves, the particles of the medium oscillate about their equilibrium
positions along the direction of the propagation of the wave.
The work done by the oscillating source is converted to the particles of a string (or a
stretched rope) in the form of potential energy stored as tension in the string and kinetic
Fig (1-16)A vibrating spring forming a
longitudinal wave
Fig (1-17)A vibrating spring forming a
transverse wave
13
Un
it 1
:
W
ave
s
Ch
ap
ter
1:
W
ave
Mo
tio
n
A wave may also be continuous (called a traveling wave train) as long as the simple
harmonic motion of the source keeps on(Fig 1-15).
The stretched rope may be replaced by a spring in which a longitudinal wave (Fig 1-16)
or a transverse wave (Fig 1-17) may be generated . We conclude that as a source oscillates ,
the particles of the medium oscillate successively in the same way. The vibration transfers
first from the source to the particle of the medium next to it, then into the one connected to
it, then into the following ones and so on. Thus, the vibration or disturbance forms a wave,
since the wave is nothing but a disturbance (or energy) on the move along which energy is
carried through .
12
Fig (1-15)A train wave spreading in a taut (stretched) rope due to a
continuous simple harmonic motion at the near end
Un
it 1: W
ave
s C
ha
pte
r 1: W
ave
Mo
tion
15
Fig (1-19)Wavelength in a transverse wave
wavelengthdirection of
propagationcrest
rest positionamplitude
direction ofvibration
Fig (1-20)Wavelength in a longitudinal wave
Thus, the wavelength is the distance between two successive points of the same phase
(Fig 1-21). Alternatively, it is the distance which the wave travels during one periodic
time (Fig 1-22).
The number of waves passing by a certain point along the wave path in one second is
called frequency.
disp
lacm
ent
Un
it 1
:
W
ave
s
Ch
ap
ter
1:
W
ave
Mo
tio
n
energy manifested in the vibration of the particles of
the string.
Referring to (Fig 1-18), the points at maximum
upward displacement in the positive direction are
called crests, while the points of maximum downward
displacement are called troughs.
Observing any part of a vibrating string carrying a
transverse wave, we find that it has one crest and one
trough during one complete oscillation .
Frequency (ν) (Hertz) and wavelength (λ) (meter):
The distance between two successive crests or two
successive troughs in a transverse wave is called
wavelength (Fig 1-19). Similarly, the distance
between two successive contractions (compressions)
or two successive rarefactions in a longitudinal wave
is called wavelength (Fig 1-20).
Thus, we may represent the wavelength by either
of the two distances (AC) and (BD)(Fig 1-19 ). It is
to be noted that the two successive pairs of points (A,C) and (B, D) move in the same
way at the same time. We say they have the same phase, i.e.,the same displacement in
the same direction.
14
Fig (1-18)A piece of foam floating on the top ofa wave (crest) or at bottom (trough)
direction of wave propagation
Un
it 1: W
ave
s C
ha
pte
r 1: W
ave
Mo
tion
The relation between frequency,wavelength and velocity of propagation:If a wave travels at velocity “v”, a distance equal to the wavelength “λ” , then the wave
takes time equal to the periodic time “T” to travel this distance.
This is a general relation for all types of waves.In all cases , within a periodic time “T” a wave travels a wavelength. Frequency is the number of oscillations in one second or the number of wavelengths
traveled by a wave propagating in a certain direction in one second.
v = T
T = 1
= 1T
ν
ν
Fig (1-22b)A train of waves at velocity “v” generated by a vibrator
v = λν
17
Un
it 1
:
W
ave
s
Ch
ap
ter
1:
W
ave
Mo
tio
n
Fig (1-22a)The distance which a wave moves
in a periodic time T is the wavelength
Fig (1-21)The distance after each one full vibration
completed in one period T is the wavelength
rest position
at the end of 1/4 period
at the end of 1/2 period
at the end of 3/4 period
at the end of one periodamplitude
at time t + T
at time t
16
Un
it 1: W
ave
s C
ha
pte
r 1: W
ave
Mo
tion
In a Nutshell.A wave is a disturbance which spreads and carries energy along. .Displacement is the distance of an object at any instant from its rest(equilibrium) position.. The amplitude of oscillation “A” is the maximum displacement of an oscillating objectfrom its rest position, or the distance between two points along the path of the oscillatingobject where the velocity at one point is maximum and at the other is nil. . A complete oscillation is the movement of a continuously vibrating body ( e.g. a simplependulum) is the interval between the instants of time as it passes by a certain pointalong its path twice successively with motion in the same direction.. Frequency “ν” is the number of complete oscillations produced by a vibrating object inone second and is equal to the inverse of the periodic time.
Frequency =
It is also the number of waves passing by a certain point along the path of a wave in onesecond.. Periodic time “T” is the time taken by a continuously vibrating body to perform one
complete oscillation, or the time taken by a continuously vibrating body ( e.g. a simplependulum ) to pass by a point along its path twice successively with motion in the samedirection.. Mechanical waves are either:1 ) transverse waves.2 ) longitudinal waves.. Transverse waves are waves in which the particles of a medium oscillate about theirequilibrium positions in a direction perpendicular to the direction of propagation of thewave.. Longitudinal waves are waves in which the particles of a medium oscillate about theirequilibrium positions along the same path of propagation of the wave.. Transverse waves comprise crests and troughs in succession.
1Periodic time
19
Un
it 1
:
W
ave
s
Ch
ap
ter
1:
W
ave
Mo
tio
n
Examples:-1) If the wavelength of a sound wave produced by a train is 0.6 m and the frequency is
550 Hz,what is the velocity of sound in air?
Solution:-
v = λ νv = 0.6 x 550 = 330 m/s
2) If the number of waves passing by a certain point in one second is 12 oscillations andthe wavelength is 0.1 m, calculate the speed of propagation.
Solution:-
v = λ νv = 12 X 0.1 = 1.2 m/s
3) Light waves propagate in space at speed 300 000km/s (3x108m/s), and the wavelengthof light is 5000 A˚ .What is the frequency of this light ?
1 Angstrom(A0 ) =10-10 m
Solution:-
c = v = 3 x 108 m/s
λ = 5 x 103 x 10-10 = 5 x10-7 m
c = λ ν
3 x 108 = 5 x 10-7 x ν
Hz = 3 x 108
5 x 10-7 = 6 x 1014ν
18
Un
it 1: W
ave
s C
ha
pte
r 1: W
ave
Mo
tion
Questions and DrillsI) Define
Wave - Transverse Wave - Longitudinal Wave - Wavelength
II) Complete:a) Displacement is ......b) Amplitude of oscillation is ......c) Complete oscillation is ......d) Periodic time is ......e) Frequency is ......
III) Essay question:Deduce the relation between frequency, wavelength and velocity of wave propagation.
IV) Put a tick sign (√) next to the right choice in the following :1) The relation between the velocity of propagation of the waves “v” in a medium , its
frequency and wavelength is :a) v = λν b) v = ν / λc) v = d) none (there is no correct answer)
2) Transverse waves are waves consisting of :a) Compressions and rarefactions b) Crests and troughsc) Crests and troughs, where the particles of the medium move short distances about
their equilibrium positions in a direction perpendicular to the direction of propagation.d) Compressions and rarefactions, where the particles of the medium move short
distances about their equilibrium positions along the direction of propagation of thewave .
3) If the wavelength of a sound wave produced by an audio ( sound producing) source is0 . 5 m , the frequency is 666 Hz ,then the velocity of propagation of sound in air is :a) 338 m / s b) 333 m / s c) 330 m / s d) 346 m / s
21
λν
Un
it 1
:
W
ave
s
Ch
ap
ter
1:
W
ave
Mo
tio
n
. Longitudinal waves comprise compressions and rarefactions in succession .. Wavelength is the distance between two successive points along the direction ofpropagation of the wave, where the phase is the same (same displacement and samedirection).. The relation between frequency, wavelength and velocity of a wave is given by: v = λ ν
20
Un
it 1
:
W
ave
s
Ch
ap
ter
1:
W
ave
Mo
tio
n
4) If the velocity of sound in air is 340 m/s, for a sound of frequency (tone) 225 Hz , thewavelength(m) is :
a) 4/3 b) 3/4 c) 20 d) 3/2 5) Light of wavelength 6000 A˚(1A˚ = 10-10m ) propagates in space at velocity 300 x 103km/s,
its frequency is:(a) 4 x 1010 Hz (b) 4 x 1014 Hz
(c) 5 x 1014 Hz (d) 5 x 1012 Hz
6) Two waves whose frequencies are 256 Hz and 512 Hz propagate in a certain medium ,the ratio between their wavelengths isa) 2/1 b) 1/2 c) 3/1 d) 1/3
22
25
Un
it 1
: W
ave
s C
ha
pte
r 2
:
Sound
Learn at Leisure
How does a bat see ?
A bat (Fig 2-1) does not see by its eyes. It transmits ultrasonic waves and detects thesurroundings by the echo. It works in this fashion as a radar or rather sonar (radar usingultrasonic waves).
Learn at Leisure
Imaging the embryoUltrasonic waves (not heard by human ear) are used to image an embryo. They are
considered the safest. Ultrasonic imaging depends on the reflections of sound (Fig 2-2 ).
Fig (2-2)Use of ultrasonic waves for imaging an embryo
Fig (2-1)A bat (ultrasonic radar)
24
Un
it 1
: W
ave
s C
ha
pte
r 2
:
So
un
d
Overview :Sounds and tones are produced due to the vibration of objects. Such vibrations travel
through air or any other medium in all directions. When these vibrations reach the ear,they are transmitted through the auditory nerve, then the brain translates them to soundsand tones
Reflection and Refraction of Sound:Firstly: Reflection of Sound:
When a loud sound is produced at a suitable distance from a wall or a mountain, asound is heard back resembling the original one. It is generated due to the reflection fromthe wall or the mountain, appearing as if it were coming from behind the wall or themountain. This sound is called echo. Hence, echo is the repetiton of sound produced dueto reflection.
Sound waves propagate in air in the form of concentric spheres of successivecompressions and rarefactions, whose center is the source of the sound . If these waves areobstructed by a large obstacle, they are reflected back also in the form of concentricspheres of compressions and rarefactions,whose center appears as if it lay behind thereflecting surface and at a distance equal to the distance of the original source from thatsurface. According to the laws of reflection:
1) the angle of reflection is equal to the angle of incidence.2) the incident ray, the reflected ray and the normal to the surface at the point of
incidence all lie in one plane normal to the reflecting surface.Note that: the sound ray is a straight line indicating the direction of propagation of the
sound wave.
Chapter 2 Sound
27
Un
it 1
: W
ave
s C
ha
pte
r 2
:
Sound
Fig (2-4)Sound travels easier at night
Learn at Leisure
Why does sound travel easier at night than during the day ? The velocity of sound in the air depends on the temperature of the air, since sound
waves propagate in hot air faster than in cold air. When sound travels between two layersof air of different temperatures, it undergoes refraction (Fig 2 - 4). The figure illustratesthe decrease of sound intensity as heard by an observer at a certain distance due to therefraction of sound upwards (leaking away), due to the increase of temperature on a hotday. At night, the sound is heard for longer distances due to the decrease of thetemperature of the air adjacent to the surface of the Earth compared to layers above.Therefore, sound at night is refracted more toward the surface of the Earth.
source observer
cold air
hot air
at daytime
hot air
cold air
ground
ground
at night
26
Un
it 1
: W
ave
s C
ha
pte
r 2
:
So
un
d
Fig (2-3)Refraction of sound
v1
v2
Normal Normal
Hot air
Cold air
sinsin
vv
1
2
φΘ
=θ
Secondly: Refraction of soundWhen sound falls on a surface between two media, part of it is reflected back to the first
medium according to the laws of reflection, while the rest is transmitted to the secondmedium, deviating from its original path (Fig 2-3 ). Refraction of sound - upon transmittingfrom one medium to another - depends on the velocity of sound in these two media. Inother words :
This means that when the velocity of sound in the first medium v1 is greater than thevelocity of sound in the second medium v2, the sound refracts nearer to the normal, i.e., φ > θand vice versa. It is to be noted that the velocity of sound in gases decreases as their densityincreases, while in liquids and solids the velocity is affected by another factors,which ismore effective than denisty.
(2-1)
29
Un
it 1
: W
ave
s C
ha
pte
r 2
:
Sound
Fig (2-7 b)Interference pattern
Fig (2-7 a)Interference fringesbetween two waves
Such sources may be obtained, for example, using two speakers for the same electricalsource (Fig 2-5). The connected arcs in the figure represent the positions of the maxima ofcompressions, while the dashed arcs represent the maxima of rarefactions. The distancebetween any two successive arcs or any two successive dashed arcs is the wavelength.
Due to the combination of two waves of equal frequency and amplitude (Fig 2-6), somepoints or regions exist where the compressions of the first source intersect thecompressions of the second source, and the rarefactions of the first source intersect therarefactions of the second source. In both cases, the path difference must equal m λ, wherem is an integer . Therefore, at such positions we have constructive interference, where theintensity of the wave increases (Fig 2-6 a). There are also points or regions where thecompressions of the first source intersect rarefactions of the second source or vice versa.Therefore, the path difference equals (m + ) λ , where m is an integer, then we havedestructive interference and the intensity of the wave diminishes to zero (Fig 2-6 b).
Learn at Leisure
Can we see the interference of sound waves ?The experiment of sound interference is similar to the ripple tank experiment, where
two sources vibrate and produce mechanical waves. These waves interfere producingregions of constructive interference and regions of destructive interference ( Fig 2-7).
12
28
Un
it 1
: W
ave
s C
ha
pte
r 2
:
So
un
d
Fig (2-6 a)Constructive interference of the waves
resultant wave first wave
second wave of thesame phase
second wave at180˚ phase shift Fig (2-6 b)
Destructive interference of the waves
resultant wave first wave
Fig (2-5)Formation of constructive and destructive
fringes due to two sound sources
lines of constructive interference(maximum sound intensity)
lines of destructive interference (minimum sound intensity)
source 2source 1
Interference and diffraction of sound Firstly: Interference of sound
Interference is a combination of two waves or more of the same frequency, amplitude,and direction of propagation. Interference may be constructive (strengthing the intensity) ordestructive (weakening the intensity or eliminating it altogether). Fig(2-5) demonstrates theinterference of sound waves, which are longitudinal waves consisting of compressions andrarefactions. The two sources S1 and S2 emit waves of the same frequency and amplitude.
31
Un
it 1
: W
ave
s C
ha
pte
r 2
:
Sound
3) It refracts upon traveling from one medium to another due to the difference in velocity(Fig 2 - 4) :
where φ is the angle of incidence and θ in the angle of refraction, v1 is velocity of soundin the first medium and v2 is the velocity of sound in the second medium.
4) Sound waves of equal frequency and amplitude interfere to produce regions ofconstructive interference (increase of intensity) and regions of destructive interference(decrease of intensity).
5) Sound diffracts when passing through a small aperture or a sharp edge, provided that thediscontinuity is comparable to the wavelength.
Superposition of waves Waves combine such that the resultant wave has intensity equal to the sum of intensities
of the individual waves (Fig 2-9 a). When the frequencies are slightly different while theamplitude is the same, this combination (superposition) leads to beats (Fig 2-9 b , c , d ).
If we move a tight wire or rope such that one pulse is generated (Fig 2-10 a ), then thispulse continues until it reaches the far end. If this end is attached to a sliding ring, then thereflected wave is positive ,i.e., in the same direction as the incident pulse. Whereas if thisend is fixed such that it cannot move, then the reflected wave is always reversed (Fig 2-10b). When the reflected wave meets the incident wave, constructive interference is producedin the first case (Fig 2-10 c ),and destructive interference is produced in the second case
(Fig 2-10 d).
Fig (2-9 a)The resultant wave is the sum of two waves
displacement
first wave
resultantwave
distance
sinsin
vv
1
2
φΘ
=θ
second wave
30
Un
it 1
: W
ave
s C
ha
pte
r 2
:
So
un
d
Secondly: Sound Diffraction
What is meant by diffraction?Diffraction is a change (or bending) of the wave path when passing through a slit or an
aperture, small enough compared to the wavelength, or when passing by sharp edges in thesame medium.
We observe sound diffraction in our daily life. For example, if you speak in one roomand a window or a door is open, someone in the next room may overhear you , although heis not standing right next to the window or the door (Fig 2-8). The sound intensity in theneighboring room depends on the position of a listener in that room, being maximum ofcourse directly next to the opening. The spreading of sound in the neighboring room isattributed to diffraction .
Sound as a wave motionFrom above, we conclude that sound has wave properties :
1) It propagates in a medium in straight lines in all directions.2) It undergoes reflection when falling on a surface, and the angle of reflection equals the
angle of incidence.
Fig (2-8)Diffraction of sound through a door opening to a nearby room
spreading ofsound in the
form ofconcentric
spheres
door
room
audiosource
33
Un
it 1
: W
ave
s C
ha
pte
r 2
:
Sound
thirdly
Fig (2-10b)Reflection of a pulse
incident pulse
far end free tomove (attached to a
sliding ring) far end fixed tothe wall
reflected pulsereflected pulse
firstly
fourthlysecondly
secondlyfirstly
fourthlythirdly
Fig (2-10c)Combination of two positive pulsespropagating in opposite directions
Fig (2-10d)Combination of two pulses one positive and the other
negative moving in opposite directions
33
Un
it 1
: W
ave
s C
ha
pte
r 2
:
Sound
thirdly
Fig (2-10b)Reflection of a pulse
incident pulse
far end free tomove (attached to a
sliding ring) far end fixed tothe wall
reflected pulsereflected pulse
firstly
fourthlysecondly
secondlyfirstly
fourthlythirdly
Fig (2-10c)Combination of two positive pulsespropagating in opposite directions
Fig (2-10d)Combination of two pulses one positive and the other
negative moving in opposite directions
32
Un
it 1
: W
ave
s C
ha
pte
r 2
:
So
un
d
destructiveinterference
destructiveinterference
Fig (2-9c)Formation of harmonic tones with time
Fig (2-9b)Formation of harmonic tones with distance
Fig (2-9d)Harmonic tones resemble two nearly equal interleaved combs
Fig (2-10a)Formation of an incident pulse
distancetime
string and hand at rest
hand moves up pulling thestring upwards
hand moves down
middle of the pulse
hand at rest
1
0
-1
1
-1
0
2
1
0
-1
-2
32
Un
it 1
: W
ave
s C
ha
pte
r 2
:
So
un
d
destructiveinterference
destructiveinterference
Fig (2-9c)Formation of harmonic tones with time
Fig (2-9b)Formation of harmonic tones with distance
Fig (2-9d)Harmonic tones resemble two nearly equal interleaved combs
Fig (2-10a)Formation of an incident pulse
distancetime
string and hand at rest
hand moves up pulling thestring upwards
hand moves down
middle of the pulse
hand at rest
1
0
-1
1
-1
0
2
1
0
-1
-2
32
Un
it 1
: W
ave
s C
ha
pte
r 2
:
So
un
d
destructiveinterference
destructiveinterference
Fig (2-9c)Formation of harmonic tones with time
Fig (2-9b)Formation of harmonic tones with distance
Fig (2-9d)Harmonic tones resemble two nearly equal interleaved combs
Fig (2-10a)Formation of an incident pulse
distancetime
string and hand at rest
hand moves up pulling thestring upwards
hand moves down
middle of the pulse
hand at rest
1
0
-1
1
-1
0
2
1
0
-1
-2
32
Un
it 1
: W
ave
s C
ha
pte
r 2
:
So
un
d
destructiveinterference
destructiveinterference
Fig (2-9c)Formation of harmonic tones with time
Fig (2-9b)Formation of harmonic tones with distance
Fig (2-9d)Harmonic tones resemble two nearly equal interleaved combs
Fig (2-10a)Formation of an incident pulse
distancetime
string and hand at rest
hand moves up pulling thestring upwards
hand moves down
middle of the pulse
hand at rest
1
0
-1
1
-1
0
2
1
0
-1
-2
32
Un
it 1
: W
ave
s C
ha
pte
r 2
:
So
un
ddestructiveinterference
destructiveinterference
Fig (2-9c)Formation of harmonic tones with time
Fig (2-9b)Formation of harmonic tones with distance
Fig (2-9d)Harmonic tones resemble two nearly equal interleaved combs
Fig (2-10a)Formation of an incident pulse
distancetime
string and hand at rest
hand moves up pulling thestring upwards
hand moves down
middle of the pulse
hand at rest
1
0
-1
1
-1
0
2
1
0
-1
-2
35
Un
it 1
: W
ave
s C
ha
pte
r 2
:
Sound
Melde's Experiment Melde's experiment best illustrates standing waves on strings or wires. The apparatus is
shown (Fig 2-12). It consists of a vibrating source, connected to a soft string whose lengthranges from 2 to 3 meters. The other end of the string passes over a smooth pulley and isconnected at its free end to appropriate weights. When the source vibrates, a wave train isproduced in the string, which reflects upon reaching the pulley.The reflected and incidentwaves are combined (superposed or superimposed) to form standing waves. These standingwaves have nodes and antinodes (Fig 2-13), provided the source frequency has a certainvalue compared to the string (wire) length.
Fig (2-11c)A string fixed at both ends and
pulled in the middle
Fig (2-12)Melde's apparatus
waves generated at the middle andpropagating in both directions
waves reflected fromboth fixed ends tensiontension
vibratorpulley
weightsgenerator
34
Un
it 1
: W
ave
s C
ha
pte
r 2
:
So
un
d
Standing ( Stationary ) Waves:Standing waves are formed when there is a continuous train of waves in a tight wire
( string, spiral spring, or rope) in one direction and a continuous train of waves reflected inthe opposite direction. These two wave trains interfere, giving a pattern of particles of themedium which appears localized, i.e., not moving to the right, or to the left but movingperpendicularly to the wire. This effect may be visualized by moving a spiral spring, (stringor rope) up and down in reciprocity from both ends (Fig 2-11 a), or fixing it from one endand moving the other end in a simple harmonic motion (Fig 2-11 b), or pulling a string -fixed from both ends - from the middle, such that waves are transmitted in both directionsand get reflected, and hence interfere (Fig 2-11 c).
Fig (2-11a)A spiral spring oscillating from both ends in reciprocity
two waves inopposite directions
node
antinode
antinode
node
node
antinodenode
Fig (2-11b)A spiral spring oscillating in a simple harmonic
motion at one end while fixed at the other
34
Un
it 1
: W
ave
s C
ha
pte
r 2
:
So
un
d
Standing ( Stationary ) Waves:Standing waves are formed when there is a continuous train of waves in a tight wire
( string, spiral spring, or rope) in one direction and a continuous train of waves reflected inthe opposite direction. These two wave trains interfere, giving a pattern of particles of themedium which appears localized, i.e., not moving to the right, or to the left but movingperpendicularly to the wire. This effect may be visualized by moving a spiral spring, (stringor rope) up and down in reciprocity from both ends (Fig 2-11 a), or fixing it from one endand moving the other end in a simple harmonic motion (Fig 2-11 b), or pulling a string -fixed from both ends - from the middle, such that waves are transmitted in both directionsand get reflected, and hence interfere (Fig 2-11 c).
Fig (2-11a)A spiral spring oscillating from both ends in reciprocity
two waves inopposite directions
node
antinode
antinode
node
node
antinodenode
Fig (2-11b)A spiral spring oscillating in a simple harmonic
motion at one end while fixed at the other
33
Un
it 1
: W
ave
s C
ha
pte
r 2
:
Sound
thirdly
Fig (2-10b)Reflection of a pulse
incident pulse
far end free tomove (attached to a
sliding ring) far end fixed tothe wall
reflected pulsereflected pulse
firstly
fourthlysecondly
secondlyfirstly
fourthlythirdly
Fig (2-10c)Combination of two positive pulsespropagating in opposite directions
Fig (2-10d)Combination of two pulses one positive and the other
negative moving in opposite directions 33
Un
it 1
: W
ave
s C
ha
pte
r 2
:
Sound
thirdly
Fig (2-10b)Reflection of a pulse
incident pulse
far end free tomove (attached to a
sliding ring) far end fixed tothe wall
reflected pulsereflected pulse
firstly
fourthlysecondly
secondlyfirstly
fourthlythirdly
Fig (2-10c)Combination of two positive pulsespropagating in opposite directions
Fig (2-10d)Combination of two pulses one positive and the other
negative moving in opposite directions
37
Un
it 1
: W
ave
s C
ha
pte
r 2
:
Sound
Fig (2-14)Nodes and antinodes
Fig (2-15)The distance between twosuccessive nodes or two
successive antinodes is halfa wavelength 35
Unit
1:
W
aves C
hapte
r 2:
So
un
d
Melde's Experiment Melde's experiment best illustrates standing waves on strings or wires. The apparatus is
shown (Fig 2-12). It consists of a vibrating source, connected to a soft string whose lengthranges from 2 to 3 meters. The other end of the string passes over a smooth pulley and isconnected at its free end to appropriate weights. When the source vibrates, a wave train isproduced in the string, which reflects upon reaching the pulley.The reflected and incidentwaves are combined (superposed or superimposed) to form standing waves. These standingwaves have nodes and antinodes (Fig 2-13), provided the source frequency has a certainvalue compared to the string (wire) length.
Fig (2-11c)A string fixed at both ends and
pulled in the middle
Fig (2-12)Melde's apparatus
waves generated at the middle andpropagating in both directions
waves reflected fromboth fixed ends tensiontension
vibratorpulley
weightsgenerator
36
Un
it 1
: W
ave
s C
ha
pte
r 2
:
So
un
d
The node is the position where the amplitude of the vibration is zero, while the antinodeis the position where the amplitude of the vibration is maximum. Nodes and antinodes arespaced at equal distances apart (Fig 2-14).
The wavelength of a standing wave is twice the distance between any two successiveantinodes or two successive nodes. As the weight in the experiment is increased, the tensionin the wire is increased, so is the velocity of propagation, and the frequency for the samewire length is also increased (Fig 2-15).
Fig (2-13a)A vibrating string showing a standing wave pattern
Fig (2-13b)Variation of standing wave patterns with the
ratio of the string length to the wavelength as time goes on
35
Un
it 1
: W
ave
s C
ha
pte
r 2
:
So
un
d
Melde's Experiment Melde's experiment best illustrates standing waves on strings or wires. The apparatus is
shown (Fig 2-12). It consists of a vibrating source, connected to a soft string whose lengthranges from 2 to 3 meters. The other end of the string passes over a smooth pulley and isconnected at its free end to appropriate weights. When the source vibrates, a wave train isproduced in the string, which reflects upon reaching the pulley.The reflected and incidentwaves are combined (superposed or superimposed) to form standing waves. These standingwaves have nodes and antinodes (Fig 2-13), provided the source frequency has a certainvalue compared to the string (wire) length.
Fig (2-11c)A string fixed at both ends and
pulled in the middle
Fig (2-12)Melde's apparatus
waves generated at the middle andpropagating in both directions
waves reflected fromboth fixed ends tensiontension
vibratorpulley
weightsgenerator
34
Un
it 1
: W
ave
s C
ha
pte
r 2
:
So
un
d
Standing ( Stationary ) Waves:Standing waves are formed when there is a continuous train of waves in a tight wire
( string, spiral spring, or rope) in one direction and a continuous train of waves reflected inthe opposite direction. These two wave trains interfere, giving a pattern of particles of themedium which appears localized, i.e., not moving to the right, or to the left but movingperpendicularly to the wire. This effect may be visualized by moving a spiral spring, (stringor rope) up and down in reciprocity from both ends (Fig 2-11 a), or fixing it from one endand moving the other end in a simple harmonic motion (Fig 2-11 b), or pulling a string -fixed from both ends - from the middle, such that waves are transmitted in both directionsand get reflected, and hence interfere (Fig 2-11 c).
Fig (2-11a)A spiral spring oscillating from both ends in reciprocity
two waves inopposite directions
node
antinode
antinode
node
node
antinodenode
Fig (2-11b)A spiral spring oscillating in a simple harmonic
motion at one end while fixed at the other
39
Un
it 1
: W
ave
s C
ha
pte
r 2
:
Sound
Since the arc is small
(2-2)
where m is the mass per unit lengh of the sting material.Vibration of strings:
If we have a tight string fixed on both ends and pulledfrom the middle then released to vibrate freely, we notethat the particles of the string vibrate perpendicularly toits length, i.e., perpendicularly to the direction ofpropagation of the wave. Such a vibration is transverse,i.e., transverse waves propagate on both halves of thestring in bolh directions until they reach the fixed ends,then they are reflected back in the opposite direction. Bymultiple reflections, the incident and reflected waves,interfere producing standing waves, where an antinode isformed at the middle, and a node is formed at each of thefixed ends.
A string vibrating in this way may produce different tones.The fundamental tone (the lowest frequency the string mayvibrate at on its own) will produce one antinode and twonodes (as above). The string may vibrate in many other ways.We may divide the wire into segments.
For example, we may have two segments (3 nodes and 2 antinodes), or 3 segments (4 nodes and3 antinodes) (Fig 2-17). When the string has one segment (λ = 2 ), it produces the fundamentalfrequency (first harmonic). When the string has two segments, it produces the first overtone
/ 2 Rθ =
FT / mv =
2FT / 2 RFc = =∴ Mv2
RFTR
v2 = = = FT / mFTM /
FT M
=
Fig (2-17)Formation of harmonics
fundamental tone(1st harmonic)
first overtone(2nd harmonic)
second overtone(3rd harmonic)
third overtone(4th harmonic)
l
l l
ll
l
39
Un
it 1
: W
ave
s C
ha
pte
r 2
:
Sound
Since the arc is small
(2-2)
where m is the mass per unit lengh of the sting material.Vibration of strings:
If we have a tight string fixed on both ends and pulledfrom the middle then released to vibrate freely, we notethat the particles of the string vibrate perpendicularly toits length, i.e., perpendicularly to the direction ofpropagation of the wave. Such a vibration is transverse,i.e., transverse waves propagate on both halves of thestring in bolh directions until they reach the fixed ends,then they are reflected back in the opposite direction. Bymultiple reflections, the incident and reflected waves,interfere producing standing waves, where an antinode isformed at the middle, and a node is formed at each of thefixed ends.
A string vibrating in this way may produce different tones.The fundamental tone (the lowest frequency the string mayvibrate at on its own) will produce one antinode and twonodes (as above). The string may vibrate in many other ways.We may divide the wire into segments.
For example, we may have two segments (3 nodes and 2 antinodes), or 3 segments (4 nodes and3 antinodes) (Fig 2-17). When the string has one segment (λ = 2 ), it produces the fundamentalfrequency (first harmonic). When the string has two segments, it produces the first overtone
/ 2 Rθ =
FT / mv =
2FT / 2 RFc = =∴ Mv2
RFTR
v2 = = = FT / mFTM /
FT M
=
Fig (2-17)Formation of harmonics
fundamental tone(1st harmonic)
first overtone(2nd harmonic)
second overtone(3rd harmonic)
third overtone(4th harmonic)
l
l l
ll
l
36
Un
it 1
: W
ave
s C
ha
pte
r 2
:
So
un
d
The node is the position where the amplitude of the vibration is zero, while the antinodeis the position where the amplitude of the vibration is maximum. Nodes and antinodes arespaced at equal distances apart (Fig 2-14).
The wavelength of a standing wave is twice the distance between any two successiveantinodes or two successive nodes. As the weight in the experiment is increased, the tensionin the wire is increased, so is the velocity of propagation, and the frequency for the samewire length is also increased (Fig 2-15).
Fig (2-13a)A vibrating string showing a standing wave pattern
Fig (2-13b)Variation of standing wave patterns with the
ratio of the string length to the wavelength as time goes on
45
Un
it 1
: W
ave
s C
ha
pte
r 2
:
Sound
Questions and Drills
I) Essay questions 1) State the laws of reflection of sound.2) Show how to demonstrate the interference of sound.3) Explain : Sound is a wave motion.
II) Define:1) an antinode. 2) a node. 3) the wavelength of a standing wave.
III) Complete :1) The velocity of propagation of a transverse wave in a string is given by:v = ...................................... 2) The fundamental frequency produced in a string is given by:
ν = .....................................3) Keeping tension constant, the frequency of a string is.......................proportional to its length .4) Keeping the length of a string constant, the fundamental frequency is directly
proportional to .................................5) Keeping the length of the string and tension constant, the fundamental frequency is
inversely proportional to ................................
IV. Choose the right answer:1) Standing waves are formed by the combination of two waves propagating
a) in the same direction.b) in opposite directions provided they have equal frequency and intensity.c) in opposite directions without necessarily having equal frequency and intensity.d) in two perpendicular directions.
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 47
49
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
Reflection and refraction of light Light propagates in straight lines in all directions, unless met by an obstructing medium.
If so, it undergoes reflection, refraction and partial absorption depending on the nature of
the medium. When a light ray falls on a surface separating two media - which are different
in optical density - then part of light is reflected and the rest is refracted, neglecting
absorption. We note from Fig (3-3) that each of the incident ray, reflected ray and refracted
ray as well as the normal to the surface at the point of
incidence all lie in one plane perpendicular to the separating
surface.
In the case of reflection : the angle of incidence is equal to
the angle of reflection
In the case of refraction : the ratio between the sine of the
angle of incidence in the first medium to the sine of the
angle of refraction in the second medium is equal to the ratio
φ
θ
Fig (3 – 2)
An electromagnetic wave consists of an electric fieldand a magnetic field perpendicular to each otherand to the direction of propagation of the wave
angle ofincidence reflected ray
incidentray
angle ofrefraction
second medium(glass)
first medium(air)
oscillating magnetic field
oscillating electric field
Fig (3 – 3)Reflection and refraction
of light
refracted ray
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 49
48
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
Overview :Light is an indispensible form of energy. The Sun is the main natural source of energy to
us. The energy from the Sun is almost divided between heat and light. Thanks to the lightfrom the Sun, the plants perform photosynthesis, hence make their own food. Man dependson plants and animals, which in turn feed on plants.
We have seen before that sound has a wave nature. It propagates from a source causingmechanical waves in the medium. Light also has a wave nature. It is subject to the laws ofreflection, refraction, interference and diffraction. But light is different from sound in that itdoes not require a medium to propagate in. Light is part of an extensive range of wavescalled electromagnetic waves, which all travel at a constant speed in space equal to 3 x 108 m/s,while varying in frequency. This range of waves is called the electromagnetic spectrum (Fig 3-1). It includes, for example, radio waves, infrared, visible light, ultraviolet,
X- rays and Gamma rays. They all share common features. They are all transverseelectromagnetic waves, but of different frequencies (and wavelengths).
Electromagnetic waves consist of time varying electric and magnetic fields. Bothoscillate at equal frequency at the same phase, and are perpendicular to each other and tothe direction of propagation (Fig 3-2), hence called transverse waves.
Fig (3 – 1)
Electromagnetic spectrum
Chapter 3 Light
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 48
51
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
nn
2
1= sin
sinφθ
n1 sin φ = n2 sin θ
2) The refraction of light is attributed to the difference in the speed of light, when light is
transmitted from one medium to another.
where v1 is the speed of light in the first medium, v2 is the speed of light in the secondmedium. Substituting equation (3-3) in (3-1), we have:
This is Snell’s law The absolute refractive index for the medium of incidence times the sine of the angle of
incidence is equal to the absolute refractive index of the medium of refraction times thesine of the angle of refraction.3) We can use refraction in analysing a bundle of light into its components of different
wavelengths, since the absolute refractive index varies with wavelength. Therefore, whitelight may be decomposed into its components. This can be seen, for example, in soap bubbles.
Learn at Leisure
Why refraction ?If light falls from a less dense medium onto a more dense medium ,the refracted ray
approaches the normal. This resembles a car in which one of its wheels goes through amuddy soil, while the other is free on the paved road. The wheel that goes into the mudbecomes slower. Therefore, the car changes direction (Fig 3-4). The opposite is also true, asin refraction from a more dense material to a less dense one. The refracted ray deviatesaway from the normal (Fig 3-5).
cnv =
n2n1
=v1v2 ( 3 - 3 )
( 3 - 4 )
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 51
50
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
mediumrefractive index 1.002931.3330001.5010001.4610001.3610001.520001.660000
1.48500002.419000
AirWaterBenzineCarbon tetrachlorideEthyl alcoholCrown glassRock glass QuartzDiamond
of the speed of light in the first medium to the speed of light in the second medium. This
ratio is constant for these two media, and is called relative refractive index from the first
medium to the second medium, denoted by 1n2 :
Important facts
1) The speed of light in space ''c'' is one of the physical constants of the universe and is
equal to 3 x 108 m/s. It is larger than the speed of light in any medium ''v''. The ratio
= n is called the absolute refractive index for the medium and is always > 1.
The absolute refractive indices of some materials are listed below
SinSin
= v1
v2θsinsin = n
1 2
cvn = ( 3 - 2 )
( 3 - 1 )
cv
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 50
52
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
Fig (3 – 5)Refraction from a more dense
medium to a less dense medium
Examples1) If a light ray falls on the surface of a glass slab whose refractive index is 1.5 at an angle
30˚, calculate the angle of refraction.
Solution
2) If the absolute refractive index of water is and glass , find
a) the relative refractive index from water to glass
b) the relative refractive index from glass to water.
Solutiona) The relative refractive index from water to glass
43
32
1.5 = Sin 30Sinss
s∴ n = SinSin
sin = 0.51.5
sin
∴
∴
ss
= 0.333
θ
θ = 19˚ 28´
θ
θ
Fig (3 – 4)Refraction from a less dense
medium to a more dense medium
muddysoil
muddysoil
paved road reflectedray
incidentray
incidentray
reflectedray
glass
pavedroad
glass air
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 52
54
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
∆Χ= λLd
∆ y ( 3 - 5 )
reaching the double slit have the same phase, hence, are coherent (having thesame frequency, amplitude and phase). Waves emanating from S1 and S2 arecylindrical and spread toward the observation screen C. On such a screen,waves coming from S1 and S2 combine and produce an interference pattern,appearing as a sequence of bright and dark straight parallel regions, which arethe interference fringes (Fig 3-7). The distance between two successive
fringes ∆y is given by:
where λ is the wavelength of the monochromatic source, R is the distancebetween the double slit and the observation screen and d is the distancebetween S1 and S2.
Therefore, this experiment may be used to determine the wavelength forany monochromatic light source.
Learn at Leisure
Interpretation of interference in Thomas Young’s experiment If light were not to manifest interference, we would
obtain fringes as in Fig (3-8a). We may interpret theformulas of constructive and destructive interferencepattern in Young’s experiment as follows. If thedistance of the screen from the double slit R is largerelative to the distance d between the two slits of thedouble slit, then we may consider the two rays r1, r2emanating from the double slit on their way to theobservation screen as nearly parallel. If θ is theinclination angle of the two rays, then the pathdifference between these two rays is ∆r (Fig 3-9). This
Fig (3 – 7)Interference
fringes
R
monochromaticlight soure
darkfringes
bright fringes
Secondslit
Firestslit ab
Fig (3 – 8) Fringes (a) resulting from
interference (b) if there were no interference
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 54
57
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
Light DiffractionWhen a monochromatic light falls on a circular aperture in a screen, we expect that light
should form a circular bright spot on an observation screen, considering that lightpropagates in straight lines. But careful examination of the bright spot (called Airy’s disk),i.e., studying the light intensity, reveals the existence of bright and dark fringes (Fig 3-12).
Fig (3-13) demonstrates diffraction from a rectangular slit, while Fig (3-14) shows thediffraction pattern at a sharp edge of matter. In general, diffraction is evident when thewavelength of the wave is comparable to the dimensions of the aperture, and vice versa(Fig 3-15). In fact, there is no big difference between the mechanisms of interference anddiffraction. In both cases, combination (superposition) of waves is involved (Fig 3-16).
Fig (3 – 12)
Diffraction in a circular aperture
Airy’s disk
parallel rays
firstsecondary
bright spot
Fig (3 –13b) Distribution of light intensity on a screen with the
succession of fringes resulting from diffractionfrom a rectangular aperture
Fig (3 – 13a)Diffraction from a rectangular aperture
center of the central bright fringe
lightintensity
slit
light ray
central brightspot
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 57
59
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
Learn at Leisure
Interpretation of diffractionFig (3-17) shows a plane wave advancing toward a screen in which there is a
rectangular slit. At an appropriate distance, there is a white parallel observation screen.According to the wave theory, points on the wave front at the slit may be considered assources of secondary wavelets. Light emanating from these secondary sources fall on theobservation screen at a point corresponding to the center of the slit as a lens collimates therays. In this case, wavelets have the same phase. Constructive interference results in abright fringe (Fig 3-17 a).
Fig (3 –17a)
Formation of the central bright fringe
Fig (3–17b)
Succession of the fringes
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 59
58
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
circular aperture
plane edge
sphere Fig (3–14)Diffraction patterns from different obstacles
diffraction is moreevident from anarrow slit in
comparison with λ
the dimensions of theobstacle are large incomparison with λ
the dimensions of theobstacle are mediumin comparison with λ
the dimensions of theobstacle are small
incomparison with λFig (3 – 15)
Fig (3 – 16)
Diffraction is the interferenceof secondary wavelets fromdifferent points in the slit
razor edge
(b)
incidentwave
(a)
observationscreen
0.2d0.8d
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 58
61
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
Learn at Leisure
Resolving powerDiffraction places a limit on resolving details in an image. This limit is called the limit
of resolution. If we have two point sources, and light is emitted from each through acircular aperture, then each source forms separate fringes. When the distance between thetwo sources decreases, the fringes get closer, and it becomes difficult to identify one fromthe other. It is found that the angle between the centers of the two fringes under thiscondition is given by the approximate relation ∆θ = , where D is the aperture diameter.Thus, the ability to resolve two small objects is inversely proportional to the aperturediameter and directly to the wavelength (Fig 3-19). In the case of a microscope, the lenstakes the place of the aperture and the wavelength limits the ability of the microscope todistinguish between small objects. As λ decreases, we can discern details that were notseen before. This is the advantage of the electron microscope (Chapter 12).
Fig (3–19) Resolving power
a)as the two sources get closerto each other it becomesdifficult to separate them
visually because of diffraction
c) if the two sources aredrawing near to the
observer, then he canseparate them visually
d) bacteria appearing through an electron microscope and notappearing through an optical microscope
b) the resolution is nilas the objects get toosmall and too close
together.
λD
∆
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 61
62
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
θθϕ
Light as a wave motionFrom above, we conclude that light 1) propagates in straight lines .2) reflects according to the laws of reflection .3) refracts according to the laws of refraction .4) light interferes, and as a result, light intensity increases in certain positions (bright
fringes)and diminishes to zero in other positions (dark fringes).5) light diffracts if obstructed by an obstacle .These are the same general properties of waves. Hence, light is a wave motion
Total reflection and the critical angle When a light ray travels from an optically dense medium (as water or glass) to a less
dense medium (as air), then the refracted ray deviates away from the normal (Fig 3-20). Asthe angle of incidence increases in the more dense medium (of high absolute refractiveindex), the refraction angle in the less dense medium (of low absolute refractive index)increases.
A point is reached when the angle of incidence in the more dense medium approachesa critical value φc when the angle of refraction in the less dense medium reaches itsmaximum, which is 90˚. Then,the refracted ray becomes tangent to the surface.
Fig (3–20) Incidence of light from a more optically dense medium to a less optically dense medium
(a) (b) (c) (d)
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 62
64
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
= 1 x 1.331.6
= sin c 0.8313
Sin c = 1n1
= 11.33
= 0.7518sin
Fig (3–21) Optical fibers contain the
rays despite bending
Fig (3– 22) Optical fibers
Some Applications of Total Reflection I. Fiberoptics (Optical Fibers)
Fig (3-21) shows an optical fiber. It is athread-like tube of transparent material. When lightfalls at one end, while the angle of incidence isgreater than the critical angle, it undergoessuccessive multiple reflections until it emerges fromthe other end (Fig 3-22). Fig (3-23) shows a bundleof fibers which can be bent while containing lightso that light can be made to travel in non straightlines to parts hard to reach otherwise.
Fibers can be used to transmit light withoutmuch losses, and are widely used nowadays.
b) In the case of water
φc = 48˚ 45´
2)Using the information in the example above, find the critical angle for light fallingfrom glass onto water
SolutionUsing Snell’s law, n2 sin 90˚ = n1 sin φc
1.33 x 1 = 1.6 sin φc
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 64
67
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
Learn at Leisure
The Bear’s fur The bear’s fur does not provide the bear with thermal
isolation only, but the fur hairs are massive optical fibers
which reflect ultraviolet rays. The fur appears white (Fig
3-26). because visible light reflects inside the hollow
transparent optical fibers, while the skin itself absorbs all
rays reaching it. Therefore, it is actually black.
Learn at Leisure
How an optical fiber worksIf we have a hollow tube and look through it to see a bright object on the other end, then
the object is easily seen. If the tube is bent, then the object cannot be seen. Yet, we may be
able to see it, if we use reflecting mirrors in the path of rays. Similarly, by using optical
fibers, while the ray is incident at an angle greater than the critical angle, then multiple
reflections take place, until the ray emerges from the other end, despite the bending of the
fibers.
II. The reflecting prism The critical angle between glass (refractive index 1.5) and air is 42˚. Therefore, a glass
prism whose angles are 90˚, 45˚, 45˚ is used to change the path of the rays by 90˚ or 180˚.
Such a prism may be used in optical instruments, as periscopes in submarines and
binoculars in the field (Fig 3-27).
Fig (3–26)
The bear’s fur
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 67
66
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
Fig (3–24b)
Endoscopes
Fig (3–25) Optical fibers used to carry electrical signals
Fig (3–24c) Endoscope lens
Fig (3–24d) An image of esophogus
by optical fibers
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 66
69
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
This can be explained as follows. On very hot days, the air layers adjacent to the surface
of the Earth are heated, their density decreases. Hence, their refractive index becomes
smaller than that of the upper layer. If we follow a light ray reflected off a palm tree, this
ray is traveling from an upper layer to one below. Therefore,if refracts away from the
normal, and keeps deviating taking a curved path. When its angle of incidence reaches
more than the critical angle, it undergoes total reflection and the curve goes up. To the eye
of the observer, the ray appears as if coming from under the surface of the Earth. The
observer thinks that there is a pond.
Deviation of light in a prism
When a light ray such as " a b " falls on the surface xy of a prism, it refracts in the prism
taking the path " bc ", until it falls on the surface xz and emerges in the direction "cd" (Fig
3-29). We notice from the figure that the light ray in the prism refracts twice. As a result,
the ray deviates from its original path by an angle of deviation α .
The angle of deviation α is the angle subtended by the directions of the extension of the
incident ray and the emerging ray. If the angle of incidence is φ1 , the angle of refraction
on the first surface is θ1 , the angle of incidence on the second surface is φ
2 , the angle of
emergence is θ2 and the apex angle of the prism is A, we note from the geometry (Fig
3-29).
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 69
68
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
Fig (3–27) A reflecting prism
a) a reflecting prism changing thelight path by 90˚
b) a reflecting prism changingthe light path by 180˚
c) prisms in binoculars
Fig (3–28b) Reflection of the sky in the desert
gives the illusion of water
Prisms are better for this purpose than reflecting surfaces, first, because light totallyreflects from such a prism, while it is seldom to find a metallic reflecting surface whoseefficiency is 100%. Secondly, a metallic surface eventually loses its luster,and hence itsability to reflect decreases. This does not happen in a prism. The surface at which lightrays fall on a prism or the surface from which the rays emerge may be coated with non -reflective layer of material like cryolite (Aluminum fluoride and magnesium fluoride)whose refractive index is less than that of glass, to avoid any reflection losses on theprism, even little as they are.
III.MirageThis is a familiar phenomenon observable on hot days, as paved roads appear as if wet
(Fig 3-28 a). Also, an image of the sky is made on desert plains, where palm trees or hillsappear inverted giving the illusion of water (Fig 3-28 b).
Fig (3–28a)Paved roads appear as if wet
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 68
71
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
But
Substituting for φ and θ we find that the refractive index can be determined from therelation
Experiment to determine the ray path through a glass prism and its refractive index:
Tools:An equilateral triangular prism (A = 60˚), pins, a protractor, a ruler.
Procedure:1) Place the glass prism on a sheet of drawing paper with its surface in a vertical position
and mark its position with a fine pencil line. Place two pins such that one of them (a) is very close to one side and the other (b) is about
10 cm from the first. The line joining them representsthe incident ray. Look at the other side of the prism tosee the image of the two pins, one behind the other.
Place two other pins c and d between the prism andthe eye such that they appear to be in line with thetwo pins a and b,i.e., the four pins appear to be in onestraight line.Locate the positions of the four pins.2) Remove the prism and the pins, join b and c to locate the
path of the ray (a b c d) from air to glass to air again.
n = SinSinsinsin
Fig (3– 31) Determination of light
ray path in a prism
0
0
(3 - 10)
emergingray
incidentray
n = Sin + A
2
Sin A2
αsin
sin
0
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 71
73
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
Note also that the refractive index (n) depends on up the wavelength λ, then the
minimum angle of deviation depends also on the wavelength. Thus, if a beam of white light
falls on a prism set at the minimum angle of deviation, then the emerging light disperses
into spectral colors as illustrated in Fig(3-32). From this figure, it is concluded that the
violet ray has the largest deviation ( maximum refractive index). The visible spectral colors
into which the white light is dispresed are arranged by the order: red, orange, yellow,
green, blue, indigo and violet.
The thin prism:A thin prism is a triangular glass prism. Its apex angle is a few degrees and is in the
position of minimum deviation:
Since: and are small angles.
Thus,
(radians)
and (radians)
Substituting from (3-10), we find that the refractive index of the material of the thinprism is determined by
∝ = A (n - 1) (3 - 12)
Sin + A2
= + A2
Sin A2
= A2
sin ≅
≅sin
+ A2
A2
α
αα
0
0 0
n = Sin + A
2
Sin A2
sin
sin
α0
0
n = (3 - 11)α +AA0
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 73
72
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
tn =
Sin + A2
Sin A2
s
s
α0
3) Extend dc to meet extended ab .The angle between them is the angle of deviation α.
4) Measure the angle of incidence φ1, the angle of refraction θ1, the inner incidence angle
φ2, the angle of emergence θ2 and the angle of deviation (α).5) Repeat the previous steps several times changing the angle of incidence and tabulate
the results.
Find the minimum angle of deviation and the corresponding angles φ˚ and θ˚ .- Then obtain the refractive index from equation (3-10).
The dispersion of light by a triangular prism:It has been proven previously that in the case of
minimum deviation, the refractive index may bedetermined from the relation:
where (n) is the refractive index,αo is minimum angle ofdeviation, and (A) is the angle of the prism.
Since the angle of the prism is constant for a certain prism,so the minimum angle of deviation changes by changing therefractive index. As the refractive index increases, theminimum angle of deviation increases and vice versa.
Angle of theprism A
angle ofincidence φ1
Angle of innerincidence φ2
angle of refraction θ1
angle ofemergence θ2
Angle ofdeviationα
Fig (3–32) A prism disperses the
spectrum
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 72
75
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
In a Nutshell
.Laws of reflection of light :1) Angle of incidence = Angle of reflection2) The incident ray, the reflected ray, and the normal to the reflecting surface at the
point of incidence, all lie in one plane perpendicular to the reflecting surface.
.Light refracts between two media because of the different velocities of light in the twomedia v1 & v2
.Laws of refraction of light :1) The ratio between the sine of the angle of incidence in the first medium, to the sine
of the angle of refraction in the second medium is constant, and is known as therefractive index 1n2
where φ is the angle of incidence in the first medium, and θ is the angle of refractionin the second medium
2) The incident ray, the refracted ray, and the normal to the surface of separation at thepoint of incidence, all lie in one plane normal to the surface of separation.
• The relative refractive index between two media is the ratio between the velocity of lightin the first medium v1 and the velocity of light in the second medium v2
• The absolute refractive index for a medium is given by :
where c is the velocity of light in free space and v is the velocity of light in the medium.• Snell s law :
n1sinφ = n2 sinθ
sin φsin θ1n2 =
n = CVcv
1 n2 = V1
V2
v
v
,
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 75
74
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
Dispersive Power When white light falls on a prism, the light disperses into its spectrum due to the
variation of the refractive index with wavelength.(α
0)r = A (nr - 1)
(α0)b = A (nb - 1)
where nr is the refractive index for red and nb for blue. Subtracting,
(α0)b - (α0)r = A (nb - nr) ( 3 - 13)
The LHS represents the angular dispersion between blue and red. For yellow (middlebetween blue and red), the angle is :
(α0)y = A (ny - 1) (3 - 14)
where ny is the refractive index for yellow. If (α0)y is the average of (α0)r and (α
0)b, then
nyis the average of nr and nb. We define as :
where is the dispersive power, and is independent of the apex angle.
= nb - nr
ny - 1 −
( 3 - 14)( α0)b
(α0)y
( α0)rωα=
ωα
ωα
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 74
77
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
n = Sin + A
2
Sin A2
s
s
• Refractive index of the prism material is given by:
where n is the refractive index, α˚ is the minimum angle of deviation. • The minimum angle of deviation in a thin prism is :
α˚ = Α (n - 1)
• The angular dispersion for a thin prism is :
(α0)b - (α
0)r= Α (nb -nr)
• where (α0)b is the minimum deviation angle of the blue ray, and (α
0)r is the minimum
deviation angle of the red ray. • The dispersive power :
where (α0)y is the minimum angle of deviation of the yellow light, and ny is the refractive
index for the yellow light.
α0
= b r
y
= n b - n r
n y - 1
ωα(α )b0 (α )r0
(α )y0
ωα
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 77
76
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
• The distance between two successive similar fringes (either bright or dark) is :
where λ is the wavelength of light employed, R is the distance between the double slitand the screen, and d is the distance between the two slits. • Light is a wave motion. • The critical angle is the angle of incidence in the more dense medium, corresponding to
an angle of refraction in the less dense medium equal to 90˚.• The absolute refractive index is equal to the reciprocal of the sine of the critical angle
when light travels from this medium into air or vacuum.
• Total internal reflection takes place when the angle of incidence in the more densemedium is greater than the critical angle.
• The mirage is a phenomenon that can be explained by total internal reflection. • The angle of the apex of the prism is given by:
A = θ1 + φ2
• The angle of deviation is given by:
α = (φ1 + θ2) - A
where φ1 is the angle of incidence θ2 is the angle of emergence • In the case of minimum deviation :
φ1 = θ2 = φ0
θ1 = φ2 = θ 0
∆y = λRd
n = 1Sin cs
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 76
78
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
Questions and Drills
I) Essay questions1) Explain why light is considered to be a wave motion .2) Describe an experiment to demonstrate the interference of light. 3) Explain how mirage is formed.
II) Define : a) the relative refractive index between two media. b) the absolute refractive index for a medium. c) the critical angle.d) the angle of deviation.
III) Complete : a) The distance between two successive bright fringes is given by ....................................b) Snell’s law states that : ......................................................c) The angle of deviation in a thin prism is given from relation :.....................d) The dispersive power is: ......................................................
IV) Choose the right answer :1) When light reflects :
a) the angle of incidence is less than the angle of reflection.b) the angle of incidence is greater than the angle of reflection.c) the angle of incidence is equal to the angle of reflection.d) there is no right answer above .
2) When light refracts, the ratio ,where φ is the angle of incidence and θ is the angleof refraction is: a) constant for the two media.b) variable for the two media.
sin φsin θ
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 78
80
Un
it 1
: W
ave
s C
ha
pte
r 3
: L
igh
t
7) The angle of incidence in a medium is 60˚ and the angle of refraction in the secondmedium is 30˚. Then the relative refractive index from the first to the second medium is :a) 3 b) c) d) 2
8) An incident ray at an angle 48.5˚ on one of the faces of a glass rectangular block (n =1.5), the angle of refraction is :
a)20˚ b)30˚ c) 35˚ d) 40˚
9) In an experiment it was found that the minimum angle of deviation is 48.2˚ Given thatthe angle of the prism is 58.8˚, the refractive index of the material of the prism is : a) 1.5 b) 1.63 c) 1.85 d) 1/1.85
10) If the critical angle for a medium to air is 45˚, then the absolute refractive index is : a) 1.64 b)2 c)1.7 d) 2
11) A thin prism has an angle of 5˚. Its refractive index is 1.6. It produces a minimum angleof deviation equal to :
a) 5˚ b) 6˚ c) 8˚ d) 3˚
12) A ray of light falls on a thin prism at an angle of deviation 4˚ and its apex angle 8˚.Itsrefractive index is :
a) 1.5 b)1.4 c) 1.33 d) 1.6
122
cyan magenta yellow black File: ch 3 fl ˙ Æ˝ ” 80
83
Un
it 2: F
luid
Me
ch
an
ics C
ha
pte
r 4: H
yd
rosta
tics
Chapter 4 Hydrostatics
= mv
ρVol
Overview
Fluids are materials which can flow. They are liquids and gases. Gases differ fromliquids in compressibility. Gases are compressible, while liquids are incompressible. Thus,liquids occupy a certain volume, while gases can fill any volume they occupy, i.e., thevolume of the container.
Density
Density is a basic property of matter. It is the mass per unit volume (kg / m3)
where Vol is the volume
Density varies from one element to another due to:1) difference in atomic weights2) difference in interatomic or intermolecular distances or molecular spacings.We know that bodies of less density float over more dense liquids. The following tableshows density for different material.
(4 -1)
85
Un
it 2: F
luid
Me
ch
an
ics C
ha
pte
r 4: H
yd
rosta
tics
2) Measuring density is used in clinical medicine, such as measuring blood and urinedensities. Normal blood density is 1040 kg / m3 – 1060 kg / m3. High density indicateshigher concentrations of blood cells and lower concentrations indicate anemia. The normal urine density is 1020 kg / m3. In some diseases, salts increase and cause theurine density to increase.
Pressure
Pressure at a point is the average force which acts normal to unit area at that point. Ifforce F is normal to a surface of an area A, then the affected pressure P on the surface isdetermined by the following relation:
Learn at Leisure
Elephant’s foot vs human foot Because the pressure is the force per unit area, the
pressure due to a pointed high heel is greater than thepressure due to an elephant’s foot, since the area of thepointed heel is very small (Fig 4 – 1).
(4 - 3) P = FA
Meaning of pressureFig (4–1)
84
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
4:
H
yd
rosta
tics
The ratio of density of any material to that of water at the same temperature is called therelative density of the material (no units). The relative density of a material, is equal to :
=
=
Applications to density1)Measuring density is of great importance in analysis, such as measuring the density of the
electrolyte in a car battery. When the battery is discharged, the density of the electrolyte (dilutesulfuric acid) is low due to chemical reaction with the lead plates and the formation of leadsulfate. When the battery is recharged, the sulfate is loosened from the lead plates and go backto the electrolyte, and the density increases once more. Thus, measuring the density indicateshow well the battery is charged.
Material
Solids:Aluminum†
Brass CoperGlassGoldIceIronLead
PlatinumSteelSugerWax
Liquids:Ethyl Alchol
BenzeneBlood
Gasoline
2700860088902600
19300910
79001140021400
783016001800
790900
1040690
KeroseneMercuryGlycerin
Water
Gases:Air
AmmoniaCarbon dioxide
Carbon mono oxideHelium
HydrogenNitrogenOxygen
82013600
12601000
1.290.761.961.250.18
0.0901.251.43
the mass of a certain volume of matter at a certain temperaturethe mass of the same volume of water at the same temperature
(4 - 2) the density of the material at a certain temperaturethe density of water at the same temperature
Densitykg/m3Material Density
kg/m3
87
Un
it 2: F
luid
Me
ch
an
ics C
ha
pte
r 4: H
yd
rosta
tics
acted upon by two forces: its weight downwards and the force due to the pressure of theliquid around it. As the depth of the liquid increases, the pressure increases (Fig 4 – 3).
To calculate the pressure (P), we imagine a horizontal plate x ofarea A at depth h inside a liquid of density ρ (Fig 4 – 4). This plateacts as the base of a column of the liquid. The force acting on the platex is the weight of the column of the liquid whose height is h andwhose cross section is A.
Because the liquid is incompressible, the force resulting from theliquid pressure must balance with the weight of the column(of theliquid. The volume of this column is Ah and its mass Ahρ, hence itsweight Fg is given by :
Fg = Ahρgwhere g the acceleration due to gravity. The pressure due to the
liquid from under the plate x (acting upwards) must be :
Pressure increases with liquid depth
pressure Po
pressure P
Fig (4 – 3)
P = =FA
AhρgA
(4 − 4)... P = hρg
Fig (4 – 4)Calculation of the pressure
of a column of liquid
86
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
4:
H
yd
rosta
tics
c) pressure on the surface of an object is equalto the pressure on the surface of a similar sizeof the liquid of the same volume and shape
Fig (4 – 2)
a) pressure inside a liquid is perpendicularto any surface inside the liquid
b) pressure is perpendicular to the surface of animmersed body at every point
Pressure at a point inside a liquid and its measurement.
If you push a piece of foam under water and let it go, it will rise and float. This indicatesthat water pushes the immersed foam by an upward force. This force is due to the pressuredifference across this piece of foam.
At any point inside a liquid, the pressure acts in any direction. The direction of the forceon any surface is normal to that surface. The pressure on a body is the same as the pressureon a volume of the liquid that has the same shape of the body in case this body wereremoved. In other words, the liquid occupying the same size which a body would occupy is
d) in a certain size of a liquid there is equilibriumbetween two forces: the weight of the liquid andthe pressure due to the remainder of the liquid.
Pressure in a liquid
PdA
PdA
PdA
89
Un
it 2: F
luid
Me
ch
an
ics C
ha
pte
r 4: H
yd
rosta
tics
Balance of liquids in a U - shaped tubeLet us take a U - shaped tube filled with an appropriate amount of water. Let us add a
quantity of oil in the left branch of the tube, until the level of oil reaches level C at a height
ho over the separating surface AD between water and oil, noting that both liquids do not
mix. Let the height of the water in the right branch be hw above level AD (Fig 4 – 7).
Because the pressure at A = pressure at D
... Pa + ρogho = Pa + ρwghw
where Pa is the atmospheric pressure, ρo the density of oil, ρw
the density of water. Thus, ho ρo = hwρw or
Measuring ho and hw, we may determine practically the
density of oil, knowing the density of water.
Atmospheric Pressure
Torcelli invented the mercury barometer to measure the atmospheric pressure. He took a
1 m long glass tube and filled it completely with mercury and turned it upside down in a
tank of mercury. He noticed that the level of mercury went down to a certain level that
measured 0.76 m from the surface of mercury in the tank. The void above the column of
mercury in the tube is vacuum (neglecting mercury vapor) is called Torcelli vacuum.
(4 -6)
Balance of liquidsin a U - shaped tube
Fig(4 – 7)
ρoρw
hwho
=
oilwater
88
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
4:
H
yd
rosta
tics Taking into consideration the fact that the free surface of the liquid is subject to
atmospheric pressure Pa ,then the total pressure at a point inside a liquid it depth d is given
by:
P =Pa + h ρ g
Practical observations show indeed that the liquid pressure at
a point inside it increases with increasing depth and with
increasing density at the same depth.
Thus, we conclude :
1) All points that lie on a horizontal plane inside a liquid has the
same pressure.
2) The liquid that fills connecting vessels rise in these vessels to
the same height, regardless of the geometrical shape of these
vessels provided that the base is in a horizontal plane
(Fig 4 – 5).
Therefore, the average sea level is constant
for all connected seas and oceans.
3) The base of a dam must be thicker than that
the top to withstand the increasing pressure
at increasing depths (Fig 4 – 6).
(4 − 5)
Fig (4 – 6)Dams must be thicker at the base to
withstand the pressure at increasing depths
Fig (4 – 5)Water rises to the same
level in connecting vessels
90
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
4:
H
yd
rosta
tics From Fig (4 – 8), the height h of the mercury column in the tube is constant, wether the
tube is upright or inclined. Taking two points A, B in one horizontal plane (Fig 4 – 9),
such that A is outside the tube at the surface of mercury in the tank, while B is inside the
tube. The pressure at B = the pressure at A. Thus:
Pa = ρgh
This means that the atmospheric pressure is equivalent to the
weight of a column of mercury whose height is 0.76 m and cross
sectional area 1m2 at OC° at sea level. This is known as S.T.P.
(standard temperature and pressure). Since the density of mercury
at O C° is 13595 kg/m3 and g = 9.8 m/s2
Pa = 1 Atm = 0.76 x 13595 x 9.81
= 1.013 x 105 N/m2
Mercury height in a barometer is not affected by the tilting of the manometer
Fig (4 – 8)
vacuum
mercury
atmospheric pressure
Fig (4 – 9)A simple barometer
(4 - 7)
93
Un
it 2: F
luid
Me
ch
an
ics C
ha
pte
r 4: H
yd
rosta
tics
Applications to Pressure
1) Blood is a viscous liquid pumped through a complicated network of arteries and veins
by the muscular effect of the heart. This is called steady flow (chapter 5). In the case ofturbulent flow (chapter 5), there is noise which can be detected by a stethoscope. Thereare two values for blood pressure: the systolic pressure, as blood pressure is maximum(normally 120 Torr). This occurs when the cardiac muscle contracts and blood ispushed from the left ventricle to the aorta onto the arteries. The diastolic pressure is theminimum blood pressure (normally 80 Torr) when the cardiac muscle relaxes.
2) When a tire is well inflated (under high pressure) the area of contact with the road is assmall as possible, while an underinflated (low pressure) tire has large contact area. Asthe area of contact with the road increases, friction increases and consequently, the tireis heated. Air pressure in a tire can be measured by a pressure gauge (Fig 4 – 12).
Fig (4-12)Measuring tire pressure with a pressure gauge
graduated scale
intake from the tire
92
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
4:
H
yd
rosta
tics Manometer
The manometer is a U - shaped tube containing a proper amount of liquid of a knowndensity. One end is connected to a gas reservoir. The level of the liquid in the manometermay rise in one branch and go down in the other. Taking two points A,B in one horizontalplane in the same liquid (Fig 4 – 11 a), we have the pressure at B = the pressure A
When P- the pressure of the gas enclosed in the reservoir - is greater than Pa, ρgh is theweight of a column the liquid in the free end of the manometer above point B and is thedifference between P and Pa (Fig 4 -11a),
P = Pa + ρghIn the case P < Pa (Fig 4 – 11 b),
P = Pa - ρghi.e., the level of the liquid in the free end branch is lower than the level of the liquid in
the end connected to the gas reservoir by a height h. In many cases, it suffices to measurethe pressure difference,
∆ P = P - Pa = ρgh (4 - 8)Knowing the liquid density ρ in the manometer and the height difference h between the
liquid levels in the two branches and the acceleration due to gravity g,we can calculate ∆P.Knowing Pa ,we may determine P of the gas enclosed in the reservoir.
a) when gas pressure > atmospheric pressure
b)when gas pressure < atmosphericpressure
Fig (4-11)Manometer
94
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
4:
H
yd
rosta
tics Examples
1) A solid parallelepiped (5cm x 10cmx 20cm) has density 5000kg /m3 is placed on a
horizontal plane. Calculate the highest and lowest pressure. (g = 10 m/s2)
Solution
For the highest pressure it is placed on the side with the least area (5 cm x 10 cm), where
the force is the weight.
For the lowest pressure, it is placed on the side of the greatest area (10 cm x 20cm)
2) Find the total pressure and the total force acting on the base of a tank filled with salty
water of density 1030 kg/m3. If the cross-section of the base is 1000 cm2 , the height of
the water is 1cm and the surface of the water is exposed to air. Take g = 10 m/s2 and the
atmospheric pressure = 1 Atm = 1.013 x 105N/m2
Solution
Total pressure
P = Pa + ρg h
= 1.013 x 105 + 1030 x 10 x 1
= (1.013 + 0.103) x 105 = 1.116 x 105 N/m2
Total force F = P x A = 1.116 x 105 x 1000 x 10-4
= 1.116 x 104 N
P = FA
= 5 x 10 x 20 x 10-6 x 5000 x 1010 x 20 x 10-4
= 2500 N/m2
P = FA
= 5 x 10 x 20 x 10-6 x 5000 x 105 x 10 x 10-4
= 10 4Fg
Fg
N/ m 2)
97
Un
it 2: F
luid
Me
ch
an
ics C
ha
pte
r 4: H
yd
rosta
tics
If pressure P is exerted to the small piston, this pressure is transmitted to the liquid and,subsequently, to the surface of the large piston. If the force applied to the small piston is fand the force affecting the large piston is F, and because the pressure on both pistons mustbe the same at equilibrium at the same horizontal plane, then :
(4 - 9)From this relation, it is clear that if force f affects a small piston, a large force F is
generated on the large piston. The mechanical advantage of the hydraulic press η is givenby:
(4 - 10)
Thus, the mechanical advantage of a hydraulic press is determined by the ratio of thelarge piston to the small piston. Referring to Fig (4 – 15), it is clear that if the small pistonmoves down a distance y
1 under the influence of f, then the large piston moves up a
distance y2 under the effect of F. According to the law of conservation of energy, the work
done in both cases must be the same (for 100% piston efficiency),
This shows that the mechanical advantage of thepiston may alternatively be expressed as the ratio y
1/y
2
P = fa
= FA
F = Aa
f
η
(Fig 4-15)Mechanical advantage
cylinder
liquid
(4 - 11)y1
y2
F = f
f y1
= F y2
y2
y1
fF =
f
f
f
y1
y2
= =Ff
Aa
96
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
4:
H
yd
rosta
tics
Fig (4-14)Hydraulic Press
b) a weight of 1 kg to the left generates aforce equivalent to 100 kg to the right if the
ratio of the two cross sectional areas is 1:100
Pascal’s principle
Consider a glass container (Fig 4 – 13) partially filled withliquid and equipped with a piston at the top. The pressure at apoint A inside the liquid at depth h is P =P1 + hρg where P1 isthe pressure immediately underneath the piston, which resultsfrom the atmospheric pressure, as well as the weight of the pistonand the force applied on the piston. If we increase the pressure onthe piston by an amount ∆P, by placing an additional weight onthe piston. We note that the piston does not move inside becausethe liquid is incompressible. But the pressure underneath thepiston must increase in turn by ∆P. This raises the pressure atpoint A by ∆P. This make the pressure P =P1 + ρgh + ∆ P .
Pascal formulated this result as follows :When pressure is applied on a liquid enclosed in a container, the pressure is
transmitted in full to all parts of the liquid as well as to the walls of the container. This isknown as Pascal’s principle or Pascal’s rule.
Application to Pascal’s rule : hydraulic Press
The hydraulic press (Fig 4 – 14) consists of a small piston whose cross sectional area is“a” and a large piston whose cross sectional area is “A”. The space between the twopistons is filled with an appropriate liquid.
a) force to the left is transmitted tothe right
cross sectional area A
cross sectional area a
Fig (4-13)Increase of weight on the
piston increases thepressure in the liquid
1kg100kg
99
Un
it 2: F
luid
Me
ch
an
ics C
ha
pte
r 4: H
yd
rosta
tics
3) The caterpillar also uses Pascal’s rule (Fig 4 – 19).4) A diver wears a diving suit and a helmet to protect him
from pressures at large depths. At low (shallow)depths, the diver - without the helmet - blows air in hissinuses to balance the external pressure (Fig 4 – 20). Atlarge depths, the diving suit is appropriately inflatedwith air, and the helmet protects the diver’s head fromcrushing pressures (Fig 4 – 21).
ExamplesA hydraulic press has cross sectional area 10cm2 which is acted on by a force of 100N.
The large cross sectional area is 800 cm2. Taking g = 10m/s2 ,calculate :a) the largest mass that can be lifted by the press b) the mechanical advantage of the press c) the distance traveled by the small piston so that the large piston moves a distance of 1cm
SolutionThe force acting on the large piston :
Fig (4-19) A hydraulic caterpillar
Fig (4-21) Diving at large depths (500 m)
Fig (4-20)Diving at low depths
NF = 100 x 800 = 8 x 1010
3
Fa
FA=
98
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
4:
H
yd
rosta
tics
hydraulic liquid
Learn at Leisure
Applications to Pascal’s rule
1) The hydraulic brake in a car uses Pascal’s rule as the brakingsystem uses a brake fluid. Upon pushing on the brake pedal witha small force and a relativeley long stroke (distance), thepressure is transmitted in the master brake cylinder, hence, ontothe liquid and the whole hydraulic line, then to the piston of thewheel cylinder outwardly, and finally to the brake shoes and thebrake drum. A force of friction results, which eventually stopsthe car. This type of brakes is called drum brake (rear brake) (Fig4 – 16). In the case of the front (disk) brake (Fig 4 – 17), the forcesresulting from the braking action press on the brake pads whichproduce friction enough to stop the wheel. It should be noted thatthe distance traveled by the brake shoes in both cases is small becausethe force is large.
2) In another application to Pascal’s rule, a hydraulic lift uses a liquidto lift up cars in gas stations (Fig 4 – 18). Fig (4-17)
Front brakes
Fig (4-16)Rear brakes
Fig (4-18) A hydraulic lift
101
Un
it 2: F
luid
Me
ch
an
ics C
ha
pte
r 4: H
yd
rosta
tics
1) Horizontal forces cancel each other out , because each two opposite forces are equal inmagnitude and opposite in direction.
2) As to forces in the vertical direction, we find that the weight of the liquid enclosed in volumeVol in which (Fg)
L = Vol
ρg acts downwards. Since this liquid is static, the liquid must exerton the enclosed liquid an equal force Fb upwards , which is equal to the weight of theenclosed liquid. This force results from the difference of the pressure on the upper and lowersurfaces of the parallepiped which is ∆P x A
Substituting, noting that Ah is the volume:
where (Fg)l is the weight of the displaced liquid.
We see that Fb is equal to the weight of the parallepiped of theenclosed liquid. Equilibrium requires that the force Fb worksupwards, and it is named buoyancy (buoyant force-upthrust).
The relation between the weight of a body in airand the weight when immersed in a liquid
If we substitute the virtual parallepiped by a solid parallepiped of thesame shape and volume and of density ρs (Fig 4 – 22 b), the buoyancy(upthrust or buoyant force) which the liquid exerts on the solidparallepiped remains the same Fb acting upwards. The liquid isdisplaced a distance ∆h. The weight of the parallepiped representingthe immersed body (Fg)s acts downwards. The resultant force on
Fb = Volρg = (Fg)l (4 - 12)
Fig (4-22b)Archimedes’ rule using a real parallepiped
instead of a virtual parallepiped
Fb = ∆P X A∆P = P1 - P2 = h1ρg - h2ρg = (h1 - h2) ρg∆P = hρgFb = Ahρg
100
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
4:
H
yd
rosta
tics
= 8000 x 1100
= 80 Cmy1 cm
a)To calculate the largest mass that can be lifted by the large piston,
b)To calculate the mechanical advantage,
c)To calculate the distance traveled by the small piston,
Buoyancy and Archimedes’ Principle We are familiar with the following observations:
1)An object can be easily lifted if immersed under water level, whereas it might be difficultto lift at in air.
2) A piece of foam floats when immersed in water. 3) An iron nail sinks in water while a large steel ship floats. 4) Balloons filled with helium rise up.
We can interpret the above observations as follows :When an object is immersed under the liquid surfacethen the object exerts a force on the liquid.Consequently, the fluid (liquid or gas) pushes backby an equal and opposite force (Newton’s third law). Thisforce is called buoyancy. It acts in all directions,but thenet effect is upwards. The buoyancy is given by the weightof the liquid displaced by the immersed body. To showthis, let us imagine the existence of a volume Vol of theliquid as a virtual parallepiped whose cross sectional area isA and height h.This parallepiped is acted upon by forces in alldirections (Fig 4 – 22 A). This part of the liquid (like any otherpart of stable liquid) does not move, so it is in equilibrium:
m = Fg
= 8 x 103
10 = 800 Kgkg
= Ff
= Aa
= 80010
= 80
Fig (4-22a)Archimedes’ rule considering
a virtual parallepiped
η
fy F y1 2=
103
Un
it 2: F
luid
Me
ch
an
ics C
ha
pte
r 4: H
yd
rosta
tics
difference between the body’s weight in air and the buoyancy of the liquid (Fg)s = (Fg)s - Fbwhere (Fg)s is the weight of the body while totally immersed in the liquid, (Fg)s is its
weight in air and Fb is the buoyancy (Fig 4 – 24). Concluding from above, we may formulate Archimedes’ principle as follows: A body
partially or fully immersed in a fluid (liquid or gas) is pushed upwards by a forceequal to the weight of the volume of the fluid displaced partially or fully by the body.
Learn at Leisure
Archimedes’ rule and Newton’s law
The immersed body replaces an equal size of the liquid. Because liquids areincompressible, such a body displaces a volume of the liquid equal to the volume of theimmersed body (or part of it that is immersed). This displaced liquid acts as a mass placedon the surface of the liquid. Its weight presses on the surface of the liquid. Thus, thepressure on each point of the liquid increases by this amount. Because we calculate thebuoyancy on the body as the difference between two forces acting on the surface of theimmersed body, then this difference stays the same. Hence, buoyancy on the immersedbody is unchanged by the displacement. It is equal to the weight of the displaced liquid(Fig 4 – 12). We can understand what happens as that the weight of the immersed body actson the liquid as a whole, then the liquid acts back with buoyancy as a reaction equal inmagnitude and opposite in direction (Newton’s third law). In the case when the weightbalances out with buoyancy, we have equilibrium, and the body remains suspended in theliquid. If the weight of the body exceeds buoyancy, the body sinks to the bottom where itsettles there. If the weight of the body is less than buoyancy, the body floats on top of thesurface where it settles afloat, while the weight of the displaced liquid whose volume isequal to the volume of the immersed part of the body, which is then equal to the weight ofthe floating body.
105
Un
it 2: F
luid
Me
ch
an
ics C
ha
pte
r 4: H
yd
rosta
tics
(displacement of the liquid). In the case of gases, there is buoyancy force as well, and it actsupwards. But it is not mandatory that the volume of the displaced gas is equal to the volumeof the immersed body, since gases are compressible. But buoyancy must be equal to theweight of the displaced gas. This is why balloons filled with helium rise up (Fig 4 – 27).
Learn at Leisure
The story of Archimedes and the crown Archimedes was one of the most celebrated scientists of Ancient Greece. There is an
interesting story of Archimedes’ discovery of his rule. When Heron-king of Syracuse (oneof ancient Greek cities)- doubted that his new crown might not have been made of puregold, he summoned Archimedes to seek his counsel as to whether or not the crown wasrigged, without destroying the crown of course. Archimedes was first at a loss. But one day,he was bathing in a tub. He noticed that as he dipped himself in the tub the water level rose.Archimedes brought the crown and submerged it in a water filled tub and measured thedisplaced (overflown) water. He then calculated the density of the material of the crown.He then repeated the experiment on a similar block of pure gold of the same mass, andmeasured the volume of the displaced water. He found that this volume was different. Heconcluded that the maker cheated and used less dense and hence cheaper materials. It isoften told that Archimedes was euphoric with joy as he got this idea. He came out from thetub, and ran out naked shouting: “Eureka Eureka (I found it – I found it)". This expressionhas been ever since coined as a mottofor scientific discovery. Question:was the displaced water for therigged crown more or less than thatfor equal mass of pure gold? (Fig4-28).
Fig (4-28)The crown’s story
104
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
4:
H
yd
rosta
tics Application to buoyancy
1) Hydrotherapy technique is prescribed to patientswho are unable to lift limbs because of disease orinjury in the associated muscles or joints. When abody is immersed in water it becomes, in effect,nearly weightless. As a result, the force requiredto move the limb is greatly reduced, and thetherapeutic exercise becomes possible.
2) Weightlessness experiments may involveimmersion in containers filled with a liquid whoseconcentration is adjusted so that buoyancy cancelsout the weight.
3) A submarine floats when its tanks are filled with air,and submerges when those tanks are filled with water.Fish and Wales also fill air sacs with air to enablethem to float, and empty them from air when they gounder.
4) A diver breathes air under compression when divingto shallow depths, to equate the pressure. At largerdepths, the diver adjusts the pressure in the divingsuit to control the buoyancy force (Fig 4 – 26).
Learn at Leisure
Is there buoyancy for gases ?In the case of liquids, the volume of the displaced
liquid equals the volume of the immersed body,because liquids are incompressible. Then, the forceacting on the body upwards is equal to the weight ofthe displaced liquid as a reaction to the immersion
Fig (4-25)A submarine floats and sinks by
emptying or filling water tanks
Fig (4-26)A diver floats or dives depending onthe varying density of the diving suit
Fig (4-27)A balloon filled with helium
106
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
4:
H
yd
rosta
tics Learn at Leisure
Why does a ship float while an iron nail sinks ?It is noted that a ship is of large mass but contains large voids and spaces, which cause it
to displace plenty of water, hence large buoyancy pushes the ship upwards. Part of the ship(hull) remains submerged in water for the ship to be afloat. This submerged part is whatcauses the displacement of water of an equal volume causing buoyancy to balance out withthe weight of the ship (4 – 29). As the cargo on the ship increases, the submerged partincreases to build up more buoyancy enough to keep the ship afloat (Fig 4 – 30a). The ironnail sinks because the buoyancy force on it is small due to its small volume (Fig 4 – 30 b).
Learn at Leisure
The Dead Sea Have you noticed the difference between swimming in the sea, the Nile, and swimming
pools? The Dead Sea in Jordan is enclosed. It is not connected to any sea or ocean. The saltconcentration is very high. No one drowns in the dead sea (why ?)
Fig (4-29)A ship floats despite its massive weight
Fig (4-30b)A ship floats despite its massive weight
while a nail sinks
Fig (4-30a)The part of the boat that is submerged
depends on the weight
106
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
4:
H
yd
rosta
tics Learn at Leisure
Why does a ship float while an iron nail sinks ?It is noted that a ship is of large mass but contains large voids and spaces, which cause it
to displace plenty of water, hence large buoyancy pushes the ship upwards. Part of the ship(hull) remains submerged in water for the ship to be afloat. This submerged part is whatcauses the displacement of water of an equal volume causing buoyancy to balance out withthe weight of the ship (4 – 29). As the cargo on the ship increases, the submerged partincreases to build up more buoyancy enough to keep the ship afloat (Fig 4 – 30a). The ironnail sinks because the buoyancy force on it is small due to its small volume (Fig 4 – 30 b).
Learn at Leisure
The Dead Sea Have you noticed the difference between swimming in the sea, the Nile, and swimming
pools? The Dead Sea in Jordan is enclosed. It is not connected to any sea or ocean. The saltconcentration is very high. No one drowns in the dead sea (why ?)
Fig (4-29)A ship floats despite its massive weight
Fig (4-30b)A ship floats despite its massive weight
while a nail sinks
Fig (4-30a)The part of the boat that is submerged
depends on the weight
109
Un
it 2: F
luid
Me
ch
an
ics C
ha
pte
r 4: H
yd
rosta
tics
In a Nutshell
I.Definitions and Basic Concepts • The density (ρ) is the mass per unit volume (kg/m3)• The pressure P at a point is the normal force acting on a unit surface area (N/m2).• All points lying in the same plane have the same pressure. • The atmospheric pressure is equivalent to the pressure produced by the weight of a
mercury column of height about 0.76 m and base area 1m2 at 0°C• The units of atmospheric pressure are :
a) Pascal (1 N / m 2).b) Bar (10 5 N / m 2).c)cm Hg.d) Torr (mm. Hg).
• The manometer is an instrument for measuring the difference in the pressure of a gasinside a container and the outer atmospheric pressure. • Pascal’s principle : The pressure applied on an enclosed liquid is transmitted undiminished to every
portion of the liquid and to the walls of the container. • Archimedes’ principle :
A body immersed wholly or partially in a fluid experiences an upthrust force(buoyant force) in the vertical direction equal to the weight of the liquid displaced bythe body.• The weight of the volume displaced = volume of the immersed body x density of the
liquid x the acceleration due to gravity. • The immersed body in the liquid is acted upon by two forces, the upthrust force Fb and
the weight of the body (Fg)s.
If (Fg)s = Fb the body will be suspended in the liquid.
113
Un
it 2: F
luid
Me
ch
an
ics C
ha
pte
r 4: H
yd
rosta
tics
5) If the ratio between large and small piston diameters is 9:2. The ratio between the twoforces on the two pistons are :a) 9:2 b) 2: 9 c) 4:18d) 81 : 4 e) 4:81
6) A body has mass 5 kg in air, its weight when immersed in liquid becomes 40 N. If theacceleration due to gravity is 10 m / s2, then the upthrust force on the body is : a) 10 kg b) 10 N c) 35 kgd) 35 N e) 90 N
7) A piece of wood floats on water such that 1/4 its volume appears over the water surface.If the water density is 1000 kg / m3 , the wood density is then : a) 1333 kg / m3 b) 250 kg / m3 c) 750 kg/m3
d) 1000 kg / m3 e) 500 kg / m3
8) The lead density is greater than the copper density, the copper density is greater than thatof aluminum. If we immerse a number of cubes with equal volumes of these metals inwater and weigh them then, compared to their weights in air: a) the decrease of weight of the lead cube is greater than the decrease of weight of the
copper cube. b) the decrease of weight of the aluminum cube is greater than the decrease of weight of
the copper cube. c) the decrease of weight of the aluminum cube is equal to the decrease of weight of the
lead cube. d) the decrease of weight of the aluminum cube is less than the decrease of weight of the
lead cube. e) the decrease of weight of the lead cube is less than the decrease of weight of the
copper cube.
112
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
4:
H
yd
rosta
tics Questions and Drills
I) Put mark ( ) to the correct statement:1) The following factors affect the pressure at the bottom of a vessel except one, tick it:
a) the liquid depth in the vessel. b) the density of the liquid c) the acceleration due to gravity. d) the atmospheric pressure. e) the area of the vessel base.
2) Which of the following factors have no effect on the height of mercury column in abarometer?a) the density of mercury b) the cross sectional area of the tube. c) atmospheric pressure. d) the acceleration due to gravity. e) the temperature of mercury.
3) When a ship moves from river water to sea water,which of the following statements is right ? a) the water density increases and the ship sinks slightly. b) the water density increases and the ship floats upwards slightly c) the water density decreases and the ship sinks slightly.d) the water density decreases and the ship floats slightly. e) the water density does not change and nothing happens.
4) The water pressure at the bottom of the High Dam lake on the dam body depends on :a) the area of the water surface.b) the length of the dam. c) the depth of the water. d) the thickness of dam wall.e) the density of wall substance.
115
Un
it 2: F
luid
Me
ch
an
ics C
ha
pte
r 4: H
yd
rosta
tics
5) A vessel on a table contains two liquids: oil and water. A wooden cube is placed on the
top surface of the upper liquid. Describe what happens to the two liquids, the wooden
cube, the bottom of the vessel and the table.
IV) Drills :
1) The pressure on the base of a cylinder containing oil with diameter 8 m is 1.5 x 103
N/m2. Find the total force on the base.
(7.54 x 104 N)
2) A difference in pressure of 3.039 x 105 N/m2 is recommended for air in a car tire. If the
atmospheric pressure is 1.013 x 105 N/m2 ,calculate the absolute pressure of air in the
tire in atmospheric units.
(4 Atm)
3) A fish tank of cross-sectional area 1000 cm2 contains water of weight 4000N. Find the
pressure on its base.
(0.4 x 104 N/m2)
4) The large and small piston diameters of a hydraulic press are 24cm and 2cm
respectively. Calculate the force that must be applied to the small piston to obtain a force
of 2000 N on the large piston. Then calculate the mechanical advantage.
(13.9 N, 144)
5) A piece of aluminum has a mass of 250 gram in air. When immersed in water it has an
an apparent mass of 160 gram and it has 180 gram in alcohol. Calculate the densities of
aluminum and alcohol if the density of water is 1000 kg /m3 and g = 9.8 m / s 2
(2777.8 kg / m3, 777.8 kg/m3)
114
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
4:
H
yd
rosta
tics 9) If the ice density is, 900 kg / m 3 and the water density is 1000 kg /m3 , then the ratio of
the floating part of an ice cube is :a) 90 % b) 10 % c) 100 %d) 80 % e) 20 %
10) A body is immersed wholly in a liquid. If the body density is greater than the liquiddensity,then the upthrust force exerted by the liquid on the body will be :a) equal to the mass of liquid displaced by the body. b) equal to the mass of the immersed body.c) equal to the volume of the liquid displaced by the body.d) equal to the weight of the liquid displaced by the body.e) greater than the body weight.
II) Define each of the following :1. Density 2. Pressure at a point3. Pascal’s principle 4. Archimedes’ principle
III) Essay questions :1) Prove that the pressure (P) at depth (h) in a liquid is determined from the relation.
P = Pa + ρgh
where Pa is the atmospheric pressure, ρ the liquid density and g is the acceleration due togravity.2) Describe the manometer and show how it can be used for measuring a gas pressure
inside a container. 3) What is meant by Pascal’s principle ? Describe one of its applications.4) Show that the resultant forces on an immersed body is given by :F = (ρl – ρs) g Vol ,
where ρl and ρs are the densities of the liquid and the body respectively,g is theacceleration due to gravity and Vol is the volume of the body. Explain the consequences ofthis relation.
116
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
4:
H
yd
rosta
tics 6)The atmospheric pressure on the surface of a lake is 1Atm. The pressure at its bottom is
3 Atm. Calculate the depth of the lake (density of water 1000 kg/m3, 1 Atm =
1.013x105 N/m2, g = 9.8 m/s2 ).
(20.673 m)
7) A man carries a mercury barometer with readings 76 cm Hg and 74.15 cm Hg at the
lower and upper floors, respectively. Calculate the average density of air between the
two floors if mercury density is 13600 kg / m3, the building height is 200m and g =
9.8m/s2.
(1.258 kg/m3)
8) A manometer containing mercury is attached to gas enclosed in a container. If the
difference height in the manometer is 25 cm.
Calculate the pressure difference and the absolute pressure of the enclosed air in units of
N/m2 (1Atm = 1.013 x 105 N/m2, mercury density = 13600 kg/m3 and g=9.8 ms2)
(0.3332x105 N/m2, 1.3462x105 N/m2).
9) The volume of a huge balloon filled with hydrogen is 14x104m3. Find the maximum
lifting force acting on it, if the hydrogen density is 0.092 kg/m3, air density is1.29
kg/m3 and the mass of the balloon with its accessories is 8x104 kg. (g = 10m/s 2).
(87.72x104 N)
Unit 2 : F
luid Mechanics C
hapter 5: Hydrodynam
ics
cyan magenta yellow black File: CH 5 fl ˙ Æ˝ ” 117
Unit 2 : F
luid Mechanics C
hapter 5: Hydrodynam
ics
119
Un
it 2: F
luid
Me
ch
an
ics C
ha
pte
r 5: H
yd
rosta
tics
2) In steady flow, the velocity of the liquid at each point is independent of time.
3) The flow is irrotational, i.e., there is no vortex motion.4)If no forces of friction exist between the layers of the liquid the flow is nonviscous .If
there is friction it is viscous.
Turbulent flow
If the velocity of flow of a liquid exceeds a certain limit, steadyflow changes to turbulent flow, which is characterized by theexistence of vortices (Fig 5–2). The same thing happens to gasesas a result of diffusion from a small space to a large space or fromhigh pressure to low pressure (Fig 5 – 3).
Rate of flow and the continuity equation We shall focus on steady flow. Consider a flow tube such that:1) the liquid fills the tube completely.2) the quantity of the liquid entering the tube at one end equals
the quantity of the liquid emerging out from the other endwithin the same time.
3) the velocity of the liquid flow at any point in the tube doesnot change with time. There is a relation that ties the rate offlow of the liquid with its velocity and cross sectional area.This relation is called the continuity equation. Tounderstand what the continuity equation entails, we choose
two perpendicular planes normal to the streamlines at the twosections (Fig 5 - 4). The cross sectional area at the first plane isA1 and the cross sectional area at the second plane is A2. The volume rate of flow is thevolume of the liquid flowing through area A1 in unit time.We have Qv = A1v1, where v1 isthe velocity of the liquid at section A1. The mass of the liquid (of density ρ) flowing in unittime is called the mass rate of flow Qm ,which is given by:
Fig (5-3)Smoke changes from steady
to turbulent flow
Fig (5-2)Vortices due to turbulent
flow or a violent motion of abody through a liquid
cyan magenta yellow black File: CH 5 fl ˙ Æ˝ ” 119
119
Uni
t 2
: F
luid
Mec
hani
cs
C
hapt
er 5
:
Hyd
rody
nam
ics
118
Unit 2
: Flu
id M
echanic
s C
hapte
r 5:P
ropertie
s o
f Flu
id d
ynam
ics
118
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
5:
H
yd
rosta
tics HydrodynamicsChapter 5
Fig (5-1)Streamlines
Overview
Hydrodynamics (Fluid dynamics) deals, with fluids in motion. We must distinguish
between two types of fluid motion, steady flow and turbulent flow.
Steady flowIf a liquid moves such that its adjacent layers slide with respect to each other smoothly,
we describe the motion as a laminar flow or a streamline (steady) flow. In this type offlow, particles of the liquid follow continuous paths called streamlines.Thus, we mayvisualize the liquid as if it is in a real or virtual tube containing a flux of streamlinesrepresenting the paths of the different particles of the liquid (Fig 5 – 1). These streamlinesdo not intersect, and the tangent at any point along the streamline determines the directionof the instantaneous velocity of each particle of the liquid at that point. The number ofstreamlines crossing perpendicularly a unit area at a point (density of streamlines)expresses the velocity of flow of the liquid at that point. Therefore, streamlines cram up atpoints of high velocity and keep apart at points of low velocity. Conditions of Steady Flow 1)The rate of flow of the liquid is constant along its path, since the liquid is incompressible
and the density of the liquid is independent of distance or time.
cyan magenta yellow black File: CH 5 fl ˙ Æ˝ ” 118
118
Unit 2 : F
luid Mechanics C
hapter 5: Hydrodynam
ics
121
Uni
t 2
: F
luid
Mec
hani
cs
C
hapt
er 5
:
Hyd
rody
nam
ics
120
Unit 2
: Flu
id M
echanic
s C
hapte
r 5:P
ropertie
s o
f Flu
id d
ynam
ics
120
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
5:
H
yd
rosta
tics
Fig( 5-5)Basis for the continuity equation
(5-1)
V1
V2
= A2
A1
v1
v2 (5-2)
Qm=ρQv=ρ A1 v1
Similarly, the mass rate of flow through
area A2 is ρQv = ρA2v2. Since the mass rate
of flow is constant in steady flow
ρ A1 v1 = ρ A2 v2
A1 v1= A2 v2
This is the continuity equation leading to
From this relation, we see that the velocity of the liquid at any point in the tube isinversely proportional to the cross sectional area of the tube at that point. The liquid flowsslowly where the cross sectional area A1 is large and flows rapidly, when the cross sectionalarea A2 is small (Fig 5 – 5). To understand the continuity equation better, we consider a smallamount of liquid ∆m = ρ∆Vol, where ∆Vol = A1 ∆x1 ,where ∆x1 is the distance traveled by theliquid in time ∆t . Thus, ∆x1 = v1 ∆t. Then ∆Vol = A1v1 ∆t. This same value must emerge fromthe other side of the tube, since the liquid is incompressible ,i.e, ∆Vol = A2v2 ∆t . Thus, wemust emphasize that the rate of flow of the liquid is a volume rate Qv (m3/s), or a mass rate offlow (kg/s). Both of these rates are constant for any cross section. This is called theconservation of mass, which leads to the continuity equation.
Fig (5-4)Model for deducing the continuity equation
A1
A2
cyan magenta yellow black File: CH 5 fl ˙ Æ˝ ” 120
120
Unit 2 : F
luid Mechanics C
hapter 5: Hydrodynam
ics
123
Un
it 2: F
luid
Me
ch
an
ics C
ha
pte
r 5: H
yd
rosta
tics
Learn at Leisure
Hardening of the arteries The body controls the blood flow in the arteries by muscles surrounding these arteries.
When these arteries contract, the radius of the artery decreases. From the continuityequation (5 – 1), the blood velocity must hence increase. When they relax, the bloodvelocity decreases. With age, these muscles lose that elasticity, and this is called hardeningof the arteries, and hence these muscles lose the ability to control the blood flow. Ascholestrol(fats) precipitates on the inner walls of these blood vessels, the radius decreasesfurther which increases the possibility of coagulation (formation of a clot) which blocks theblood stream, leading to angina pectoris. The patient takes medication to ensure theliquidity of the blood to prevent the coagulation. However, if the dose is excessive, hemight end up with hemorrhage or (internal bleeding). It is known that one of theconstituents of blood is platelettes, which are responsible for normal coagulation to stopbleeding. Thus, viscosity of the blood and its composition play a vital role is man’s life.
Learn at Leisure
Measuring blood pressureMeasuring blood pressure is one of the ways to check on the performance of the heart.
The sphygmomanometer is a type of manometer (Fig 5 – 9). It consists of a cuff in the
form of an air bag wrapped around the patient’s arm. A hand pump is used to inflate thebag and a mercury manometer is used to measure the pressure in the bag. The pressure is
increased in the air bag, until the blood flow ceases momentarily in the brachial artery. A
stethoscope is used to indicate the soundness of the artery’s muscle in pushing the blood.
cyan magenta yellow black File: CH 5 fl ˙ Æ˝ ” 123
123
Uni
t 2
: F
luid
Mec
hani
cs
C
hapt
er 5
:
Hyd
rody
nam
ics
122
Unit 2
: Flu
id M
echanic
s C
hapte
r 5:P
ropertie
s o
f Flu
id d
ynam
ics
122
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
5:
H
yd
rosta
tics Learn at Leisure
Blood circulation in human bodyThe circulatory system consists of a huge network of blood vessels including arteries,
veins, down to capillaries (Fig 5– 8) .The heart pumps blood through this network at a rateof 5 liters per minute (or 8.33x10-5 m3/s) with a normal pulse rate of 70 pulse per minute.The pumping rate may reach 25 liters per minute or 180 pulse per minute with excessiveactivity .Calculating the velocity of flow in the aorta (2 cm diameter),we find that the bloodvelocity is 26.5 cm/s (check the calculation). If we add up all the capillaries, we find thatthe collective cross section is 0.25m2 (what is the velocity?).
brain
arms andshoulders
lungs
heart
trunkand
organs
legs
arteries
aortaartery
pulmonaryartery
veins
veins
veins
veins
Fig (8-5)A simplified diagram for blood circulation
cyan magenta yellow black File: CH 5 fl ˙ Æ˝ ” 122
122
Unit 2 : F
luid Mechanics C
hapter 5: Hydrodynam
ics
125
Uni
t 2
: F
luid
Mec
hani
cs
C
hapt
er 5
:
Hyd
rody
nam
ics
124
Unit 2
: Flu
id M
echanic
s C
hapte
r 5:P
ropertie
s o
f Flu
id d
ynam
ics
124
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
5:
H
yd
rosta
tics The pressure in the air bag is decreased gradually, until
the pressure exerted by the heart is sufficient to push
the blood through, and open up the brachial artery.
The sudden flow of the blood in the nearly closed
artery causes a turbulent flow which produces a hiss,
which the doctor can hear with the stethoscope, while
the doctor monitors the reading on the manometer in
mm Hg. This reading is the systolic pressure (normally
120 mm Hg). As the pressure in the air bag decreases,
the hiss remains audible in the stethoscope, until the
pressure in the air bag is equal to the lowest pressure
exerted by the heart when the brachial artery is fully
open (diastolic pressure). Then,blood flows steadily and
the hiss disappears. The reading on the manometer in this case (when the heart is in rest or
relaxation) is normally 80 mm Hg. If the systolic pressure exceeds 150 mm Hg, the patient
has hypertension which might cause brain hemorrhage, and hence stroke. If the diastolic
pressure exceeds 90 mm Hg, the heart-which is supposedly then at rest -has extra pressure,
causing fatigue and eventually fibrosis in the heart muscle, leading to heart failure or
cardiac arrest.
Fig (5 - 9)Sphygmomanometer
cyan magenta yellow black File: CH 5 fl ˙ Æ˝ ” 124
124
Unit 2 : F
luid Mechanics C
hapter 5: Hydrodynam
ics
127
Un
it 2: F
luid
Me
ch
an
ics C
ha
pte
r 5: H
yd
rosta
tics
Solution
The aorta cross section is given by
The collective cross section for 30 main arteries is given by
Thus, the velocity of the blood in the main arteries is 0.044 m/s. Consequently, the
blood velocity in capillaries is much smaller, which gives time for the tissues to exchange
oxygen and carbon dioxide as well as nutrients and excretion products. Divine wonder is
countless.
Viscosity
We observe viscosity as follows :
1) We hang two funnels each on a stand and put a beaker under each. We pour alcohol inone funnel and a similar volume of glycerine in the other, and observe the velocity offlow of each. We notice that the flow velocity of alcohol is higher than that of
glycerine.
2) Take two similar beakers, one containing a certain volume of water and the other anequal volume of honey .Stir the liquid in both beakers with a glass rod .Which of the two
liquids is easier to stir ? Then, we remove the rod .We notice that :
(0.007) 2
v 0.3330
0.44m/s2 =×
=4
(0.33) = (0.0035) 2 (30) v22
A1 = r 21 = (0.007) 2 m2
A2 = r 22 x 30
= (0.0035) 2 x 30 m22
A1 v1 = A2 v2
0.044 m/s
cyan magenta yellow black File: CH 5 fl ˙ Æ˝ ” 127
127
Uni
t 2
: F
luid
Mec
hani
cs
C
hapt
er 5
:
Hyd
rody
nam
ics
126
Unit 2
: Flu
id M
echanic
s C
hapte
r 5:P
ropertie
s o
f Flu
id d
ynam
ics
126
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
5:
H
yd
rosta
tics Examples
1) A water pipe 2 cm diameter is at the entrance of an apartment building. The velocity ofthe water in it is 0.1m/s. Then, the pipe tapers to 1cm diameter. Calculate
a) the velocity of the water in the narrow pipe.
b) quantity of the water (volume and mass) flowing every minute across any section ofthe pipe (density of water = 1000 kg/m3) .
Solution
a) A1 v1= A2 v2
π (0.01m)2 (0.1 m/s) = π (0.005m)2 v2
b)The volume rate of flow (m3/s) is given by the relation
Thevolume rate of flow (m3/min) is given by
The mass flowing per minute
The mass rate of flow (kg/min) is given by
2) The average velocity of blood in aorta ( radius 0.7 am ) for an adult is 0.33 m/s From theaorta, blood is distributed to main arteries (each radius 0.35cm). If we have 30 main arteries,calculate the velocity of blood in each.
Q x 60 = 3.14 x 10-5 x 60 = 188.4 x 10-5 m3 /min
= 3.14 x 10-5 x 103 = 3.14 x 10-2
= 3.14 x 10-5 x 103 x 60 = 1.884 K
v 10 10
0.4m/s2
-4
-5= π × ×π × ×
=0 12 5
..
= x 10-4 x 0.1 x 2.5 x 10-5 x 0.4
= 3.14 10 m /s -5 3×
or
Qm
Qm
cyan magenta yellow black File: CH 5 fl ˙ Æ˝ ” 126
126
Unit 2 : F
luid Mechanics C
hapter 5: Hydrodynam
ics
129
Un
it 2: F
luid
Me
ch
an
ics C
ha
pte
r 5: H
yd
rosta
tics
a) Friction forces exist between the lower plate and the liquid layer in contact with it. Thisforce is due to the adhesive forces between the molecules of the solid surface and thecontacting liquid molecules. This leads to zero velocity of the layer in contact with thestationary plate. Similarly, the upper layer moves at the same velocity of the upper plate.
b) The existence of another friction (shear) force between each liquid layer and the adjacentone, which resists the sliding of the liquid layers with respect to each other. Thisproduces a relative change in velocity between any two adjacent layers. Thus, viscosityis the property responsible for resisting the relative motion of liquid layers .This type offlow is called nonturblent viscous laminar flow (or viscous steady flow), since no
vortices occur.
Coefficient of Viscosity
Referring to Fig (5 – 6), we find that for the moving plate to maintains its constant
velocity, a force F must exist. This force is directly proportional to both velocity and area
of the moving plate A, and inversely proportional to the distance between the plates d.
where (Eta) is a constant of proportionality called viscosity coefficient, given by
The coefficient of viscosity ( Ns/m2 or kg/m s ) may be defined as :
the tangential force acting on unit area, resulting in unit velocity difference between
two layers, separated by unit distance apart .
F AVsv
d
(5 - 3)
(5 - 4)
F Avdvs
=
vsFdAv
FAv/d
=
vs
cyan magenta yellow black File: CH 5 fl ˙ Æ˝ ” 129
129
Uni
t 2
: F
luid
Mec
hani
cs
C
hapt
er 5
:
Hyd
rody
nam
ics
128
Unit 2
: Flu
id M
echanic
s C
hapte
r 5:P
ropertie
s o
f Flu
id d
ynam
ics
128
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
5:
H
yd
rosta
tics
Force acting on the upper layer of a liquid
liquid
Layers of a liquid slide with respect to each other
Fig (6-5)Friction between layers of a liquid
a) Water is easier to stir, which means that water resistance to the glass rod is less thanthe resistance of honey.
b) The motion in honey stops almost immediately after we remove the rod, while itcontinues for a little while longer in water .
3) We take two long similar measuring cylinders and fill them to the end, one with water andthe other with glycerine. Then , take two similar steal balls and drop one in each liquid, andrecord the time each ball takes in each liquid to hit the bottom. We observe that the time inwater is less. Thus, the glycerine resistance to the ball motion is greater .
We, thus, conclude :-
1) Some liquids such as water and alcohol offer little resistance to the motion of a body inthem, and are easy to flow. We say they have low viscosity
2) Other liquids such as honey and glycerine are not as easy to move through ,i.e.,theyoffer high resistance to body motion, and are said to have high viscosity.To interpret viscosity, imagine layers of liquid trapped between two parallel plates, one
stationary and the other moving with velocity v (Fig 5 –6). The liquid layer next to thestationary plate is stationary, while the layer next to the moving plate is moving at v. The layersin between move at velocities varying from o to v. The reason for this is as follow :
d
cyan magenta yellow black File: CH 5 fl ˙ Æ˝ ” 128
128
Unit 2 : F
luid Mechanics C
hapter 5: Hydrodynam
ics
131
Un
it 2: F
luid
Me
ch
an
ics C
ha
pte
r 5: H
yd
rosta
tics
Learn at Leisure
Syphon:can a liquid go up ?
Suction of air from one end of a hose while the other
end is submerged in a liquid causes the liquid to rise up
to a certain head (vertical distance), then to flow
downwards (Fig 5 – 13). People often use this technique
to draw gasoline from a car tank. This seems contrary to
gravity. This phenomenon is called syphon. It can be
explained as follows. The liquid molecules attract each
other as beads in a chain (Fig 5 –14). The molecules may
go up to a certain distance overcoming gravity, then
come back down and flow from the free end of the hose.
Liquids have another property, called surface tension,
when the molecules of the liquid at the surface attract
each other as a membrane. This is the same theory of the
formation of air bubbles, in which the internal pressure
balances out with the external pressure and the surface
tension, so they do not blow up, unless this balance is
disturbed. Another property is capillarity, where
molecules of the container pull the molecules of the liquid by forces of adhesion. The
surface of the liquid is curved due to the surface tension in the capillary. This property is
responsible for drawing water in the stem of plants through the capillaries, so that the plant
can obtain its water and nutrients from the soil and even out to the foliage.
Fig (5-13)Syphon
Fig (5-13)Beads of liquid pulling each other
cyan magenta yellow black File: CH 5 fl ˙ Æ˝ ” 131
131
Uni
t 2
: F
luid
Mec
hani
cs
C
hapt
er 5
:
Hyd
rody
nam
ics
130
Unit 2
: Flu
id M
echanic
s C
hapte
r 5:P
ropertie
s o
f Flu
id d
ynam
ics
130
Un
it 2
:
Flu
id M
ech
an
ics
Ch
ap
ter
5:
H
yd
rosta
tics Applications of Viscosity
1-Lubrication :Metallic parts in machines have to be lubricated from time to time. This process leads to:
a) reduction of heat generated by friction. b) protecting machine parts from corrosion (wear).
Lubrication is carried out using highly viscous liquids. If we use water (low viscosity),it will soon seep away or sputter from the machine parts due its low adhesive forces.Therefore, we must use liquids with high adhesive (high viscosity), so they remain incontact with the moving machine parts.
2-Moving vehicles When a car attains its maximum speed, the total work done by the machine which is
supplied by the burnt fuel, acts most of the time against air resistance and the forces offriction between the tires and the road. At relatively low and medium velocities, airresistance to moving bodies resulting from air viscosity is directly proportional to thevelocity of the moving body. When the velocity exceeds a certain limit, then the airresistance is proportional to the square velocity rather than the velocity,leading to anoticeable increase in fuel consumption. Therefore, it is advisable not to exceed such alimit ( 80 – 90 km/h )
3-In medicine :Blood precipitation rate : when a ball undergoes a free fall in a liquid, it is under three
forces : its weight, buoyancy of the liquid and friction between the ball and the liquid due to
viscosity. It is found that such a ball attains a final velocity which increase with its radius. This is applied in medicine by taking a blood sample and measuring its precipitation rate. The
doctor may then decide if the size of red blood cells is normal or not. In the case of rheumaticfever and gout, red blood cells adhere together, and therefore, their volume and radius increaseand the sedimentation ( precipitation ) rate increases. In the case of anemia, the precipitation rateis below normal, since the red cells break up. Hence, their volume and radius decrease .
cyan magenta yellow black File: CH 5 fl ˙ Æ˝ ” 130
130
Unit 2 : F
luid Mechanics C
hapter 5: Hydrodynam
ics
133
Un
it 2: F
luid
Me
ch
an
ics C
ha
pte
r 5: H
yd
rosta
tics
Questions and Drills
I) Define
1) fluid 2) viscosity 3) coefficient of viscosity
II) Essay questions
1) Prove that the velocity of a liquid at any point in a tube is inversely proportional to thecross sectional area of the tube at that point.
2) Explain the property of viscosity.
3) Illustrate some applications of viscosity.
III) Drills
1- Water flows in a horizontal hose at a rate of 0.002 m3/s, calculate the velocity of the
water in a pipe of cross sectional area 1cm2 . (20 m/s)
2- Water flows in a rubber hose of diameter 1.2 cm with velocity 3 m/s. Calculate the
diameter of the hose if the velocity of the emerging water is 27 m/s . (0.4cm)
3- A main artery of radius 0.035 cm branches out to 80 capillaries of radius 0.1 mm. If the
velocity of blood through the artery is 0.044 m/s ,what is the velocity of blood in each
of the capillaries? (0.0067 m/s)
4- The cross sectional area of a tube at point A is 10 cm2 and at point B is 2cm2. If the
velocity of water at A is 12 m/s, what is the velocity at B ? (60 m/s)
5- The cross sectional area of a water pipe at the ground floor is 4x 10-4m2. The
velocity of the water is 2 m/s. When the pipe tapers to a cross sectional area of 2 x
10-4m2 at the end, calculate the velocity of the flow of water at the upper floor.
(4m/s)
cyan magenta yellow black File: CH 5 fl ˙ Æ˝ ” 133
133
Uni
t 2
: F
luid
Mec
hani
cs
C
hapt
er 5
:
Hyd
rody
nam
ics
132
cyan magenta yellow black File: Ch 6 fl ˙ Æ˝ ” 135cyan magenta yellow black File: Ch 6 fl ˙ Æ˝ ” 134
cyan magenta yellow black File: Ch 6 fl ˙ Æ˝ ” 134
137
Un
it 3
: H
ea
t C
ha
pte
r 6
: G
as L
aw
s
It can be concluded that :- Gas molecules are in a state of continuous random motion.- In their motion, they collide with each other and collide with the walls of the container- The distance between the molecules is called intermolecular distance (more or less
constant for different gases at the same conditions).
The evidence of the existence of intermolecular distances can be shown as follows:
When a graduated cylinder filled with ammonia gas is placed upside down on another
cylinder filled with hydrogen chloride gas (Fig 6-2), a white cloud of ammonium chloride is
formed, then it grows and diffuses until it occupies all the space within the two cylinders.
This can be explained as follows. Hydrogen chloride gas molecules - in spite of their
higher density - diffuse upwards, through spaces separating ammonia gas molecules, where
they combine together forming ammonium chloride molecules, which diffuse to fill the
Fig (6 – 2)Presence of gas intermoleculor distances
diffusion of awhite cloud of
ammoniumchloride
remove thepaper
a b c
ammonia gas
paper
Hydrogenchloride gas
(NH3)
(HCl)
A cloud ofAmmonium
chloridediffuses to fill
the twocylinders(NH4Cl)
cyan magenta yellow black File: Ch 6 fl ˙ Æ˝ ” 137
136
Un
it 3
: H
ea
t C
ha
pte
r 6
: G
as L
aw
s
Chapter (6) The Gas Laws
Overview It can be shown that gas molecules are in continuous random motion called Brownian
motion as follows:If we examine candle smoke through the microscope,we notice that the smoke
particles move randomly. The motion of the carbon particles is caIled Brownian motion,after Brown, an English botanist who discovered for the first time in 1827 that tiny pollengrains suspended in water moved randomly.
Interpretation of Brownian motionAir (gas) molecules move in a haphazard (random) motion in all directions with
different velocities. During their motion, they collide with each other and collide withsmoke particles. Due to the resultant force on a smoke particle, it will move in a certaindirection through a short distance and so on, always moving, colliding, and changingdirection. The reason for this is that the gas molecules are in a free motion (due to heat)and in constant collision, so they change their direction randomly (Fig 6-1).
Fig (6 – 1)Motion of molecules materials
c. solid molecules undergo avibrational motion
b. liquid molecules undergo atranslational and vibrational
motion
a. gas molecules undergo arandom translational motion
cyan magenta yellow black File: Ch 6 fl ˙ Æ˝ ” 136
139
Un
it 3
: H
ea
t C
ha
pte
r 6
: G
as L
aw
s
Firstly: the relation between the volume and pressure of a gas atconstant temperature (Boyle’s law) :
To study the relation between the volume of a fixed mass of gas and its pressure at
constant temperature, the apparatus shown in Fig (6- 3) is used. It consists of a burette (A)
connected by a length of rubber tube to a glass reservoir (B) containing a suitable amount
of mercury. (A) and (B) are mounted side by side onto a vertical stand attached to a base
provided by three screws with which the stand is adjusted vertically. The reservoir (B) is
movable along the stand either upwards or downwards and can be fixed at any desired
position.Procedure:1- The tap (A) is opened and the reservoir (B) is
raised until the mercury level in burette A is
about half full, taking into account that the
mercury levels are the same in both sides.
(Fig 6-3a) .
2- The tap (A) is then closed. The volume of the
enclosed air is measured, let it be (Vol)1. Its
pressure is also measured, let it be P1, which
equals the atmospheric pressure Pa (cmHg)
which may be determined using a
barometer.
3- The reservoir (B) is then raised a few
centimetres and the volume of the enclosed
air is measured (Vol)2. The difference
Fig (6 – 3)Boyle’s apparatus
(a)
(b)
(c)
cyan magenta yellow black File: Ch 6 fl ˙ Æ˝ ” 139
138
Un
it 3
: H
ea
t C
ha
pte
r 6
: G
as L
aw
s
upper cylinder. Also, ammonia gas molecules - in spite of their lower density - diffuse
downwards through spaces separating hydrogen chloride gas molecules, where they
combine forming ammonium chloride molecules, which diffuse to fill the lower cylinder.
Accordingly, we can conclude that there are large spacings separating the gas molecules,
known as intermolecular spacings. This is to be tied to the compressibility of gases. These
large intermolecular spacings allow gas molecules to get packed together when pressed.
Thus, a volume occupied by a gas decreases with increased pressure.
Gas Laws
Experiments performed to evaluate the thermal expansion of a gas are complicated. The
volume of a gas is affected by changes in pressure as well as by temperature. This
difficulty does not arise in the case of solids or liquids, as these are very much less
compressible.
In order to make a full study of the behavior of a gas, as regards volume, temperature and
pressure, three separate experiments have to be carried out to investigate the effect of each
pair, respectively, i.e., we study the relation between two variables only, keeping the third
constant.These experiments are
1- The relation between the volume and pressure at constant temperature (Boyle’s law).
2- The relation between the volume and temperature at constant pressure (Charle’s law).
3- The relation between the pressure and temperature at constant volume (Pessure law or
Jolly’s law).
We are going to study each of these three relations.
cyan magenta yellow black File: Ch 6 fl ˙ Æ˝ ” 138
141
Un
it 3
: H
ea
t C
ha
pte
r 6
: G
as L
aw
s
The effect of temperature on the volume of a gas at constant pressure:
We have already known that gases contract by cooling and expand by heating. But, doesthe same volume of different gases at constant pressure expand by the same amount?
To show this, let us do the following experiment:1- Take two flasks of exactly equal volume, each fitted with a cork through which a tube
bent 90˚ is inserted. In each tube, there is athread of mercury of length 2 or 3 cm. Fill one ofthe flasks with oxygen and the other with carbondioxide or air. Submerge the two flasks in avessel filled with water as shown in Fig (6 - 5).
2- Pour hot water into the vessel and notice thedistance moved by the mercury thread in bothtubes. You will find that these distances areequal. This indicates that equal volumes ofdifferent gases expand equally when heated through the same temperature rise . Inother words, they have the same volume expansion coefficient.
Volume expansion coefficient of a gas at constant pressure αv is defined as :"It is the increase in volume at constant pressure per unit volume at 0˚C for 1˚Crise in temperature ".
O2CO2
Fig (6 – 5)Effect of temperature on the volume
of a gas at constant pressure
cyan magenta yellow black File: Ch 6 fl ˙ Æ˝ ” 141
140
Un
it 3
: H
ea
t C
ha
pte
r 6
: G
as L
aw
s
between the two levels of mercury in both sides (h) is determined . In this case, thepressure of the enclosed air (cmHg) is P2 = Pa + h (Fig 6 - 3 b).
4- Repeat the previous step by raising the reservoir (B) another suitable distance andmeasure (Vol)3 and P
3 in the same manner.
5- The reservoir (B) is then lowered until the mercury level in (B) becomes lower than itslevel in (A) by a few centimeters. Then, the volume of the enclosed air is measured(Vol)4 and its pressure (P4) is determined P
4 = Pa - h, where h is the difference between
the two levels of mercury in both sides (Fig 6-3c).6- The previous step is repeated once more by lowering (B) another suitable distance. Then
(Vol)5 and P5 are measured in the same manner.
7- Plot the volume of the enclosed air (Vol) and the reciprocal pressure ( ).We obtaina straight line (Fig 6 - 4) Thus, we can conclude that:
,i.e.,the volume of a fixed mass of gas is inverselyproportional to the pressure, provided that thetemperature remains constant. This is "Boyle’s law". Boyle’s law can be written in another form, as:
PVol = Const. i.e., is : at a constant temperature, the productPVol of any given mass of a gas is constant.
1P
Fig (6 – 4)Relation between volume and
reciprocal pressure of gas
Vol1P
V = constP
constVol
(6 - 1)
cyan magenta yellow black File: Ch 6 fl ˙ Æ˝ ” 140
142
Un
it 3
: H
ea
t C
ha
pte
r 6
: G
as L
aw
s
Secondly: the relation between the gas volume and its temperature atconstant pressure (Charle’s law) :
To investigate the relation between the gas
volume and its temperature at constant pressure,
the apparatus shown in Fig(6 - 6a) is used. It
consists of a capillary glass tube 30 cm long and
about 1 mm diameter with one end closed. The
tube contains a short pellet of mercury enclosing
an amount of air inside it whose length is
measured by a ruler stand. The apparatus is
equipped with a thermometer inside a glass
envelope. We follow the folowing procedure:
1- The glass envelope is packed with crushed ice
and water. It is then left until the air inside the
glass tube has fully acquired the temperature of melting ice(0˚C).
2- The length of the enclosed air is then measured, and since the tube has a uniform
cross-section, the length of the encloscd air is taken as being proportional to its volume
(Vol )o˚c3- The ice and water are removed from the envelope and steam is passed through the top
and out at the bottom for several minutes to be sure that the temperature of air becomes
100˚C . Then, the length of the enclosed air is measured. It is taken as a measure of its
volume (Vol )100˚C.
4- A relation between Vol and t˚C is plotted (Fig 6-6b). We see that such a relation is astraight line,which if extended will intersect the abscissa at -273˚C .
Charle’s apparatus Fig (6 – 6a)
Corksteam outlet
pellet ofmercury
steam inlet
glassenvelope
capillarytube
cyan magenta yellow black File: Ch 6 fl ˙ Æ˝ ” 142
145
Un
it 3
: H
ea
t C
ha
pte
r 6
: G
as L
aw
s
5- Repeating this experiment several times for different gases and measuring the amount ofincrease of gas pressure at constant volume for the same rise in temperature, we find:
a- At constant volume, the pressure of a given mass of gas increases by increasingtemperature.
b- At constant volume, equal pressures of gases increase equally, when heated throughthe same range of temperatures.We define the pressure expansion coefficient of a gasat constant volume (βp) as :
“lt is the increase in gas pressure at constant volume per unit pressure at 0˚C forcm degree rise in temperature”. It is found to be the same for all gases.
Thirdly: the relation between the pressure and temperature of a gas atconstant volume (pressure law or Jolly’s law):
It was found experimentally that the increase
in gas pressure is directly proportional to the
initial pressure at 0˚C (P0˚C) as well as to the
rise in its temperature, (∆t˚C). This is expressed
as follows:
∆P P0˚C ∆ (t˚C)
∆P = βP P0˚C ∆ (t˚C)
where βp is a constant value. It is the pressure Fig (6 – 8)
thermometer
mercury
Jolly’s apparatus
∝
(6-3)β = ∆∆P
P tO
βP (t˚C)∆P
0˚C∆O
cyan magenta yellow black File: Ch 6 fl ˙ Æ˝ ” 145
144
Un
it 3
: H
ea
t C
ha
pte
r 6
: G
as L
aw
s
The effect of temperature on the pressure of a gas at constant volume:1- To investigate how the pressure of a gas depends on temperature, the apparatus shown
(Fig 6-7a) may be used. The gas under test is confined in a flask by mercury in a Utube. The flask is fitted with a cork. The surfaces of mercury in the two branches (A)and (B) have the same level at x,y. Thus, the pressure of the enclosed air isatmospheric. We then determine the temperature of air. Let it be t1˚C.
2- Submerge the flask in a vessel containing lukewarm water at t2˚C. You will notice thatthe level of mercury decreases in branch A, while it rises in branch B (Fig 6-7b).
3- We pour mercury in the funnel C, until the level of mercury in branch A returns to themark x then the volume of the enclosed air in the flask at t2˚C is equal to the volumeat t1˚C (Fig 6-7c).
4- We notice that the surface of mercury in branch B exceeds that in branch A by anamount” h” (cm). This means that the pressure of the enclosed air has increased as a resultof the temperature rise from t1˚C to t2˚C by an amount equal h (cmHg)( Fig 6-7c).
Fig (6 – 7)Effect of the temperature on the
pressure of a gas at constant volume
C
(c)(b)(a)
cyan magenta yellow black File: Ch 6 fl ˙ Æ˝ ” 144
146
Un
it 3
: H
ea
t C
ha
pte
r 6
: G
as L
aw
s
expansion cofficient of a gas with temperature, at constant volume.It is the same for all
gases.
To measure βp of a gas at constant volume, Jolly’s apparatus shown in Fig (6-8) is used.
It consists of a glass bulb (A). The bulb is joined to a capillary tube (B) bent in the form of
two right angles. The bulb and the tube are mounted on a vertical ruler attached to a board
which is fixed on a horizontal base provided with 3 leveling screws.
The capillary tube (B) is connected to a mercury reservoir (C) by means of a rubber
tube.
We follow the following procedures:
1- Determine the atmospheric pressure (Pa) using a barometer.
2- Pour mercury in (A) to 1/7 of its volume to compensate for the increase in its volume
when heated, so that the volume of the remaining part is still constant, (the volume
expansion coefficient of mercury is seven times the volume expansion coefficient of
glass).
3- Submerge reservoir (A) in a beaker filled with water and pour mercury in the free end
(C), until it rises in the other branch to mark (X).
4- Heat water in the vessel to the boiling point and wait until the temperature settles, and
the mercury level in the branch connected to the reservoir stops decreasing.
5- Move the free end (C) upwards until the the mercury level in the other branch rises to
the same mark X. Then, measure the difference in height between the mercury levels in
the two branches (h). From this, determine the pressure of the enclosed air P, which is
equal to the atmospheric pressure (cm Hg) plus h, i.e., P=Pa+h6- Move the branch (C) downwards and stop heating. Then let the reservoir cool down to
nearly 90˚C. Then move the branch (C) upwards until the mercury level in the branchconnected to the reservoir rises to mark X.
cyan magenta yellow black File: Ch 6 fl ˙ Æ˝ ” 146
150
Un
it 3
: H
ea
t C
ha
pte
r 6
: G
as L
aw
s
Other Forms of Charle’s and Jolly’s (pressure) laws :1- Refering to (Fig 6-9),
we note that the triangles ABC and ADE are similar. Therefore: BC = (Vol)1DE = (Vol)2AC = T1AE = T2Vol ∝ T
= const.
Thus, at constant pressure, the volume of a fixed mass of gas is directly proportional to
its temperature on the Kelvin scale. This is another formulation of Charle’s law.
2- Using Fig.(6-10), the following relation can be obtained in a similar way:
That is
or P ∝ TThus, at constant volume, the pressure of a fixed mass of gas is directly proportional to
its temperature on the Kelvin scale. This is another form of pressure (Jolly’s) Iaw.
VolT
(6 - 6)
P1
1
= P2
2
P = const
(6 - 7)
(Vol)1T1
(Vol)2T2
=∴
cyan magenta yellow black File: Ch 6 fl ˙ Æ˝ ” 150
157
Un
it 3
: H
ea
t C
ha
pte
r 6
: G
as L
aw
s
2- The temperature of a normal human body on the Kelvin sca1e is about:a) 0˚K b) 37˚K c) 100˚K d) 373˚K e) 310˚K
3- The volume of a given mass of a gas is :a) inversely proportional to its temperature at constant pressure.b) inversely proportional to its pressure at constant temperature.c) directly proportional to its pressure at constant temperature.d) directly proportional to its temperature at variable pressure.e) inversely proportional to its pressure at variable temperature.
4- The pressure of a gas at 10˚C is doubled if it is heated at constant volume to :a) 20˚C b) 80˚C c) 160˚Cd) 293˚C e) 410˚C.
5- If we press a gas slowly to half of its original volume: a) its temperature is doubled.b) its temperature is decreased to half its value.c) its pressure will be half of its original value.d) the velocity of its molecules is doubled.e) the pressure of the gas is doubled.
III) Eassy questions1- How can you show experimentally that the volume coefficient of expansion at
constant pressure is the same for all gases?2- Describe an experiment to find the pressure coefficient of a gas at constant volume
and that it is the same for all gases.3- How can you verify Boyle’s law experimentally?4- How can you show that pressure of a gas increases by raising temperature at constant
volume?
cyan magenta yellow black File: Ch 6 fl ˙ Æ˝ ” 157
156
Un
it 3
: H
ea
t C
ha
pte
r 6
: G
as L
aw
s
Questions and Drills
I) Complete (Fill in the spaces) :Which phrase (a-e) completes each of the next following statements (1-3)?
a) increases by a small value.b) decreases by a small value.c) remains constant. d) doubles.e) dereases to its half value.
1- If the pressure of a gas is doubled at constant temperature. So its volume...............2- If a barometer is transferred to the top of a mountain above the sea level, the length
of mercury in the barometer .............3- If the absolute temperature of a gas is decreased to be half its original value at
constant pressure, so its volume ...............
II) Choose the correct answer:1- The increase of the temperature of a car’s tire during motion leads to :
1) an increase in air pressure inside the tire. 2) an increase of air volume inside the tire.3) a decrease of the contact area of the tire with the road.
Choose the correct letter (a-e) a) (1, 2, 3) are correct. b) (1, 2) only are correct.c) (1, 3) only are correct. d) 3 only is correct. e) 1 only is correct
cyan magenta yellow black File: Ch 6 fl ˙ Æ˝ ” 156
158
Un
it 3
: H
ea
t C
ha
pte
r 6
: G
as L
aw
s
5- How can you determine experimentally the absolute zero?6- Explain the meaning of zero Kelvin and the absolute temperature scale. 7- Deduce the general gas law.
IV) Drills:1- The temperature of one liter of gas is raised from 10˚C to 293˚C at constant pressure,
find its volume. (2 liters)
2- A container containing air at 0˚C is cooled to (-91˚C). Its pressure becomes 40 cmHg. Find the pressure of the gas at 0˚C. (60 cm
Hg.)
3- The volume of a quantity of oxygen at 91˚C under 84 cm Hg is 760 cm3 (S.T.P). Findits volume at 0˚C under a pressure of 76 cm.Hg. (630 cm3)
4- A flask containing air is heated from 15˚C to 87˚C. Find the ratio between the volumeof air that goes out from it to its original volume. (25%)
5- A tire contains air under pressure 1.5 Atm at temperature (-3˚C). Find the pressure ofair inside the tire if the temperature is raised to 51˚C, assuming that the volume isconstant. (1.8 Atm)
6- An air bubble has a volume of 28cm3 at a depth of 10.13 m beneath the water surface.Find its volume before reaching the surface of the water, assuming that thetemperature at a depth of 10.13 m, is 7˚C and that at the surface is 27˚C.
(Let g = 10ms-2 , Pa = 1.013 × 105 N/m2, ρ = 1000 kg/m3) (60 cm3)
cyan magenta yellow black File: Ch 6 fl ˙ Æ˝ ” 158
cyan magenta yellow black File: ch7 fl ˙ Æ˝ ” 159
161
Un
it 3: H
ea
t Ch
ap
ter 7
: Th
e K
ine
ticT
he
ory
of G
ase
s
Avogadro’s number
Different quantities of substances - even as small as 1cm3 - contain a huge number of
atoms or molecules. It is convenient to express such numbers in terms of a unit called
mole or gram mole. It is agreed upon that a mole (or gram mole) of any substance
contains the number of atoms or molecules equal to the number of atoms in 12 gram of
carbon .It is found experimentally that 12 gram of carbon contains 6.023x1023 carbon
atoms. This is known as Avogadro’s number. NA The mole is, thus, introduced as a
measuring unit of a quantity of matter in the international system of units. Although the
original definition of mole was associated with carbon, yet the concept of the mole is
generalized to any ensemble of particles, such that one mole contains Avogadro’s number
of these particles. Thus, a mole of iron contains 6.023 x 1023 iron atom, and one mole of
water contains 6.023 x 1023 water molecules. In general, the mass of one mole of any
substance equals numerically the atomic or molecular mass (in grams) of this substance,
i.e., the mass of one mole of carbon is 12 gram, one mole of oxygen is 32 gram and one
mole of water is 18 gram. Each of these quantities contains the same number of atoms or
molecules, which is Avogadro’s number.
Thus, the mole of any substance is defined as the quantity of this substance in grams,
which equals the atomic or molecular mass of the substance. Oxygen gas has atomic mass
16, i.e the molecular mass of a mole of Oxygen is 32.
Avogadro’s law
Avogadro’s law states that:
Equal volumes of different gases contain equal number of molecules under the
same conditions of temperature and pressure. Alternatirely, Each mole of any gas at
cyan magenta yellow black File: ch7 fl ˙ Æ˝ ” 161
160
Unit 3
: H
eat C
hapte
r 7: T
he K
ineticT
heory
of G
ases
The KineticTheory of Gases
Overview
To study the behavior of gases and explain their different laws, one can make use of the
kinetic theory of gases. This theory is based on postulates given below:
1) A gas is composed of molecules which we shall regard as very minute perfectly elastic
spheres obeying Newton’s law of motion.
2) The intermolecular distances are relatively large, hence, the volume of the gas
molecules is negligible compared to the volume occupied by the gas itself(the volume
of the container).
3) The forces of intermolecular attraction between the gas molecules are very weak due to
the large intermolecular distances, so they are negligible. Therefore, the potential energy
of the molecules is zero. This means that the gas molecules do not interact with each
other. Thus, the mean distances which a molecule moves before colliding with another
(called mean free path) does not depend on the mass or type of molecule, and is
statistically the same for all gases under the same conditions. Therefore, a certain volume
of any gas at S.T.P. contains the same number of molecules regardless of the gas type.
4) Gas molecules are in continuous random motion due to the collisions between each
other and due to their collisions with the walls of the container. The molecules move
between any two successive collisions in straight lines.
5) The collisions between the gas molecules are perfectly elastic, i.e., the total kinetic
energy of the gas molecules remains constant before and after the collisions.
6) The gas is in thermal equilibrium state with the walls of its container.
Chapter 7
cyan magenta yellow black File: ch7 fl ˙ Æ˝ ” 160
163
Un
it 3: H
ea
t Ch
ap
ter 7
: Th
e K
ine
ticT
he
ory
of G
ase
s
In reality, a gas contains a huge number of molecules moving randomly. Studying the gas
on the level of the molecules is called the microscopic point of view. This has led to the
kinetic theory of gases. We start by the following postulates:
1) A gas contains a huge number of molecules in random motion.
2) The size of the molecule is much smaller than the total volume of the gas.
3) Collisions among the molecules and with the walls are elastic, and hence no energy is
lost.
4) Interactive forces among the molecules are negligible, except at collision. Hence, there
is no potential energy involved.
5) Molecules obey Newton’s laws. Consider one of the gas molecules in a box in the form
of a cube whose side is l (Fig 7-1). The mass of the molecule is m, its average velocity
is v and the x component of velocity is vx. The pressure exerted by the gas on the walls
of the box originates from the collision of the gas molecules with the walls. The
pressure P is the force per unit area P = F/A, where A= l2, and F is the force which the
molecule applies to the wall. The linear momentum is PL .The change in the linear
momentum for a molecule PL is the difference between the linear momentum before and
after collision with the wall:
Because the collision is elastic, the velocity after collision in the x direction is -vx . The changein linear momentum transmitted to the wall PL is opposite to the change in the linear
F = PL
t
cyan magenta yellow black File: ch7 fl ˙ Æ˝ ” 163
162
Unit 3
: H
eat C
hapte
r 7: T
he K
ineticT
heory
of G
ases 0˚C and 1 Atm (S.T.P) occupies a volume of 22.4 liters. Therefore, each mole of any
gas contains the same number of molecules at S.T.P. This number is given by NA =
6.023 x 1023 molecules/Mole.
To understand this, let us say we have NA molecules of oxygen (molecular mass is 32 x
mH where mH is the mass of the hydrogen atom). The Mass of one mole of oxygen is mH
x 32 x NA. Thus, the mass is a constant (NAmH) times 32. If we have NA hydrogen
molecules, then their mass is 2 x (NAmH). In other words, 32 gram oxygen (one mole
oxygen) and 2 gram hydrogen (one mole hydrogen) have the same number of molecules
(NA). Also, at the same temperature and pressure, the interatomic or intermolecular
distances are the same on average, due to the random motion of gas molecules and non
existence of attractive forces or potential energy between them, which are the postulates of
the perfect gas. Consequently, any mole of any gas at S.T.P. occupies the same volume,
which is found experimentally to be 22.4 liters.
Gas density:
Knowing the number of molecules (N) in a given volume of gas Vol and the mass of one
molecule(m), we can calculate the density of the gas (ρ) from the relation:
Gas pressure
Our study of the properties of gases in terms of pressure, volume and temperature have
led to the deduction of the gas laws. Such a study is called the macroscopic point of view.
(7 - 1)ρ = NmV
kg/mol
3
cyan magenta yellow black File: ch7 fl ˙ Æ˝ ” 162
164
Unit 3
: H
eat C
hapte
r 7: T
he K
ineticT
heory
of G
ases momentum of the molecule.
PL = -mvx - (mvx) = - 2mvx
The change in the linear momentum delivered to
the wall PL is opposite to the change in the linear
momentum of the molecule.
PL = 2mvx
The force with which the molecule acts on the
wall is given by :
where t , is the time of contact between the molecule and the wall upon impact. The
impulse Iimp delivered by the molecule to the wall is given by:
Iimp = F t
= PL
= 2mvx
Because the time interval t is very small and indeterminate, we can take the time
between collisions tav as a substitute measure. In this case, the force is the average force
acting all the time such that:
Fav tav = F t
where tav is the average time between for collisions of a molecule with the walls:
Number of gas molecules in cube Fig (7 – 1 )
F = PL
t=
2mvX
t
t 2
vavx
=l
l
ll
Z
Y
cyan magenta yellow black File: ch7 fl ˙ Æ˝ ” 164
167
Un
it 3: H
ea
t Ch
ap
ter 7
: Th
e K
ine
ticT
he
ory
of G
ase
s
23
uT = 12
mvav2
nRT = 23
nNAmvav
2
2
23
RNA
T = mvav2
232
mv2
mv2
.
where is constant for every molecule and is called Boltzmann constant(k) :
From equations (7-7) and (7-8) ,
This relation ties the macroscopic theory of a gas to the microscopic model .It is to be
noted that temperature T (a macroscopic quantity) measures the average kinetic energy of a
molecule (a microscopic quantity). As the temperature increases, the kinetic energy
increases.
It is to be noted from equation (7-9) that each direction of motion (dimension or degree
of freedom) has associated with kinetic energy kT. In fact this relation does not apply
solely to gas molecules but also to electrons in a metal , and even to any ensemble of
particles in random motion.
We might be tempted to believe that at T = 0˚K, the kinetic energy, and hence the
velocity is zero, i.e., everything stops at absolute zero. In reality, we cannot claim so,
because at absolute zero the equations of the ideal gas are no longer valid. It is known that
gases are transformed in turn to liquids at low temperatures (chapter 8). Einstein showed
that even at absolute zero, there is still energy called rest energy. In such a case, the above
equations become inapplicable.
u = RNA
= 1.3γ 10−23 J / k˚k 38× (7 - 8)J/˚K
kT (7 - 9)32 mv2
RNA
12
(7 - 7)
cyan magenta yellow black File: ch7 fl ˙ Æ˝ ” 167
166
Unit 3
: H
eat C
hapte
r 7: T
he K
ineticT
heory
of G
ases
(7 - 3)∴ p = 13
2 vavρP
p = 13
2Nmv
vavP Vol
(7 - 2)
Referring to equation (7-1),
where ρ is the gas density and v2 is the mean square velocity of the gas molecules.
The scientific concept of temperature :
From equation (7-2), multiplying the numerator and
denominator by 2 :
We note that the number of gas molecules N is the number
of moles times Avogadro’s number NA :
From the laws of the ideal gas, the macroscopic relation is given by :
where R is the universal gas constant = 8.314 J/mole˚K, n is the number of moles in the
substance. This relation is based on experimental observations, while the microscopic
relation is based on theoretical deduction. We must equate the right hand side of both
equations (7-4) and (7-6).
Boltzman
23
N mvav2
2PVol =
(7 - 5)
(7 - 6)
N = nNA
Pv = 23
nNAmvav
2
2
PVolmv2
(7 - 4)
Pv = nRTPVol
=
= nRPV
olT
= nRT
cyan magenta yellow black File: ch7 fl ˙ Æ˝ ” 166
169
Un
it 3: H
ea
t Ch
ap
ter 7
: Th
e K
ine
ticT
he
ory
of G
ase
s
and occupies 22.4 liters at S.T.P., and that Avogadro’s number equals 6.02 x 1023, Boltzmann’sconstant equals 1.38 x 10-23 J/˚K, and the atmospheric pressure is 1.013 x 105 N/m2
Solution
There are two methods to solve this problem. The first method:
=
where M is the mass of one mole of the gas and Vol (0˚C), Po(0˚C) and the values of the
volume and pressure are those at S.T.P. and ρo is the density of the gas and vo is average
velocity at S.T.P.
The second method:
P = 13
v2= 1
3MV
v2Po(0˚C)o o olρo
M V12
V = 3 x 0.76 x 13600 x 9.8 x 22.0.028
3 x 1.013 x 105 x 22.4 x 10-3
0.028= 493 m/svVo
v = 3KTN A
M = 3 x 1.38 x 10 -23 x 273 x 6.02 x 10 23
0.028
12
MN A
v 2 = 32
KT
12
m v 2 = 32
KT
NA
NA
k
kT
Vo
o
k
2
o2
= 493 m/s
3Po(0oC) (Vol)0oC
M
3Po(0oC)
ρο
cyan magenta yellow black File: ch7 fl ˙ Æ˝ ” 169
171
Un
it 3: H
ea
t Ch
ap
ter 7
: Th
e K
ine
ticT
he
ory
of G
ase
s
Questions and Drills
I) Essay questions :
1. State the main postulates of the kinetic theory of gases.
2. On the basis of the postulates of the kinetic theory of gases, show how to prove that the
gas pressure P is given by the relation:
where ρ is the gas density and v2 is the mean - square speed of its molecules.
3. Using the previous relation, show how to find expressions for each of the following:
a) the root - mean - square speed of the gas molecules.
b) the concept of the gas temperature.
c) the average kinetic energy of a free particle.
4. A uniform cubic vessel of side length l has gas whose molecule has mass m moving in
the x direction with velocity vx, and collides with the walls of the vessel in perfectly elastic
collisions.
a) What is the linear momentum of the molecule before collision?
b) What is the linear momentum of the molecule after collision?
c) What is the change in linear momentum of the molecule on collision?
d) What is the distance traveled by the molecule before the next collision with the walls
of the vessel?
e) What is the number of the collisions with the walls of the vessel per second made by
ρ v2 P= 13
cyan magenta yellow black File: ch7 fl ˙ Æ˝ ” 171
170
Unit 3
: H
eat C
hapte
r 7: T
he K
ineticT
heory
of G
ases In a Nutshell
• The mole of any substance equals the molecular mass number in grams.
• Avagadro’s number is the number of molecules in one mole and equals 6.023 x 1023
• The density of a gas is given by :
where N is the number of gas molecules in a certain volume Vol and m is the mass of one
molecule.
• The pressure of a gas in a container is calculated from the relation:
where ρ is the density of the gas and v2 is the mean - square speed of the molecules.
• The average kinetic energy of one of the gas molecules is directly proportional to its
absolute temperature in ˚K and the relation between them is:
where k is Boltzmann’s constant = 1.38 x 10-23 J/˚K
ρ = NmV
kg/mol
3
vav2P= 1
3ρ
12
m v 2 = 32
KTk
cyan magenta yellow black File: ch7 fl ˙ Æ˝ ” 170
172
Unit 3
: H
eat C
hapte
r 7: T
he K
ineticT
heory
of G
ases the molecule?
f) What is the total change in linear momentum of one molecule per second due to its
successive collisions with the walls of the vessel?
g) What does the above quantity represent?
h) If NA is the number of the gas molecules in the container, what will be the total force
acting on the internal surface of the vessel?
II) Drills:
1. Hydrogen gas in a vessel is at S.T.P. Calculate the root mean square speed of its
molecules, given that a hydrogen mole = 0.002 kg and Avogadro’s number = 6.02 x
1023, 1 Atm = 1.013 x 105N/m2 (1844 m/s)
2. What is the change in linear momentum of the hydrogen molecule in the above problem
on each impact perpendicular to the walls of the vessel? (1.224 x 10-23kg.ms-1)
3. Calculate the average kinetic energy of a free electron at 27˚C, given that Boltzmann’s
constant k = 1.38 x 10-23 J˚K-1. (6.21 x 10-21J).
4. Using the data given in the previous problem, find the root mean square speed of a free
electron if its mass is 9.1 x 10-23 kg. (1.168 x105m/s)
5. Find the ratio between the root mean square speed of the molecules of a certain gas at
temperature 6000˚K (Sun’s surface) and that at temperature 300˚ K (Earth’s surface).
(4.472)
cyan magenta yellow black File: ch7 fl ˙ Æ˝ ” 172
175
Unit
3:
Heat
Chapte
r 8:
C
ryogenic
s (L
ow
T
em
pera
ture
P
hysic
s
Mechanism of achieving low temperatures
Low temperatures may be achieved by drawing or removing energy out of the matter,This can be done in various ways. The simplest is to establish contact with anotherprecooled substance. Ice or dry ice ( solid CO2) or liquid air may be used. Temperatures of77°K ( liquid nitrogen temperature) have been widely used. Liquid helium temperature(4.2°K) has even been reached . From the concept of latent heat of vaporization, the liquidgas draws energy from the material to be cooled in order for the liquid gas to evaporate tobe gaseous again. This results in the cooling of the substancerequired.
Superfluidity
Some liquid gases have the property of superfluidity, i.e., theycan flow without resistance (or without friction) at temperaturesclose to obsolute zero. Helium liquid has this property. In otherwords, it loses viscosity completely at such low temperatures. It caneven flow upwards uninterruptedly against gravity or friction alongthe walls of its container (Fig 8–1). It also has very low specificheat and is one of the best thermal conductors.
Dewar’s Flask
It is a glass or metallic container evacuated to prevent heattransfer. It is used to store liquid gases, since it is designed toprevent heat losses by conduction, convection and radiation. Itconsists of a double walled pyrex container with silver plated wallsto minimize heat transfer by radiation. The spacing between thewalls is evacuated to prevent conduction and convection, e.g., as in
Fig (8-1)Superfluidity
Fig (8-2)Dewar’s flask
vacuum
reflectingsurfaces
hot or coldliquid
174
Un
it 3
: H
ea
t C
ha
pte
r 8
: C
ryo
ge
nic
s (L
ow
T
em
pe
ratu
re P
hysic
s
Cryogenics (Low Temperature Physics)
Overview
Cryogenics (or low temperature physics) is a branch of physics dealing with the cases
when temperature approaches absolute zero (-237°C) or 0˚K . The temperature scale used
in low temperature physics is the Kelvin temperature scale (the absolute temperature scale)
which is based on the behavior of the ideal gas.
Van Der Waals’ Effect:
One of the postulates of the kinetic theory of gases uponwhich the ideal gas laws are based is neglecting the interactive(attractive) forces among the molecules of the gas, as well asneglecting the size or the volume of the gas molecule incomparison with the volume of the container. The properties ofa real gas differ from those of the ideal gas, as the gas densityincreases. Interaction among gas molecules can no longer beneglected. This interaction is called van der Waals’ effect . It isunlike chemical interaction between atoms leading to theformation of molecules. The attractive forces among the molecules become important asthey lead to the liquefaction of the gas under high pressure. Due to the high pressure, vander Waals’ interaction takes over, where two molecules approaching each other attracttogether, and eventually attract more molecules, until the gas switches to the condensedstate of matter (liquid or even solid).
This mechanism explains the liquefaction of gases, which has led to achieving very lowtemperatures approaching near absolute zero.
Van der Waals
Chapter 8
176
Un
it 3
: H
ea
t C
ha
pte
r 8
: C
ryo
ge
nic
s (L
ow
T
em
pe
ratu
re P
hysic
s a thermos bottle (Fig 8–2). It is used to store liquid nitrogen (boiling point 77°K) and liquid
oxygen (boiling point 90°K). As to helium (boiling point 4.2° K and low specific heat), it is
stored in a double Dewar’s flask, one inside the other. The spacing between the two flasks is
filled with liquid nitrogen (due to the low specific heat and boiling point of helium).
How does a refrigerator work ?
From the law of conservation of energy, if a gas acquires thermal energy Qth it is used upin one of two ways:
1) an increase in internal energy U which is manifested by an increase in temperature.
2) work done by the gas molecules W. There are two types of heat transfer. One is at
constant temperature with the surroundings, i.e. , at constant internal energy ( U = 0).
In this case, the acquired energy is transformed in full into mechanical work done by
the gas. This is called an isothermal process. The second type is performed when the
gas is thermally isolated with its surroundings. So, it can neither acquire nor lose heat.
In this case, Qth = 0 The work done by the gas must be at the expense of its internal
energy. If W is positive, the gas does the work and the internal energy decreases ( U is
negative), i.e., the gas cools down. If work is done on the gas, then W is negative, so the
internal energy increases and its temperature rises. The process when Qth = 0 (W is
positive or negative) is called adiabatic process. A refrigerator is an application to both
isothermal and adiabatic processes , and the coolant (refrigerant) or the cryogenic liquid
used is freon (boiling point- 30°C) or its substitutes.
179
Unit
3:
Heat
Chapte
r 8:
C
ryogenic
s (L
ow
T
em
pera
ture
P
hysic
sFig (8-5)Meissner's effect
Fig (8-6)Magnetically levitated train
metallic compounds) becomes very high, i.e., the electrical
resistance vanishes. This occurs at a critical (transitional)temperature (Fig 8–4). Materials having this property arecalled superconductors. If current happens to flow in asuperconductor, it continues to flow even if the voltage
difference is removed. It will continue to flow for years, as it
is met by nearly zero resistance.
Such a metal will not be heated by current flow. No
energy is consumed in compensating electrical energy
associated with an electric current, as in ordinary resistors.
Superconductors can be used to pick up weak wireless signals. Therefore, they are used
in the electric circuits of satellites. It is interesting to notice that if a permanent magnet is
placed over a disk of a superconducting material, then the current in the superconductor
generates a magnetic field, which is always repulsive with the external magnet, so the
permanent magnet remains hanging in the air. This is called Meissner effect (Fig 8 – 5).
Fig (8-4)Superconductivity
criticaltemperature
178
Un
it 3
: H
ea
t C
ha
pte
r 8
: C
ryo
ge
nic
s (L
ow
T
em
pe
ratu
re P
hysic
s through a condenser (an apparatus outside the refrigerator). Heat exchange occurs in whichheat energy in the gas is radiated out to the surroundings being at a lower temperature. Thehigh pressured gas condenses and becomes a liquid at constant temperature (isothermalprocess). The liquid refrigerant is returned to the refrigerator once more. Before it entersinto the freezer compartment, the refrigerant is forced to expand in an adiabatic processthrough the expansion valve. In this case, the liquid molecules diffuse from a high pressureregion into a low pressure region. The refrigerant’s volume increases and does work indoing so against the spring in the valve. Thus, work is done at the expense of the internalenergy of the refrigerant. Since no external heat exchange is allowed (Qth= 0 or W ispositive so ∆U is negative). The internal energy of the liquid decreases so does itstemperature. Thus, the refrigerant is now back to be a liquid as it was when we started thecycle. The cycle repeats. The final result is the expulsion of thermal energy from the cabinto the condenser outside the refrigerator. Therefore, the refrigerator cools down. Of course,the refrigerator must be well insulated. It should be noted that the electric energy needed bythe refrigerator throughout the cycle is the energy consumed in operating the piston. Theinternal energy one throughout complete cycle must remain unchanged (∆Unet = 0). Thecompressor is, thus, a heat pump that transfers the heat from the refrigerator to the outsidethrough the work done by the compressor Wnet. Thus, the electric energy E required foroperation: E = Wnet = (Qth)
net
Superconductivity
In 1911, i.e., 3 years after the liquefaction of helium, Onnes and his
assistants discovered superconductivity. When the temperature
reaches a few degrees above absolute zero, the electrical conductivity
of some metals (platinum, aluminum, zinc, lead, mercury and some onnes
181
Unit
3:
Heat
Chapte
r 8:
C
ryogenic
s (L
ow
T
em
pera
ture
P
hysic
sFig (8-7)MRI image
Learn at Leisure
Magnetic resonance imaging ( MRI )One of the most popular and safest tools in medical diagnosis nowadays is magnetic
resonance imaging (MRI). In this method, a magnetic field affects the nuclei of hydrogen inthe body. Exciting these nuclei with an alternating magnetic field, waves are radiated fromthe excited hydrogen nuclei, which are indicative of water ( hydrogen ) localization(oedemas and tumors).An internal image of the body can be made (Fig 8–7), which helpsidentify such lumps.
Superconducting magnets are used to counteract the huge energy losses in normalmagnets, hence heating effects are reduced.
180
Un
it 3
: H
ea
t C
ha
pte
r 8
: C
ryo
ge
nic
s (L
ow
T
em
pe
ratu
re P
hysic
s The reason for this is that superconductors belong to a class of materials called
diamagnetic materials in which the magnetic field inside the material is zero. Therefore,
an external magnet induces current in the superconductor which creates a magnetic field
inside the superconductor in an opposite direction, so that the net magnetic field inside
the superconductor is zero . This phenomenon has been put to use by designing a high
speed (magnetically levitated) train. The train carries coils of a superconducting material.
When the train moves, it induces current in fixed coils, which produces a magnetic field
repelling the inducing field. The train is raised above the rails for a few centimeters . This
levitation eliminates friction (Fig 8 – 6), hence increases the train velocity. The levitated
train may reach a velocity of 225 km/h . The discovery of room temperature
superconductive materials will lead to expansion in the applications of superconductivity,
since no cooling is then needed . Superconductors are also used in electric power plants
and in transmission lines, where voltage losses are eliminated due to the vanishing
resistance .
183
Unit
3:
Heat
Chapte
r 8:
C
ryogenic
s (L
ow
T
em
pera
ture
P
hysic
s
Questions
1) Explain each of the following phenomena :
a) Van Der Waals’ effect .
b) low temperature phenomena .
c) superfluidity of some liquefied gases .
d) superconductivity .
2 ) Give reasons :
a)the use of two Dewar's flasks to store helium .
b) the spacing between the double walls in a Dewar's flask is evacuated .
c) helium can flow upwards along the walls of its container without stopping .
d) a levitated train has been designed with a very high speed (225 km/hr ) .
e) a magnet may remain hanging up above a superconductor regardless of the polarity .
3) State the most important applications for each of the following :
a) Dewar's flask .
b) superconductors .
4) Illustrate the difference between :
a)chemical reaction and Van Der Waals’ reaction .
b) the helium liquid and the nitrogen liquid .
182
Un
it 3
: H
ea
t C
ha
pte
r 8
: C
ryo
ge
nic
s (L
ow
T
em
pe
ratu
re P
hysic
s In a Nutshell
• Low temperature physics deals with the study of materials at temperatures near absolute zero
• Van der Waals’ effect expresses the mutual interaction between molecules and is different
from chemical interaction, which leads to the formation of molecules.
• The mechanism for achieving very low temperatures depends on drawing energy from the
material. This may be done by putting the material to be cooled in contact with a cooler
material such as a liquefied gas.
• Superfluidity :
Some liquefied gases can flow without resistance or without friction at temperatures
close to absolute zero. Helium is a superfluid , i.e. its viscosity vanishes. It can also
flow up along the walls of the container against gravity and friction and has low
specific heat.
• Dewar's flask is a glass or metallic container evacuated to prevent heat transfer. It is used
to store liquefied gases such as nitrogen, oxygen and helium and so on.
• Superconductivity:
Some metals have excessive electrical conductivity (zero resistance) at very low
temperatures.
• Meissner effect :
If a permanent magnet is placed above a superconductor, the current induced in the
superconductor generates a magnetic field which repels the permanent magnet so the
permanent magent remains hanging in the air.
187
Unit 4
: Dynam
ic E
lectric
ity Chapte
r 9: E
lectric
Curre
nt a
nd O
hm
’s L
aw
sectional area (m2) and (ρe) is the resistivity (Ωm).The electrical conductivity of a certain
material σ (Ω-1m-1) is the reciprocal of the resistivity σ = ( ).
6) Ohm’s Law:
The current intensity in a conductor is directly proportional to the potential
difference across its terminals at a constant temperature : V = IR
7) as a convention, the direction of the electric current always goes from the positiveterminal to the negative terminal outside the source into a closed electric circuit. It isopposite to the direction of motion of electrons. It is called the conventional direction ofcurrent.
Connecting resistorsFirstly: series connection
Resistors are connected in series to obtain a higher resistance (Fig9–1). The equivalent
resistance of a group of resistors connected in series can be obtained in connecting these
resistors in an electric circuit comprising a battery,an ammeter,a rheostat (variable resistor)
and a switch (Fig 9-2). The circuit is closed and the rheostat is adjusted so that an
appropriate current I is passed. The voltage difference across each resistor is measured (V1
across R1, V2 across R2, V3 across R3) as well as the total voltage (V), which is equal to thesum of the voltage differences across the resistors in the series circuit and this is calledKirchhoff,s law
Fig (9 – 1)Connection in series
1ρe
189
Unit 4
: Dynam
ic E
lectric
ity Chapte
r 9: E
lectric
Curre
nt a
nd O
hm
’s L
aw
Secondly: Parallel connection
The purpose of connecting resistors in parallel is to obtain a small resistance out of abunch of large resistances (Fig 9 – 3). To obtain the equivalent resistance for a parallelconnection, the combination is included in an electric circuit comprising a battery, anammeter and a rheostat all connected as shown (Fig 9 – 4).
We close the circuit and adjust the rheostat to
obtain an appropriate current in the main circuit
of intensity I (A), which can be measured by the
ammeter. The total voltage difference can then
be measured across the terminals of the
resistances by a voltmeter (V). The current in
each branch is measured ( I1 in R1, I2 in R2, and
I3 in R3). In a parallel connection, the total
current is determined by the smallest resistance.
This case is similar to the flow of water in pipes.
Fig (9 – 3)Connection in parallel
Fig (9 - 4)Measuring the equivalent resistance
in a parallel connection
188
Un
it 4
: D
yn
am
ic E
lectr
icity C
ha
pte
r 9
: E
lectr
ic C
urr
en
t a
nd
Oh
m’s
La
w
V = V1 + V2 + V3
But ... V = IR
V1 = IR1
V2 = IR2
V3 = IR3
IR = IR1 + IR2 + IR3
R = R1 + R2 + R3 (9-1)
Thus, the equivalent resistance R of a group of resistors connected in series equals the sum of
these resistances. It is to be noted that the largest resistance in the combination determines the
total resistance in a series connection. If N resistances are connected in series each equal r then :
R = Nr
We conclude that if we want a large resistance out of a bunch of small resistances, wesimply connect them in series.
Fig (9 – 2)Measuring the equivalent resistance in a
series connection
191
Unit 4
: Dynam
ic E
lectric
ity Chapte
r 9: E
lectric
Curre
nt a
nd O
hm
’s L
aw
I = ER + r
VB
Ohm’s Law for a closed circuitWe know that the emf of an electric cell (battery - source) is the total work done inside
and outside the cell to transfer an electric charge of 1C in the electric circuit. If we denotethe emf of a battery by VB, the total current in the circuit by I, the external resistance by Rand the internal resistance of the cell by r, then
VB = IR + IrVB = I (R + r)
This is known as Ohm’s law for a closed circuit, from which we find that the currentintensity in a closed circuit is the emf of the total source divided by the total (external plusinternal) resistance of the circuit.Relation between emf and voltage across a source
From Fig (9 – 5), we find V = VB - Ir
From this relation, we see that as I is decreased gradually in the circuit shown (Fig 9– 5),-by decreasing the external resistance R- the voltage difference across the source increases.
When the current vanishes, the voltage difference across the source becomes equal to theemf of the source. Hence, we may define the emf of a source as the voltage differenceacross it when the current ceases to flow in the circuit.
190
Un
it 4
: D
yn
am
ic E
lectr
icity C
ha
pte
r 9
: E
lectr
ic C
urr
en
t a
nd
Oh
m’s
La
w
VR
= VR1
+ VR2
+ VR3
1R
= 1R1
+ 1R2
+ 1R3
The smallest pipe ( the highest resistance) determines the flow rate in a seriesconnection, while the widest pipe (the least resistance) determines the rate of flow in aparallel connection, since it draws most of the water current.
It is to be noted that :
where R is the equivalent resistance, and V is the voltage difference across resistorsconnected in parallel. The total current I is the sum of the branch currents. I1 + I2 + I3 .
Thus:
Hence, the reciprocal of the equivalent resistance R is the sum of the reciprocal ofresistances in the case of a parallel connection. In the case of two resistors in parallel, theequivalent resistance R is given by :
R = R1 R2R1 + R2
When N resistances are connected in parallel each equal to r,
Therefore, if we wish to obtain a small resistance out of a bunch of resistors, we simplyconnect them in parallel.
(9 - 4)
I = VR
, I 1 = VR1
, I 2 = VR2
, I 3 = VR3
1R
= Nr
(9 - 5)R = rN
193
Unit 4
: Dynam
ic E
lectric
ity Chapte
r 9: E
lectric
Curre
nt a
nd O
hm
’s L
aw
V2 = IR2 = 0.25 x 70 = 17.5V
V1 = IR1 = 0.25 x 25 = 6.25V
V3 = IR3 = 0.25 x 85 = 21.25V
2) If the resistors in the previous example are connected in parallel to the same battery,calculate:a) the current flowing in each resistor.b) the total resistance. c) the current through the circuit.
solution :a) the voltage difference across each resistor = 45V, since they are connected in parallel
and the battery is of negligible internal resistance. The current flowing through eachresistor is calculated separately as follows :
b)The total (equivalent or combined) resistance R is calculated as follows :
R = 15.14 Ωc) The current flowing through the circuit I is :
I = VR
= 4515.14
= 2.972 A
I2 = VR2
= 4570
= 0.643 A
I3 = VR3
= 4585
= 0.529 A
I1 = VR1
= 4525
= 1.8 A
1R
= 1R1
+ 1R2
+ 1R3
= 125
+ 170
+ 185
194
Un
it 4
: D
yn
am
ic E
lectr
icity C
ha
pte
r 9
: E
lectr
ic C
urr
en
t a
nd
Oh
m’s
La
w
It can be calculated also as the sum of the currents I1 , I2 , I3 flowing through allresistors:
I = 1.8 + 0.643 + 0.529 = 2.972 A
3) In the figure shown above two resistors A and B are connected in parallel. The combinationis connected in series with a resistor C and a 18 volt battery of negligble internal resistance.If the resistances of A,B and C are 3Ω,6Ω,7Ω, respectively, calculate:
a) the total resistance.
b) the current flowing through the circuit.
c) the current through each of A and B.Solution :
The equivalent resistance for the combination (A, B) is :
The equivalent resistance for the combination (A,B)and( C) is :
R = R´ + R3 = 2 + 7 = 9 Ω
The current I flowing through the circuit is :
R´=R1 R2
R1 + R2
= 3 x 63 + 6
= 2
197
Unit 4
: Dynam
ic E
lectric
ity Chapte
r 9: E
lectric
Curre
nt a
nd O
hm
’s L
aw
• In a parallel connection:
• For N equal resistances each r :
• Ohm’s law for a closed circuit:
where VB is the emf of the source, r is its internal resistance and R is the externalresistance.
R = rN
I = ER + r
VB
1R
= 1R1
+ 1R2
+ 1R3
199
Unit 4
: Dynam
ic E
lectric
ity Chapte
r 9: E
lectric
Curre
nt a
nd O
hm
’s L
aw
6) In the circuit shown :
a) the ammeter reading is .....................
b) the voltmeter reading is .....................
7) In the circuit shown
a) the ammeter reading A1 is .....................
b) the ammeter reading A2 is ......................
II) Choose the right answer:
Four lamps 6 each are connected in parallel. The combination is connected to a 12Vbattery with a negligible internal resistance:
1) The current in the battery equals..........
a) 8A b) 6A c) 4A d) 2A e) 72A
2) The total charge leaving the battery in 10s is...............
a) 80C b) 60C c) 40C d) 20C e)2C
3) The current in each lamp is .........
a) A b) 8A c) A d) 1A e) 2A
4) The voltage difference across each lamp is..........
a) 3V b) 12 V c) 6 V d) 2 V e) 4 V
5) The total resistance of the four equal lamps is...........
a) b) 24 c) d) 6 e) 12
6) If the 4 lamps are connected in series, the total resistance is.............
a) b) 24 c) d) 6 e) 12
32
32
23
32
32
23
201
Unit 4
: Dynam
ic E
lectric
ity Chapte
r 9: E
lectric
Curre
nt a
nd O
hm
’s L
aw
3) The circuit shown in Fig (9 – 5) consists of a 15 V battery, an external resistance 2.7and a switch. If the internal resistance of the battery is 0.3 , determine :
a) the reading of the voltmeter when the switch is open, assuming that the voltmeterresistance is infinite. (15 V)
b) the reading of the voltmeter when the switch is closed. (13.5)
4) A student wound a wire of a finite length as a resistor. Then, he made another of thesame material but half the diameter of the first wire and double the length. Find theratio of the two resistances.
5) A copper wire 30 m long and 2x10-6m2 cross sectional area has a voltage difference of3V across. Calculate the current if the copper resistivity is 1.79 x 10-8 .m
(11.17 A)
6) A 5.7 resistor is connected across the terminals of a battery of 12 emf and 0.3internal resistance. Calculate:
a) the current in the circuit . (2. A)
b) the voltage difference across the resistor. (11.4 V)
(1:8)
203
Un
it 4: D
yna
mic E
lectricity C
ha
pte
r 10
: Ma
gn
etic E
ffects o
f Ele
ctric Cu
rren
t an
d M
ea
surin
g In
strum
en
ts
Oersted
Overview
In 1819, Hans Christian Oersted- a Danish physicist-brought a compass near a wirecarrying an electric current. He noticed that the compass was deflected. When he turnedthe current off, the compass assumed its original position. The deflection of the compass-while current was flowing through the wire- indicated that it was being acted upon byan external magnetic field.This discovery started a chain of events that has helped shapeour industrial civilization.
In this unit we are going to study the magnetic field of current- carrying conductors inthe form of:
a) a straight wire. b) a circular loop. c) a solenoid.
Magnetic field due to current in a straight wire:
We can examine the pattern of the flux density surrounding a long straight wirecarrying a direct current using iron filings sprinkled on a paper surrounding the wire in avertical position.It will be noted that they become aligned in concentric circles aroundthe wire, as shown in Fig (10-1).
Fig (10 – 1)Pattern of iron filings
around a wire carryingcurrent
Magnetic Effects of Electric Current and Measuring InstrumentsChapter 10
204
Un
it 4: D
yna
mic E
lectricity C
ha
pte
r 10
: Ma
gn
etic E
ffects o
f Ele
ctric Cu
rren
t an
d M
ea
surin
g In
strum
en
ts
The figure shows that the circular magnetic flux lines are closer together near the wireand farther apart from each other as the distance from the wire increases.
As the electric current in the wire increases, the iron filings rearrange themselves aftergently tapping the board such that the concentric circles become more crowded.
This indicates that the magnetic field due to the electric current passing through astraight wire increases with increasing the current intensity and vice verse.
The magnetic flux density β measured in Weber/m2 or Tesla( where φm is themagnetic flux, A is the area) a point near a long straight wire carrying current I can bedetermined using the formula:
This relation is called Ampere's circuital law, where d is the normal distance between
the point and the wire, and µ is the magnetic permeability of the medium (in air it is 4π
x 10-7 Weber/Am). Thus, B is inversely proportional to d and directly proportional to I.
This is why it is advisable to live away from high voltage
towers.
Ampere's right hand rule:
To determine the direction of the magnetic field resulting
from an electric current in a wire, imagine that you grasp the
wire with your right hand such that the thumb points in the
direction of the current. The rest of the fingers around the wire
give the direction of the magnetic field due to the current
(Fig 10-2).
BAm= φ
B = µ I2 d
(10-1)
Fig (10 – 2)Right hand rule
207
Un
it 4: D
yna
mic E
lectricity C
ha
pte
r 10
: Ma
gn
etic E
ffects o
f Ele
ctric Cu
rren
t an
d M
ea
surin
g In
strum
en
ts
Thus circular loop carrying current may be considered as a bar magnet (Fig 10-3)
Examples:
Determine the magnetic flux density at the center of a circular loop of radius 11cmcarrying a current of 1.4 A. if the wire loop consists of 20 turns and µ
air = 4 π x 10-7
Weber/Am
solution:
= 4 x 22 x 10-7 x 20 x 1.47 x 2 x 0.11
= 16 x 10-5
B = µ NI2r
= 4 x 10-7 x 20 x 1.42 x 0.11
Fig (10 – 5)A circular loop carrying current in the direction
of screwing.
Fig (10 – 4)Right hand screw
direction of screwing.
Tesla
206
Un
it 4: D
yna
mic E
lectricity C
ha
pte
r 10
: Ma
gn
etic E
ffects o
f Ele
ctric Cu
rren
t an
d M
ea
surin
g In
strum
en
ts
To study the magnetic field due to a circular loop (or a coil),iron filings are sprinkledon the board as shown in Fig (3 -10). Tapping it gently, the filings arrange themselves asshown in figure, from which we can notice that:1. the flux lines near the center of the loop are no longer circular.2. the magentic flux density changes from point to point.3. the magnetic flux lines at the center of the loop are straight parallel lines
perpendicular to the plane of the coil. This means that the magnetic field in this regionis uniform.The flux density at the center of a circular loop of N turns and radius r carrying current
I is given by :
From this relation, the magnetic flux density at the center of a circular loop dependson three parameters:1. number of turns of the circular loop where B ∝ N.2. current intensity passing through the circular loop where B ∝ I.
3. radius of circular loop where B ∝ .
- Right-hand screw rule:To determine the direction of the magnetic field at the center of a circular loop or coil,
imagine a righthand screw being screwed to tie along the wire in the direction of the
current. The direction of fastening of the screw gives the direction of the magnetic flux at
the center of the loop (Figs.10 -4, 10 -5 )
Thus, a circular loop carrying current acts as a magnetic dipole or a bar magnet.
It is to be noted that no single poles exist in nature. They always exist in N - S pairs.
1r
B= µ N I2 r (10-2)
208
Un
it 4: D
yna
mic E
lectricity C
ha
pte
r 10
: Ma
gn
etic E
ffects o
f Ele
ctric Cu
rren
t an
d M
ea
surin
g In
strum
en
ts
Magnetic field due to current in a solenoid:
When an electric current is passed through a solenoid ( a long spiral or cylindrical coil) as
shown in Fig(10-6), the resultant magnetic flux is very similar to that as a bar magnet. As
shown in Fig(10-6A), the magnetic flux lines make a complete circuit inside and outside the
coil,i.e., each line is a closed path. The side at which the flux emerges is the north pole, the
other side where the magnetic flux reenters is the south pole. The magnetic flux density in the
interior of a solenoid carrying an electric current depends on :
1) the current intensity passing through the coil where B∝ I.
2) the number of turns per unit length where B ∝ n :
∴ B ∝ nI
B = µ nIwhere µ is the permeability of the core material. In this case, it is airThis relation may be rewritten as follow:
a) field pattern b) polarity of the field usingAmpere’s right hand rule.
Fig (10 – 6)Magnetic field due to a solenoid
210
Un
it 4: D
yna
mic E
lectricity C
ha
pte
r 10
: Ma
gn
etic E
ffects o
f Ele
ctric Cu
rren
t an
d M
ea
surin
g In
strum
en
ts
Force due to magnetic field acting on a straight wire carrying current.If we place a straight wire carrying current
between the poles of a magnet,a force resultswhich acts on the wire and is perpendicular toboth the wire and the field (Fig 10-7). Thedirection of the force is reversed if we reversethe current or the magnetic field. In all cases,the force is perpendicular to both electriccurrent and the magnetic field. In case thewire is allowed to move due to this generatedforce, the direction of motion is perpendicularto both the electric current and the magneticfield. The direction of the force with which amagnetic field acts on a current- carrying wireperpendicular to the field can be obtained byapplying Fleming’s left hand rule.
Fleming’s left hand rule
Form your left hand fingers as follows: the
pointer and thumb perpendicular to each other
and to the rest of the fingers. Make the pointer
point to the direction of the magnetic flux, and
the rest of the fingers- except the thumb- in the
direction of the current. Then, the thumb points
to the magnetic force or motion (Fig 10-8).
It is found that the force acting on a wire carrying current flowing perpendicularly to a
Fig (10 – 7)Force due to a magnetic field acting on a
straight wire carrying current, mark”x”denotes the direction into the paper.
Fig (10 – 8) Fleming's left hand rule.
the rest of the fingersare in the directionof the current.
the pointer isin the direction
of themagnetic flux
the thumbis in the
direction ofmotion or
force
212
Un
it 4: D
yna
mic E
lectricity C
ha
pte
r 10
: Ma
gn
etic E
ffects o
f Ele
ctric Cu
rren
t an
d M
ea
surin
g In
strum
en
ts
You can imagine what the direction of the force will be in different cases. Themark means out of the page, and the mark means into the page.
The force between two parallel wires each carrying current.When a current I1 passes in a wire and a current I2 passes in another parallel wire, a force results
between the two wires. This force is attractive if the two currents flow in the same direction. The forceis repulsive if the two currents flow opposite to each other. We can calculate this force as follows:
. x
Fig (10 – 10)Force between two parallel wires each carrying current
(b)
Fig (10 – 9) A wire carrying current in a direction inclinded by
an angle θ to the magnetic field.
a) the two currents are in the same direction. b) the two currents are in opposite directions.
a) the force vanishes when θ = 0(wire and magnetic field are parallel)
b) a force exists for θ other than zero
215
Un
it 4: D
yna
mic E
lectricity C
ha
pte
r 10
: Ma
gn
etic E
ffects o
f Ele
ctric Cu
rren
t an
d M
ea
surin
g In
strum
en
ts
c) top view (plan) of the rectanglewhen the magnetic dipole moment is
perpendicular to the field.
a) the coil is parallel to themagnetic field.
d) top view (plan) of the rectanglewhen the magnetic dipole moment
makes an angle θ with the field.
e) top view (plan) when the rectangleis perpendicular to the field or the
magnetic dipole moment is parallel
to the field and the couple is zero.
b) top view (plan)of the rectanglewhen the coil is parallel to the field.
Fig (10 – 11)A torque acting on a coil carrying a
current
217
Un
it 4: D
yna
mic E
lectricity C
ha
pte
r 10
: Ma
gn
etic E
ffects o
f Ele
ctric Cu
rren
t an
d M
ea
surin
g In
strum
en
ts
The essential parts of this device are shown in Fig(10-12). It consists of a rectangle ofa thin wire coil wrapped around a light aluminum frame mountend on a soft iron core.The frame is pivoted on agate bearings. The assembly rotates between the poles of a Ushaped (horse shoe) magnet. Its rotational motion is restrained by a pair of spiral controlsprings, which also serve as current leads to the coil. Depending upon the direction ofthe current being measured, the coil and pointer rotate either in clockwise orcounterclockwise direction. The permanent magnet's poles are curved so that themagnetic flux lines are radially directed. Thus, the magnetic flux density is constant andperpendicular to the side of the rectangle irrespective of the angle of the coil. thedeflection of the pointer is proportional to the current in the coil.
The current flows in the coil from the right side upwards, and emerges from the otherside. Then the magnetic force generates a torque which makes the coil rotate clockwise.The pointer deflects until it settles at a certain reading when the torque is balanced with thespring torsion which is counterclockwise. Thus, at balance, we can read the current value.When the current is reversed, the pointer deflects in the opposite direction.
The galvanometer sensitivity:
The galvanometer sensitivity is defined as the scale deflection per unit current
intensity passing through its coil i.e, sensitivity = degree/micro ampere (deg/µA) .
Direct current (DC) ammeter :
An ammeter is a device which- through calibrated scales- is used to measure directlythe electric current. A galvanometer is an ammeter of limited range due to its movingcoil sensitivity. To extend the range of the galvanometer, it is necessary to add a verylow resistance, called a shunt Rs to be connected in parallel with the galvanometer coilRg as shown in Fig (10-13).
θΙ
216
Un
it 4: D
yna
mic E
lectricity C
ha
pte
r 10
: Ma
gn
etic E
ffects o
f Ele
ctric Cu
rren
t an
d M
ea
surin
g In
strum
en
ts
Applications: Measuning InstumentsThe sensitive moving coil galvanometer
A sensutive moving coil galvanometer is an apparatus used to detect very weakcurrents in a circuit, measure their intensities and determine their polarities. Its principleof operation depends on the torque that is generated in a current -carrying coil moving ina magnetic field.
a) a simplified view of agalvanometer when the pointer is inthe middle of the graduated scale
graduatedscale
pointer spiralspring
ironcore
pointer
magnet
aluminum frame
(Fig 10 – 12)A moving coil galvanometer
b) top view
c) a galvanometer converted to a milliammeter
d) top view
radial magneticfield
soft iron core
pointer
aluminumframe
permanentmagnet
control spring
coil
falcrum
permanentmagnet
permanentmagnet
central spring
soft iron core
uniform scale
218
Un
it 4: D
yna
mic E
lectricity C
ha
pte
r 10
: Ma
gn
etic E
ffects o
f Ele
ctric Cu
rren
t an
d M
ea
surin
g In
strum
en
ts
Placing the parallel shunt assures that the ammeter as a whole will have a very lowresistance, which is necessary if the current in the circuit is to be unaltered afterconnecting the ammeter in series.
Most of the current in the circuit passes through the shunt Rs, while only a smallcurrent Ig passes in the galvanometer coil Rg. If the maximum current to be measured isI, which is the full scale deflection (FSD), then
I = Ig + IsIs = I - Ig
Because Rs and Rg are connected in parallel, the voltage difference across each is the same.
The two equations can be solved simultaneously to find Rs. Thus,
I s Rs = Ig Rg
Rs = Ig Rg
I s
.
b) use of a shunt resistance
a) a DC ammeter is a galvanometerwhose pointer deflects in one direction
Fig (10 – 13)The DC ammeter
pointermagnet coil
Rs = I g Rg
I - I g
221
Un
it 4: D
yna
mic E
lectricity C
ha
pte
r 10
: Ma
gn
etic E
ffects o
f Ele
ctric Cu
rren
t an
d M
ea
surin
g In
strum
en
ts
Ohmmeter
Measuring a resistance depends on measuring the current passing through it by anammeter and the voltage drop across it by a voltmeter. If the current is I and the voltagedrop is V, the resistance R from Ohm's law is R =V/I
If the voltage is fixed and known, we may remove the voltmeter from the circuit andcalibrate the galvanometer to give the value of the resistance directly (Fig10-15).As theresistance is increased, the current in the circuit decreases,and consequently, thegalvanometer reading.The Ohmmeter shown (Fig10-15) is actually a microammeterwhich reads 400µA as a full scale deflection (FSD).Its resistance is 250Ω connected inseries with 3000Ω, a variable resistance whose maximum value is 6565Ω, and a 1.5 Vbattery of negligible internal resistance.When we short circuit (sc) the terminals of theinstrument (RX =0), current flows in the circuit.For this current to give FSD, the resistance
in the circuit must be: Ω
The variable resistance must be adjusted to give FSD, when the variable resistance is500 since 250+3000+500=3750Ω.
Now, if the unknown resistance is introduced into the circuit, the current flowing willbe less, and the pointer will deflect short of FSD.
6565Ω
3000Ω
Rg=250Ω
Rx
Fig (10 – 15)A circuit for calibrating an ohmmeter
galvanometervariableresistance
instrumentterminals
standardresistance
battery
1.5400 x 10-6
= 3750
220
Un
it 4: D
yna
mic E
lectricity C
ha
pte
r 10
: Ma
gn
etic E
ffects o
f Ele
ctric Cu
rren
t an
d M
ea
surin
g In
strum
en
ts
voltage,it must be converted to a high-resistance instrument.The voltmeter must draw anegligible current, so that it will not affect the voltage drop to be measured.To do this,alarge multiplier resistor is connected in series with the galvanometer as shown in Fig.(10-14).The voltmeter is connected parallel across the two points between which thevoltage difference is to be measured. Let us call the resistance of the galvanometer coilRg and the multiplier resistance Rm which is connected in series parrallel with Rg. Themaximum current that passes through it is Ig, which is the current needed for the fullscale deflection (FSD) voltage V.
The voltage difference across the coil is : Vg = Ig Rg
The maximum voltage drop to be measured is: V = Ig Rg + Ig Rm = Vg + Ig Rm
Example
A galvanometer has an internal resistance of 0.1Ω and gives a full scale deflectionfor a current of 1mA. Calculate the multiplier resistance necessary to convert thisgalvanometer to a voltmeter whose maximum range is 50V.Solution
Vg = Ig Rg = 0.001 x 0.1 = 1 x 10-4 V
= 49999.9 ΩThe total resistance of the voltmeter is :
Rtotal = 49999.9 + 0.1 = 50000 Ω
Rm = V - V g
I g
(10-8)
Rm = =V - Vg
Ig
50 - 1 x 10-4
1 x 10-3
223
Un
it 4: D
yna
mic E
lectricity C
ha
pte
r 10
: Ma
gn
etic E
ffects o
f Ele
ctric Cu
rren
t an
d M
ea
surin
g In
strum
en
ts
Fig (10 – 17)Analog multimeter
Fig (10 – 18)Digital multimeter
They depend on digital electronics (Chapter 15). All the above instruments measurevoltage or current in one direction (DC). Therefore, they are called DC/multimeters. Butif the current or voltage is AC, the instrument used is called AC/ multimeters.
222
Un
it 4: D
yna
mic E
lectricity C
ha
pte
r 10
: Ma
gn
etic E
ffects o
f Ele
ctric Cu
rren
t an
d M
ea
surin
g In
strum
en
ts
Thus, we may calibrate the instrument in terms of the resistance to be measured. If Rx= 3750Ω, the current in the instrument is 200 µA, which is 1/2 the maximum current,and hence the deflection is 1/2 FSD.
If the resistance is doubled, i.e., 7500Ω., the deflection will be FSD. For three timesthe total resistance, i.e., 11250Ω, the deflection will be FSD corresponding to a currentof 100µA. It is to be noted that the graduated scale used to measure the resistance(Fig 10-16) is opposite to the graduated scale for the current .
This means that the maximum deflection corresponds to zero resistance (short circuitor sc). As the resistance increases, the deflection decreases.
It is to be noted also that the scale is not linear. The spacings between the readings ofthe scale to the right are further apart than the readings to the left.
The instruments using a point, are called analog instruments. A combined instrumentcalled multimeter can be switched around to measure voltage, curent and resistance(Fig 10-17).
Another set of instruments now exist which depend on reading numerals, denotingvoltage, current a resistance on a small LCD (liquid crystal display) without the need fora pointer. Such instruments are called digital multimeters (Fig 10-18).
131
4
IµARx(Ω)
400
200
100
0
0
3750
11250
Fig (10 – 16)An ohmmeter has a nonlinear graduated scale
225
Un
it 4: D
yna
mic E
lectricity C
ha
pte
r 10
: Ma
gn
etic E
ffects o
f Ele
ctric Cu
rren
t an
d M
ea
surin
g In
strum
en
ts
a) the length of the wire.b) the current intensity.c) the magnetic flux density.d) the angle between the wire and the direction of the magnetic field.
• A moving coil galvanometer is an instrument used to detect, measure and determinethe polarity of very weak electric currents.
• The operation of a moving coil galvanometer is based on the torque acting on a currentloop in the presence of a magnetic field.
• The sensitivity of a galvanometer is defined as the scale deflection per unit currentintensity flowing through its coil.
• The ammeter is a device which is used through a calibrated scale to measure directlythe electric current.
• To extend the range of the galvanometer, a low resistor known as a shunt is connectedin parallel with the coil of the galvanometer.
• The total resistance of the ammeter (with the shunt) is very small, therefore, it does notappreciably change the current to be measured in a closed circuit.
• The voltmeter is a device used to measure the potential difference across two points ofan electric circuit. It is basically a moving coil galvanometer having a very highresistance called a multiplier resistance connected in series with its coil.
• Since the total resistance of the voltmeter is very great, it does not affect much the flowof current through the element across which it is connected to measure its potentialdifference.
• The ohmmeter is an instrument which is used to measure an unknown resistance.• An ohmmeter is basically a microammeter connected in series with a constant cell
resistance, a variable resistance and a 1.5 volt battery. If its terminals are in contact(sc), the pointer gives full-scale deflection (FSD). If a resistor is inserted between itsterminals, the current flowing decreases. Hence, the pointer's deflection decreases,and indicates directly the value of the inserted resistor through a calibrated scale.
224
Un
it 4: D
yna
mic E
lectricity C
ha
pte
r 10
: Ma
gn
etic E
ffects o
f Ele
ctric Cu
rren
t an
d M
ea
surin
g In
strum
en
ts
In a Nutshell
- Definitions and Basic Concepts:• A megnetic field is produced around a current-carrying wire.• The intensity of the magnetic field produced around a current-carrying wire,
increases, by :a) getting closer to the wire.b) increasing the current.
• The direction of the magnetic field produced around a current-carrying straight wire isdetermined by Ampere’s right-hand rule.
• The lines of force around a current-carrying wire forming a circular loop, resemble to agreat extent those of a short bar magnet.
• The magnetic flux density at the center of a current-carrying circular loop depends on:a) the number of loop turns.b) the current intensity in the loop.c)the radius of the loop.
• The direction of the magnetic field at the center of a current-carrying loop isdetermined by the right-hand screw rule.
• The magnetic field produced by a current flowing through a solenoid (coil of severalclosely spaced loops) resembles to a great extent that of a bar magnet.
• The magnetic flux density at any point on the axis of a current-carrying a solenoiddepends on :a) the current intensity.b) the number of turns per unit length.
• Right-hand screw rule is used to determine the polarity of a solenoid carrying a current.• The unit of magnetic flux density is Web / m2, (Tesla or N/Am). • The force exerted by a magnetic field on a current-carrying wire placed in the field
depends on:
229
Un
it 4: D
yna
mic E
lectricity C
ha
pte
r 10
: Ma
gn
etic E
ffects o
f Ele
ctric Cu
rren
t an
d M
ea
surin
g In
strum
en
ts
5) What is the magnetic flux density at a point on the axis of a solenoid of length 50 cm
carrying a current of 2A and has 4000 turns? (0.02 Tesla).
6) A rectangular loop (12 x 10 cm) of 50 turns, carrying a current of 3 A, is placed in a
magnetic field of 0.4 Tesla flux density, such that the plane of the loop is parallel to
the field. Calculate the torque acting on the loop. (0.72 Nm)
7) A galvanometer's loop of 5 x 12 cm2 and 600 turns is suspended in a magnetic field of
0.1 Tesla flux density. Calculate the current required to produce a torque of 1 Nm.
(2.78 A).
8) A loop of cross-sectional area 0.2 m2 and 500 turns,carrying a current of 10 A is
placed at 30˚ between the normal to its plane and a magnetic field of 0.25 Tesla flux
density. Calculate the torque acting on the loop.
(125 Nm).
9) The coil of an ammeter is capable of carrying current up to 40 mA. If the resistance of
the coil is 0.5 , and it is desired to use the ammeter for measureing a current of 1 A,
What is the resistance value of the required shunt? (0.021 )
10) A galvanometer gives full scale deflection at current 0.02 A, and its terminal voltage
is 5 V. What is the value of the multiplier resistance required to make it valid to
measure potential differences up to 150 V? (7250 )
11) A voltmeter reads up to 150 V at full scale deflection. If the resistance of its coil is
50 and the current flowing is 4 x 10-4 A. Calculate the resistance of the potential
multiplier connected to the coil? (374950 )
12) A galvanometer reads up to 5A and has a resistance of 0.1 . If we want to increase
its reading 10 times, what is the value of the required shunt resistor? (0.0111 ).
13) An ammeter has resistance 30 . Calculate the value of the required shunt resistor to
228
Un
it 4: D
yna
mic E
lectricity C
ha
pte
r 10
: Ma
gn
etic E
ffects o
f Ele
ctric Cu
rren
t an
d M
ea
surin
g In
strum
en
ts
b) the coil of the moving coil galvanometer is attached to a pair of spiral springs.
c) when the moving coil galvanometer is used as a voltmeter, a resistor of high
resistance is connected in series with its coil.
d) an ammeter is connected in series with a circuit, but the voltmeter is connected
parallel to it.
e) connecting a constant resistor inside the ohmmeter.
f) the cell connected to the ohmmeter should have a constant emf.
9) What is meant by each of: potential multiplier and shunt? What is the use of each?
Deduce the rule related to each.
10) Explain how you can use the moving coil galvanometer to measure each of the
electric current, the electromotive force and the electrical resistance.
II) Drills:1) A coil of cross sectional area 0.2 m2 is placed normal to a regular magnetic flux of
density 0.04 Weber/m2. Calculate the magnetic flux which passes through this coil.
(0.008 Weber).
2) A wire of 10 cm length, carrying a current 5 A, is placed in a magnetic field of 1Tesla
flux density. Calculate the force acting on the wire, when:
a) the wire is at right angles to the magnetic field. (0.5 N)
b) the angle between the wire and the field is 45˚. (0.356 N)
c) the wire is parallel to the magnetic flux lines. (0)
3) A straight wire of diameter 2 mm carries a current of 5A. Find the magnetic flux
density at a distance of 0.2 m from the wire. (5x10-6 Tesla).
4) A circular loop of radius 0.1 m carries a current of 10 A. What is the magnetic flux
density at its center? (the loop has one turn). (2 x 10-5 Tesla)
230
Un
it 4: D
yna
mic E
lectricity C
ha
pte
r 10
: Ma
gn
etic E
ffects o
f Ele
ctric Cu
rren
t an
d M
ea
surin
g In
strum
en
ts
decrease the ammeter FSD to one third (decrease the sensitivity), and determine also
the total resistance of the ammeter and the shunt resistor. (15 , 10 ).
14) A galvanometer of resistance 54 , when connected to a shunt (a), the current
flowing through the galvanometer is 0.1 of the total current. But if connected to a
shunt (b), 0.12 of the total current flows through the galvanometer. Find the
resistances of a and b. (6 , 7.364 )
15) A moving coil galvanometer of resistance 50 ohms gives full scale deflection at
current 0.5A. How could it be converted to measure:
a) potential differences up to 200V? (350 in series).
b) electric currents up to 2A? (16 2/3 in parallel)
16) A milliammeter of resistance 5 has a coil capable of carrying a current of 15 mA.
It is desired to use it as an ohmmeter using an electric cell of 1.5V having internal
resistance 1 . Calculate the required standard resistor, and calculate the external
resistance needed to make the pointer deflect to 10mA? Calculate the current that
flows through it when connected to an external resistor of 400 ?
(94 , 50 , 3mA)
233
Unit 4 : Electricity D
ynamic Elictricity and Electrom
agnetism C
hapter 11: Electromagnetic Induction
the coil, a deflection of the pointer was noticed in the opposite direction. This phenomenon iscalled "electromagnetic induction". According to this phenomenon, an electromotive force andan electric current are induced in the coil, when the magnet is plunged into or removed from thecoil. As a result, Faraday concluded that the induced electromotive force and also the inducedelectric current were generatd in the circuit as a result of the time variation of the magnetic fluxlinked with the coil during the motion of the magnet.
Moreover, the action of the magnet is met by a reaction from the coil.If the magnet isplunged into the coil, the induced magnetic field acts in a way to oppose the motion of themagnet. If the magnet is pulled out, the induced magnetic field acts to retain (or keep) themagnet in. Faraday concluded that the induced emf and current were generated in thecircuit as a result of the time variation of magnetic field lines as they cut the windings ofthe coil while the magnet was in motion.
Faraday’s laws:From the above Faraday’s observations, one can conclude the following:
1) the relative motion between a conductor and a magnetic field in which there is timevariation of the magnetic flux linked with the conductor, induces an electromotive forcein the conductor. Its direction depends on the direction of motion of the conductorrelative to the field.
2) the magnitude of the induced electromotive force is proportional to the rate by which theconductor cuts the lines of the magnetic flux linked with it, i.e.,
where ∆φm is the variation in the magnetic flux intercepted by the conductor through thetime interval ∆t3) the magnitude of the induced electromotive force is proportional to the number of turns
N of the coil which cut (or link with) the magnetic flux., i.e.,emf N
Thus, from the analysis of the above mentioned results, one can conclude the following relation: ∝
t
∝emf m
(11 - 1) = - N t
emf m
232
Uni
t 4 :
Elec
trici
ty D
ynam
ic E
lictri
city
and
Ele
ctro
mag
netis
m
Cha
pter
11:
Ele
ctro
mag
netic
Indu
ctio
n
OverviewIt has been noticed that the passage of an electric current in a conductor produces a magnetic
field. Soon after Oersted's discovery that magnetism could be produced by an electric current,aquestion arose, namely, could magnetic field produce an electric current ?
This problem was addressed by Faraday through a series of experiments which led to oneof the breakthroughs in the field of physics, namely, the discovery of electromagneticinduction. On the basis of such a discovery, the principle of operation and function of mostof the electric equipment - such as the electrical generators (dynamos) and transformers -depend.
Faraday’s Experiment: Faraday made a cylindrical coil of insulated copper wire, such that the coil turns were
separated from each other. He connected the two terminals of the coil to a sensitivegalvanometer having its zero reading at the mid point of its graduated scale, as shown in Fig(11-1) . When Faraday plunged a magnet into the coil, he noticed that the pointer of thegalvanometer was deflected momentarily in a certain direction. On removing the magnet from
Fig (11-1a)The magnet is plunged into the coil
Fig (11-1b)The magnet is pulled out of the coil
Electromagnetic Induction Chapter 11
235
Unit 4 : Electricity D
ynamic Elictricity and Electrom
agnetism C
hapter 11: Electromagnetic Induction
This is known as Farady's law of electormagnetic induction. The negative sign in theabove relation indicates that the direction of the induced etectromotive force or the inducedcurrent tends to oppose the cause producing it. This rule is known as Lenz's rule.
Lenz’s ruleThe induced current must be in a direction such as to oppose the change producing it.Fig (11-2) illustrates a direct application of Lenz’s rule :
The direction of the induced current in a straight wire:In one of his several experiments, Faraday showed that the induced current in a straight
wire flowed in a direction perpendicular to the magnetic field. Many years later. Flemingconcluded a simple rule:
Fleming’s right hand rule Extend the thumb,pointer and the middle finger of the right hand, mutually perpendicular
to each other. Let the pointer points to the direction of the field, and the thumb in thedirection of motion, then the middle finger (with therest of the fingers) will point to the direction of theinduced current or voltage as shown in Fig (11-3).
Fig (11 - 3)Fleming’s right hand rule
thumb(motion)
point to (field)
rest of the fingers(induced current
or voltage)
234
Uni
t 4 :
Elec
trici
ty D
ynam
ic E
lictri
city
and
Ele
ctro
mag
netis
m
Cha
pter
11:
Ele
ctro
mag
netic
Indu
ctio
n
(S)
Fig (11 – 2)Lenz's law
237
Unit 4 : Electricity D
ynamic Elictricity and Electrom
agnetism C
hapter 11: Electromagnetic Induction
where M is the coefficient of mutual induction (mutual inductance) of the two coils. Itsunit is VsA-1 and is equivalent to what is calIed "Henry". Thus, the henry is the unit used tomeasure the inductance in general. The negative sign in equation (11-2) follows fromLenz's rule, namely, that the direction of the induced electromotive force (or the directionof the induced current) is such as to oppose the cause producing it. The coefficient ofmutual inductanc between two coils depends on the following factors.
1. the presence of an iron core inside the coil. 2. the volume of the coil and the number of its turns.3. the distance separating them.The transformer is considered as a clear example of mutual induction
Experiment to study mutual induction One can study experimentally the mutual induction as follows:
Connect one of the two coils in a circuit which contains a battery, a switch and a rheostat. Onecoil is called the "primary coil", while the other coil - connected to a sensitive galvanometer with itszero point at the middle of its scale - is known as the "secondary coil". Fig( 11-5). Let us do theexperiment as follows:1) Close the circuit of the primary coil, while plunging the primary coil into the secondary coil.
One notices a deflection in the galvanometer in a certain direction, indicating the generationof an induced electromotive force in the secondary coil due to the variation of the number ofmagnetic flux lines linked with the turns of the secondary coil. On taking away the primarycoil from the secondary coil, one notices that the pointer of the sensitive galvanometer isdeflected in the opposite direction.
2 = - MI 1
t(emf)
(emf)
(11 - 2)
I1
t
1
236
Uni
t 4 :
Elec
trici
ty D
ynam
ic E
lictri
city
and
Ele
ctro
mag
netis
m
Cha
pter
11:
Ele
ctro
mag
netic
Indu
ctio
n
Fig (11 - 4A)in case there is no current in the first
coil, there is no emf in the second coil.
Fig (11- 4b)at the instant of closing the ciruit of the firstcoil, an emf is generated in the second coil.
Fig (11 - 4c)after the current in the primary is steady (the flux is steady) emf in the secondary coil: 0
Mutual induction between two coils:
If the two stationary coils are arranged such that one coil surrounds the other., i.e., onecoil is plunged into the second one, or even one is placed in the neighborhood of the otheras shown in Fig.(11-4), then the variation in the intensity of the electric current in one of thetwo coils (opening and closing the switch) will induce an electromotive force in the othercoil, according to Faraday’s law. This induced electromotive force is proportional to therate of change in the magnetic flux linked with the other coil. Since the magnetic flux isproportional to the intensity of current in the first coil.
239
Unit 4 : Electricity D
ynamic Elictricity and Electrom
agnetism C
hapter 11: Electromagnetic Induction
magnetic field will be in a direction as to resist the increase in the affecting magnetic field.II. The pointer of the galvanometer deflects in the opposite direction in the following cases:
a) on the withdrawal of the primary, or taking it far away from the secondary coil. b) on decreasing the intensity of the current in the primary.c) on switching off the primary circuit. In the above cases, he intensity of the magnetic field affecting the secondary coil
decreases and the magnetic flux linkage decreases. The induced emf in the secondary coildecreases as the affecting field decreases with time. The direction of the inducedelectromotive force (and the induced current) is in the forward direction, so as to produce amagnetic field in the same direction as the current in the primary. This in turn resists thedecrease in the affecting magnetic field. All these observations clarify Lenz's rule, wherethe direction of the induced current is such as to resist (or to oppose) the time variationcausing it.
Self induction of a coil:One can understand what is meant by self induction of a coil by connecting the coil of a
strong electromagnet (a coil of largenumber of turns) in series with a 6Vbattery, and a switch as shown in Fig(11-6). Current passes in the consideredcoil, due to which a strong magneticfield is formed, since each turn acts as asmall magnet. The magnetic flux linkswith the neighboring turns.
On switching off the circuit, it is noticedthat an electric spark is passed between thetwo terminals of the switch. This isexplained as follows.
Fig (11 - 6)Effect of self induction in a coil
battery switch
neon lamp
electromagnetcoil
VB
238
Uni
t 4 :
Elec
trici
ty D
ynam
ic E
lictri
city
and
Ele
ctro
mag
netis
m
Cha
pter
11:
Ele
ctro
mag
netic
Indu
ctio
n 2) Plunge the whole primary coil to reside in thesecondary one, then increase the intensity of thecurrent in the primary coil. Notice the deflectionof the pointer of the galvanometer in acertain direction. Decrease the current in theprimary, and notice that the deflection of thepointer takes place in the opposite direction. Thisindicates the generation of an inducedelectromotive force in the secondary coil onincreasing or decreasing the intensity of thecurrent in the primary coil.
3) With the primary coil inside the secondary one,close the circuit of the primary coil, a deflectionis noticed in the galvanometer in a certaindirection. Open the primary circuit, and noticethat the deflection is in the opposite direction.This indicates that an electromotive force isinduced in the secondary coil upon switching on or switching off the primary circuit.The analysis of the above mentioned observations leads to the following conclusions:
I. The pointer of the sensitive galvanometer deflects in a certain direction in the following cases:a) bringing the primary coil close to the secondary coil or when the primary coil is
plunged inside the secondary one.b) increasing the intensity of the current in the primary coil. c) switching on the primary circuit.In all cases above, there is a positive increase in magnetic flux linkage and the induced
emf in the secondary coil increases as the affecting magntic field increases with time. Theinduced current is in opposing direction to that in the primary. In such a case, the induced
battery
switch rheostat
primarycoil
secondarycoil
galvanometer
Fig (11 - 5)Mutual inductance between
two coils
241
Unit 4 : Electricity D
ynamic Elictricity and Electrom
agnetism C
hapter 11: Electromagnetic Induction
The Henry :It is the self-inductance of a coil in which an emf of one volt is induced when the current
passing through it changes at a rate of one Ampere per second (vsA)-1.The self inductance of a coil depends on:
a) its geometry.b) its number of turns.c) the spacing between the turns.d) the magnetic permeability of its core.
Among the applications of self induction isthe fluorescent lamp, where magnetic energy isstored in the coil. This energy is discharged inan evacuated tube filled by an inert gas,causing collisions of its atoms and theirsubsequent ionization and collision with thewalls of the tube.
The inner walls are coated with a fluorescent material which causes visible light to beemitted upon the collision of the inert gas ions with it.
Electromagnetic induction is also used in Ruhmkorff coil, which is used as an ignitioncoil in internal combustion engines (such as a car).
Eddy Currents:If the magnetic flux changes with time through a solid conductor ,currents will be induced
in closed paths in the conductor. Such currents are called "eddy currents". The change in theintercepted magnetic flux is effected either by moving the solid in a suitable magnetic field orby subjecting the metallic solid to an alternating magnetic field( for example field due to anAC current). The eddy currents are associated with heating effects. Thus, they are useful inmelting metals in what is called the induction furnaces.
Henry
240
Uni
t 4 :
Elec
trici
ty D
ynam
ic E
lictri
city
and
Ele
ctro
mag
netis
m
Cha
pter
11:
Ele
ctro
mag
netic
Indu
ctio
n
∝
L = - / t
eemf
When the coil circuit is switched off, the current ceases to pass in it, and this isassociated with a decrease of the magnetic field of the neighboring turns to zero. This inturn is accompanied by a time variation of the flux linkage, i.e., each turn cuts thediminishing lines of the magnetic flux, and thus, an induced electromotive force isgenerated in the coil.
The induced electromotive force is formed in the turns of the coil as a whole as a resultof the self induction of the coil itself. This induced electromotive force is generated due tothe self induction of the coil on switching off or switching on the circuit following Lenz'srule. Thus, an induced electric current is generated in the same direction as the originalcurrent. When the circuit is switched off, to retain the existing current, a spark is formedbetween the two terminals of the switch. When the number of turns of the coil is large, theinduced emf on switching off the circuit will be much larger than that of the battery, Thiscauses a neon lamp connected in paralle1 between the two terminals of the switch to glow(aneon lamp requires a potential difference about 180V to glow).
Since the induced electromotive force is proportional to the rate of change of the currentin the coil, then the emf induced by self induction is directly proportional to the rate ofchange of the current in the coil. That is :
where L is a constant of proportionality known as the coefficiont of self induction (selfinductance) of the coil, and the negative sign in equation (11-3) indicates that the inducedelectromotivc force opposes the change causing it (Lenz's rule).
Thus, the self inductance of a coil is defined as: It is the electromotive force induced in the coil when the current passing through it changes
at a rate equals one Ampere per second. The self inductance is measured in the unit henry.
I1
t(emf)1
= - L I 1
t(11 - 3)e(emf)1∴
243
Unit 4 : Electricity D
ynamic Elictricity and Electrom
agnetism C
hapter 11: Electromagnetic Induction
Alternating current generator: The AC generator (or the dynamo) is a device which converts the mechanical energy into
electrical energy. In a generator, a coil rotates in a magnetic field, and the resulting induced
current can be transferred (or transmitted) by wires for long distances.
The simple electric generator consists as shown in Fig (11-8) of four main parts :
a) a field magnet.
b) an armature.
c)two slip rings.
d) two brushes.
The field magnet may be a permanent
magnet or an electromagnet. The armature
consists of a single loop of wire or coil of
many turns suspended between the two poles
of the field magnet. A pair of slip rings are
connected, one to each end of the loop. They
rotate with the loop in the magnetic field.
The induced current in the coil passes to the
external circuit through two graphite
brushes, each touching one of the two
corresponding slip rings. Fig (11-9) shows
the direction of rotation of the armature between the poles and the direction of the induced
current at a certain instant. The loop rotates around its axis in a circle of radius r. Its linear
velocity is
v = ω r
where ω is the angular velocity equal to 2πf, (where f is the frequency). Substituting for
slip rings
directionof motion
brushes
permanent field magnetic
armature
Fig (11 - 8)A simplified schematic for an
AC generator (dynamo)
245
Unit 4 : Electricity D
ynamic Elictricity and Electrom
agnetism C
hapter 11: Electromagnetic Induction
From Fig (11-10), we see that the induced currentchanges direction every half a revolution. It followsa sine wave. From figure, we can also understandthe meaning of f.
Throughout a complete revolution, the currentincreases from zero to a maximum, then decreasesto zero, then reverses direction, and increases in thenegative direction up to a negative maximum. Then,it heads back to zero. In one complete revolution,one complete oscillation has occurred. The numberof oscillations per second is the frequency f. Thefrequecy of home use power is 50Hz
Example:The coil of a simple AC generator consists of 100 turns, the cross sectional area of each
is 0.21 m2. The coil rotates with frequency 50 Hz (cycles/second) in a magnetic field ofconstant flux density B = 10-3 Weber/m2. What is the maximum induced emf generated?and what is the instantaneous value at θ = 30˚?Solution:
(emf)max = NBA ω = NBA (2 πf)
Thus, the maximum induced emf generated equals 0.6 volts.
It is worth remembering that the induced current is directly proportional to the inducedemf. Thus, the instantaneous value of the induced current is given by :
I = Imax sin (2 πf t )
= 100 x 10-3 x 0.21 x 2 x 227
x 50 = 6.6 V
graphitebrushes
slip rings
armature
Fig (11-10a)AC generator
e = max sin = 6.6 x sin 30 = 6.6 x 12
= 3.3 V˚emf = (emf)max
246
Uni
t 4 :
Elec
trici
ty D
ynam
ic E
lictri
city
and
Ele
ctro
mag
netis
m
Cha
pter
11:
Ele
ctro
mag
netic
Indu
ctio
n This induced current reaches its maximum value when the induced emf reaches itsmaximum value, and it vanishes as the induced emf is zero.
Effective value of the alternating current:It is worth mentioning that the average value
of an AC current equals zero, because the ACcurrent changes from (Imax) to (-Imax). Neverthless,the electric energy is consumed as thermalenergy due to the motion of electric charges,and the rate of the electric energy consumed isproportional to the square of the intensity ofthe current.
The effective value of the intensity of thealternating current is the value of the directcurrent which generates the same rate of thermaleffect in a resistance (or the same power) as thatgenerated by the considered AC current.
Ieff = 0.707 Imax
The value Ieff is called the "effective value of the alternating current". There is a similar relation for the effective electromotive force, that is :
(emf)eff = 0.707 (emf)max
Veff = 0.707 Vmax
Example:If the effective intensity of current in a circuit equals 10 A, and the effective voltage is
240 volts,what is the maximum value for current and voltage ?
Fig (11-10b) The relation between current and
angle of rotation(sine wave)
maximum positive current
maximumnegative current
coil position
current
onecomplete
revolution
(11 - 11)
(11 - 12)
θ
249
Unit 4 : Electricity D
ynamic Elictricity and Electrom
agnetism C
hapter 11: Electromagnetic Induction
Continuing the rotation, the brush Fl acts as a positive pole, while F2 acts as thenegative pole of the dynamo. Accordingly, the current in the external circuit will bealways in one direction as shown. It is noticed that using the commutator renders theinduced emf in Fig (11-11d) in one direction, but its value changes from zero up to amaximum value, then decreases again to zero during each half cycle of the coil rotation,but it is always in one direction.
To obtain a uni-directional current ofapproximately constant value, i.e., to obtain a nearlyDC (value), many coils separated by small angles areused. A cylinder is used which is split into a numberof segments, double the number of coils. Thus, thecurrent in the external circuit is almost constant. Thisis the way to obtain a DC generator (Fig 11-12).
The transformer: The electric transformer is a device whose
function is based on the mutual inductionbetween two coils, and is used to step up or tostep down an AC voltage. Transformers areused to transfer the electric energy fromgenerators at electric power stations. Suchtransformers are called step - up transformers,while the transformers used at the zoneswhere the energy has to be distributed among buildings are called step-down transformers.The transformer as shown in Fig (11-13) consists of two coils: a primary coil and asecondary coil. The two coils are wound around a soft iron core made of thin iron sheets
Fig (11-12)Nearly DC current
Fig (11-13a)Step UP transformer
primary coil
secondary coil
output
input
soft iron core( laminas)
248
Uni
t 4 :
Elec
trici
ty D
ynam
ic E
lictri
city
and
Ele
ctro
mag
netis
m
Cha
pter
11:
Ele
ctro
mag
netic
Indu
ctio
n commutator consists of two halves 1 and 2 of ahollow metallic cylinder split in between, and are wellinsultated from each other as shown in Fig(11-11).Two brushes F
1and F
2 touch the two halves
during the rotation of the coil. The external circuit isconnected to the two brushes Fl and F2. It is necessarythat the two brushes F1 and F2 touch the insulatorbetween the two halves at the moment when the planeof the coil is perpendicular to the magnetic field, i.e,at the instant when the generated electromotive forcein the coil is zero.
Let us consider that the coil starts rotation in thedirection shown (Fig.11-11c).During the first halfrotation, brush F1 touches the half cylinder (1),while brush F2 touches the other half (2) of thecylinder. The current in such a case will pass in thecoil in the direction w x y z. As a result, the currentpasses in the external circuit in the direction from Flto F2 during the first half of the cycle. In the secondhalf of the cycle, the electric current reverses itsdirection in the coil, i.e., the current passes in the coilin the direction z y x w. At the same time, brush F1 will be in contact with the half(2), while F2will be in contact with the half(1), i.e., the two halves of the commutator reverse their positionrelative to the two brushes. In such a case, the current in the external circuit passes from, F1 to F2,which is the same direction as that in the first half of the cycle.
direction ofrotation
brushes
split cylinder
Fig (11-11c) Use of a split cylinder
rectifies the current
Fig (11-11d)Unidirectional current versus θ (sine wave)
current
number ofrevolutions
251
Unit 4 : Electricity D
ynamic Elictricity and Electrom
agnetism C
hapter 11: Electromagnetic Induction
This equation shows the interrelation between the emf Vs in the secondary and Vp in theprimary. If Ns is larger than Np, one has a step-up transformer, where the emf in thesecondary coil will be larger than the emf in the primary one. For example, if the number ofturns of the secondary coil is twice that for the primary coil, one gets Vs = 2VP.
While, for the case when Ns is less than Np one gets a step-down transformer, where, insuch a case Vs will be less than Vp.
The relation between the current intensities in the two coils of the transformer: Let us assume that there is no loss in the electric energy in the transformer (almost zero
resistance), then according to the law of conservation of energy, the electric energy madeavailable by the source in the primary coil must equal that delivered to the load in thesecondary coil.
Vp Ip t = Vs Ist
From which the input power is equal to the output power, i.e, Vp Ip = Vs Is
Thus,From the equations (11-11) and (11-12),
This shows that the intensity of the electric current in either of the two coils is inverselyproportional to the number of its turns.
For example: if the number of turns of the secondary coil is twice that of the primary coil,then the intensity of current in the secondary coil equals half that in the primary coil.
From this argument, we see the importance of the use of the step-up transformer at the
I s
I p =
Np
Ns
V s
V p =
Ip
Is
250
Uni
t 4 :
Elec
trici
ty D
ynam
ic E
lictri
city
and
Ele
ctro
mag
netis
m
Cha
pter
11:
Ele
ctro
mag
netic
Indu
ctio
n (laminas) insulated from each other, to minimize the effectof eddy currents and to minimize the dissipated electricenergy. When an electric current passes in the primary coil,a magnetic field is generated. The core makes the lines ofsuch a field pass through the secondary coil.
The relation between the two emfs in the twocoils of the transformer:
When the primary coil is connectcd to a source of AC
voltage, the variation in the magnetic field linked with the primary current generates aninduced emf in the secondary coil having the same frequency. The induced emf in thesecondary is determined from the relation:
where, Ns is number of turns of the secondary coil and is the rate of change of themangetic flux linked between the primary through the secondary coil. The electromotiveforce in the primary is in turn related to the rate of change of the magnetic flux and isdetermined from the relation :
where, Np is the number of turns of the primary coil. Assume that the wasted magneticenergy is negligible, i.e., there is no considerable loss in the magnetic flux, i.e., the wholeresulting magnetic flux passes through the secondary coil (no stray lines). Dividing theabove two relations one can get the following formula :
Vs = Ns t
Vs= - t
Vp = Np t
Vp= -
Fig (11-13b)Transformer symbol
(11 - 13)V s
V p =
N s
N p
m
m
255
Unit 4 : Electricity D
ynamic Elictricity and Electrom
agnetism C
hapter 11: Electromagnetic Induction
Solution:
Learn at Leisure
AC/DCThere are two types of current or voltage AC and DC .In the case of DC, Ohm’s law
dictates that what determines the current is the resistance.In the case of AC, what determines the current are three elements,the resistor, the
inductor and the capacitor. Household appliances use 220 V AC of frequency 50 HZ. Often,we need to convert this to a lower DC voltage, as in the case of the mobile charger andsome other appliances.To do this, we use a down transformer and a rectifier. It is to benoted that a low AC current is more hazardous to man than a low DC current (why?).
η= x100
η= x x100Vs NpVp Ns
80 = x x1008 1100220 Ns
Ns = 50
IsIp
= NpNs
Is0.1
= 110050
Is = 2.2 A
turns
Vs IsVp Ip
254
Uni
t 4 :
Elec
trici
ty D
ynam
ic E
lictri
city
and
Ele
ctro
mag
netis
m
Cha
pter
11:
Ele
ctro
mag
netic
Indu
ctio
n an audio signal (Fig 11-16). The same thing happens in the hard disk in the computer,where data is stored by magnetization. In this way, the data is not lost from the hard diskwhen the power of the computer is switched off.
Examples:1- A trandformer connected to a 240 V AC power source gives 900 V output emf with
current intensity 4A. What is the intensity of the source current assuming that the efficiencyof the transformer is 100%?Solution:
2) An electric bell is connected to a transformer of efficiency 80% which gives 8 V output,while the input household voltage is 220 volts. What is the number of turns of the secondarycoil if the number of turns of the primary coil is 1100 ? and what is the intensity of current inthe secondary coil if the current in the primary coil is 0.1 A ?
. . .Vs
Vp
= Ip
I s
I p = 900 x 4240
= 15 A
900240
= Ip
I s4
Fig (11-13b)Use of electromagnetic induction in recording
input audio
unmagnetizedtape
iron core
magnetized tape
magnetic field
257
Unit 4 : Electricity D
ynamic Elictricity and Electrom
agnetism C
hapter 11: Electromagnetic Induction
Operation of a DC motor through one complete revolution:Starting from a position at which the plane of the coil is parallel to the lines of the
magnetic flux, and the brush F1
- connected to the positive terminal of the battery - touchesthe half cylinder x, while F
2 - connected to the negative terminal of the battery - touches the
half cylinder y as shown in Fig (11-17). Thus, current passes in the coil in the direction dcba.Applying Fleming's left hand rule, one concludes that the wire ab is affected by a force inthe upward direction, while the wire cd is affected by a force in the downward direction. Thetwo produced forces (couple) form a torque, and the coil begins to rotate in the directionshown in the figure. As the coil rotates, the moment of the couple decreases gradually till itvanishes, when the coil plane becomes perpendicular to the lines of the magnetic flux. Butthe coil having gained a momentum will continue motion due to its inertia, which in turnpushes the coil to the other side. The two halves x and y of the commutator interchangeposition, such that the half cylinder x will be in touch with the brush F2,while the brush F
1will touch to other half cylinder y. Thus, the current in the coil will reverse direction andpass in the direction abcd. Applying "Fleming's left hand rule" for the new position of thecoil shows that the force acting on the wire ab will be downward, while the force acting onthe wire cd will be upwards. The obtained torque enables the coil to continue rotation in thesame circular direction. The torque increases gradually to its maximum value when the planeof the coil becomes parallel to the lines of the magnetic flux. Then, it decreases to zero whenthe plane of the coil is perpendicular to the lines of magnetic flux. The inertia of the coilthen causes it to continue rotating to the other side. This permits the two halves tointerchange positions and with respect to the two brushes F1 and F2, and thus, the current inthe coil is reversed once more. The coil continues rotating in the same circular directionmaking one complete revolution, and so on.
In order to increase the power of the motor, a number of coils may be used with equal
256
Uni
t 4 :
Elec
trici
ty D
ynam
ic E
lictri
city
and
Ele
ctro
mag
netis
m
Cha
pter
11:
Ele
ctro
mag
netic
Indu
ctio
n DC motor:It is a device which converts electric energy to
mechanical energy. It operates on a DC source(battery) (Fig 11-17). It consists in its simplestform of a rectangular coil abcd comprising a largenumber of turns of insulated copper wire woundaround a soft iron core made of thin insulatedsheets to cut down on eddy currents.
The core and the coil can rotate between thetwo poles of a strong horseshoe (U-shaped) fieldmagnet. The two terminals of the coil areconnected to two halves of a split cylinder(commutator). The two halves (x,y) are insulated from each other and capable of rotatingaround the axis of the coil.
The plane separating the two halves is perpendicular to the plane of the coil and the lineconnecting the two brushes is parallel to the lines of magnetic flux.
To operate the motor, the two brushes must be connected to the battery.
The motor and the galvanometer:The principle of operation of the electric motor and that of the moving coil galvanometer
are alike. The main difference is that the electric motor must rotate continuously in thesame direction. The design of the electric motor necessitates that the two halves x,y of thecylinder must interchange positions relative to the two brushes F1 and F2 each half cycle.As a result, the electric current passing in the motor must reverse direction in the coil eachhalf revolution.
Fig (11 - 17)DC motor
ficld magnet
brushes
variable resistance
splitcylinder
259
Unit 4 : Electricity D
ynamic Elictricity and Electrom
agnetism C
hapter 11: Electromagnetic Induction
Learn at Leisure
The microphone and the loud speakerThe operation of a moving coil microphone depends
on the vibration of an armature according to the soundwaves. The magnetic field in an iron core changes,which results in the generation of an emf in a coilwound around the iron core. This emf is of variableamplitude and frequency according to the individualsound. Thus, a sound signal (mechanical wave) isconverted to an electrical (audio) signal.
In a moving coil loud speaker, the reverse takesplace.
Fig (11 - 19)A microphone
Learn at Leisure
Search for Metals A metal detector is used in the search for
metals. Its operation depends on measuring
the change in the self inductance L of a coil
due to its proximity to a metal. The current
in the detector changes giving away the
hidden matal (Fig. 11-18).
Fig (11 - 18)A metal detector
magnet
coil
Soundwaves
258
Uni
t 4 :
Elec
trici
ty D
ynam
ic E
lictri
city
and
Ele
ctro
mag
netis
m
Cha
pter
11:
Ele
ctro
mag
netic
Indu
ctio
n angles between their planes. The two terminals, of each coil are connected to two oppositesplits of a cylinder. The cylinder is split into a number of segments twice that of the numberof the coils. During rotation, each two opposite segments touch the two brushes F1 and F2
when their corresponding coil is in position of largest torque.
Learn at Leisure
Motor and generator at the same time:While operating the motor, its coil cuts the line of magnetic flux of the field magnet.
There is a rate of change of cutting magnetic lines. Therefore, an emf is induced oppositeto the source, reducing the current, and hence, the speed.This back emf in the motor coilacts to the stabilize the speed of rotation of the coil. If the speed of the coil tends toincrease, the back emf increases. The difference between the back emf and the externalsource is the voltage drop across the coil resistance. Therefore, as the back emf increases,the current decreases, and so does the speed of rotation. Conversely, if the coil speed tendsto decrease, the back emf decreases, and the current increases. So the speed of rotation ofa DC motor remains constant.But the speed can be changed by changing the source(battery) voltage. It should be noted that if we try to stop the motor by force while it isconnected to the source, the motor coil burns out, because the back emf would disappearand the battery voltage is applied in full across the small coil resistance so it burns out.Also, if a transformer works on no load (secondary is open circuited), the secondarycurrent is zero and the primary current should be zero. However, a back emf in the primaryalmost balances out with the input voltage due to self-induction. A small current in theprimary exists even at no load to produce the flux linkage. So, an ideal transformer doesnot really exist. However, at full load, we may consider - as an approximation - that theideal transformer model works.
261
Unit 4 : Electricity D
ynamic Elictricity and Electrom
agnetism C
hapter 11: Electromagnetic Induction
Fig (11 - 21b)A section in ignition circuit components
Fig (11 - 21a)A schematic for the ignition circuit in a car
copacitor
point contact
platinumpoint contacts car battery
primary coil
secondary coil
cam
iron core
spark plug
distributer
spark
rotating distributer
discharge arm
ignition coil
point contact
contact
spark plug
260
Uni
t 4 :
Elec
trici
ty D
ynam
ic E
lictri
city
and
Ele
ctro
mag
netis
m
Cha
pter
11:
Ele
ctro
mag
netic
Indu
ctio
n The variable audio (electrical) signal producesvariations in the coil current. The coil isconnected to a diaphragm which vibrates dueto the force generated in the presence of amagnetic field. Mechanical (sound) wavesresult, resembling the original audio signal.
Thus, sound is heard back (Fig 11-20).
paper core
soft iron cylinder
movingcoil
soft ironsheet
permanentmagnet
flexible wires tothe coil
diaphragm
Fig (11 - 20)A loud speaker
Learn at Leisure
lgnition circuit in a carThe ignition coil (Fig 11-21) consists of two coils one inside the other, both wound
around a soft iron core. The primary coil and the core comprise an electromagnet. theprimary circuit is opened regularly by a distributer cam as it rotates, thus, opening andclosing the point contacts. The secondary coil contains thousands of turns. A large emf isgenerated from time to time at the same rate of opening the primary circuit. This large emfgenerates a spark across the air gap across the spark plugs. The spark plugs are connectedalternately to the secondery coil as the distributer rotates. A capacitor is used to protect thepoints from corrosion due to the spark.
Electronic ignition system works on the same principle, but the cam is replaced bytransistors as a switch (Chapter 15).
263
Unit 4 : Electricity D
ynamic Elictricity and Electrom
agnetism C
hapter 11: Electromagnetic Induction
• Self-induction: It is the electromagnetic effect induced in the same coil when theintensity of the current increases or decreases.This effect acts to resist such a change inthe intensity of current.
• Coefficient of self-induction : It is measured numerically by the electromotive forcegenerated by induction in the coil when the intensity of the current passing through itchanges at a rate of 1A/s.
• The unit of measuring the self induction (Henry): It is the self induction of a coil inwhich an emf of 1V is induced when a current passes through it which changes at a rateof 1A/s.
• The self-induction of a coil depends on :a) its geometry. b) its number of turns.c) the spacing between its turns. d) the magnetic permeability of its core.
• The Dynamo (AC Generator): It is a device used to convert the mechanical energy toelectric energy(AC current and voltage) when its coil rotates in a magnetic field.The simple dynamo (AC generator) consists of :a) field magnet (strong magnet).b) a coil of insulated copper wire suspended between the two poles of the magnet.c) two metallic rings in contact with two graphite brushes connected to an external
circuit.• A commutator: (cylinder split into a number of insulated segments) is used to obtain aDC current and voltage (DC generator).• The alternating current: It is current which changes periodically its intensity and
direction with time according to a sinusoidal curve. • The electric transformer: It is an electric device used to step up or step down an emf
through mutual electromagnetic induction.
1H 1V.SA
=s
262
Uni
t 4 :
Elec
trici
ty D
ynam
ic E
lictri
city
and
Ele
ctro
mag
netis
m
Cha
pter
11:
Ele
ctro
mag
netic
Indu
ctio
n In a Nutshell
Definitions and Basic Concepts:- Electromagnetic induction : It is a phenomenon in which an induced electromotiveforce and also an induced current are generated in the coil on plunging a magnet intoorwithdrawing a magnet out of a coil.
• The presence of a soft iron core inside a coil concentrates the lines of magnetic flux thatlink with the coil. This in turn increases the induced electromotive force and also theinduced current.
• Faraday's law for the induced emf : The induced emf generated in a coil byelectromagnetic induction is proportional to the time rate by which the conductor cutsthe lines of magnetic flux and is also proportional to the number of turns of the coil.
• Lenz's rule: the direction of the induced current generated by induction is such thatto oppose the change in the magnetic flux producing it.
• Fleming's right hand rule: Place the thumb, the pointer and the middle finger(withthe rest of the fingers) of the right hand mutually at right angles. If the pointerpoints in the direction of the magnetic field and the thumb in the direction ofmotion then the middle finger (with the rest of the fingers) will point in thedirection of the induced current.
• Mutual induction: It is the electromagnetic interaction between two coils kept close toeach other (or one inside the other).An electric current with time varying intensitypassing in one coil (primary coil)will produce in the second one (secondary coil) aninduced current in a direction such that to oppose the variations of the current intensity inthe primary coil.
264
Uni
t 4 :
Elec
trici
ty D
ynam
ic E
lictri
city
and
Ele
ctro
mag
netis
m
Cha
pter
11:
Ele
ctro
mag
netic
Indu
ctio
n • The efficiency of the transformer: It is the ratio between the output electric energygiven in the secondary and that available to the primary.• The electric motor: It is an electric device used to convert the electric energy intomechanical energy .
Basic laws:• The induced emf genrated in a coil of N turns as a result of time variation of magnetic
flux φm linked with the coil in an interval of time is given by the relation:
The nagative sign indicates that the direction of the induced emf (and thus the current) issuch as to oppose the cause producig it.• The emf induced in a secondary coil due to the time variation in the lines of magnetic flux
resulting from a primary coil linking with the secondary coil in a time interval t is givenby the relation :
emf = where M is the coefficient of mutual induction.
• The emf induced by self induction as a result of the current ∆I passing through the coil ina time ∆t is given by the relation :
where M is the coefficient of self induction of coil.
= - N t
vVemf
- M
emf = - L
It
It
∆∆
∆
∆
267
Unit 4 : Electricity D
ynamic Elictricity and Electrom
agnetism C
hapter 11: Electromagnetic Induction
4) A current passes in the primary coil, then this coil is plunged into a secondary coilwhose terminals are connected to a galvanometer. The deflection of its needle will be ina direction:a) opposite to the current in the primary coil.b) points to zero readingc) increasing.d) same as the current in the primary coile) variable
5) Opening the primary circuit while the primary coil is inside the secondary one, leads tothe generation of :a) an induced forward current.b) an electric fieldc) an induced back current.d) an AC current.e) a magnetic field.
6) The slow rate of growth of the current in the solenoidal coil is due to the:a) production of forward current.b) production of a magnetic field.c) production of a back induced current opposing (resisting ) the original one.d) production of a magnetic flux.e) production of an electric field.
7) The ohmic resistors are made of double wound wires:a) to decrease the resistance of the wire.b) to increase the resistance of the wire.
266
Uni
t 4 :
Elec
trici
ty D
ynam
ic E
lictri
city
and
Ele
ctro
mag
netis
m
Cha
pter
11:
Ele
ctro
mag
netic
Indu
ctio
n Questions & Drills
I) Put ( ) against the right answer:1) The pointer of a galvanometer whose terminals are connected to a solenoidal coil will be
deflected if one withdraws the magnet quickly from the coil because: a) the number of the coil turns is very large.b) the coil intercepts the lines of the magnetic flux.c) the number of the turns of the coil is small.e) the number of turns of the coil is suitable.
2) The needle of the galvanometer whose terminals are connected to a solenoidal coildeflects on the withdrawal of the magnet in a direction opposite to that which occurs onplunging the magnet into the coil because:a) an induced current is generated in a direction opposite to that on plunging the magnet.b) an electric current is generated.c) the number of the lines of magnetic flux decreases.d) the number of the lines of the magnetic flux changes.e) the number of flux lines remains constant.
3) The emf induced in a coil on plunging a magnet into or withdrawing it out of a coildiffers according to the difference in :a) [the intensity of the current - the length of the wire - the number of the lines of flux]. b) [magnet strength - the velocity with which the magnet moves- the number of turns of
the coil].c) [the cross sectional area of the coil - the mass of unit length - the material from which
the wire is made]. d) [the length of the wire - the number of turns - the type of the magnet].e) [the magnetic flux density - time - the intensity of the current].
269
Unit 4 : Electricity D
ynamic Elictricity and Electrom
agnetism C
hapter 11: Electromagnetic Induction
c) several magnetsd) an insulated copper wire. e) a current rectifier.
12) The ratio between the electric energy in the secondary to that in the primary is called:a) the lost energy. b) the given energy.c) the efficiency of the transformer.d) the working strength of the transformer.e) the gained energy.
II) Define the following :1- Electromagnetic induction.2- Faraday's law of induction3- Lenz's rule.4- Fleming 's right hand rule.5- Mutual induction.6- Unit of measuring the mutual inductance.7- Self induction.8- Coefficient of self induction.9- The Henry.10- The induction coil.11- The AC current.12- The dynamo. 13- The electric motor.14 - The transformer.15- The efficiency of the transformer.16- The back emf in the motor.
268
Uni
t 4 :
Elec
trici
ty D
ynam
ic E
lictri
city
and
Ele
ctro
mag
netis
m
Cha
pter
11:
Ele
ctro
mag
netic
Indu
ctio
n c) to avoid self-induction.d) to eliminate the resistance of the wire.e) to facilitate the connection process.
8) The direction of the current produced in the dynamo coil can be determined using:a) Fleming's left hand rule. b) Lenz's rule. c) Fleming's right hand rule.
9) The rate with which the coil intercepts the lines of magnetic field in the dynamo ismaximum when: a) the plane of the coil is perpendicular to the flux lines.b) the plane of the coil is inclined to the lines by an angle 30˚ c) the face area of the coil is minimum.d) the face area of the coil is maximum. e) the plane of the coil is parallel to the lines of the magnetic flux.
10) The intensity of the current in the two coils of the transformer is : a) directly proportional to the number of the turns.b) inversely proportional to the number of the turns. c) depending on the temperature of the wire.d) depending on the substance of the wire. e) depending on the temperature of the air (ambient temperature).
11) The power of an electric motor to rotate increases on using:a) larger number of turns. b) several coils with angles between their planes.
271
Unit 4 : Electricity D
ynamic Elictricity and Electrom
agnetism C
hapter 11: Electromagnetic Induction
IV) Give reasons 1) The core of an electic transformer is made of thin sheets insulated from each other.2) A bar of soft iron will not be magnetized if a double wound wire carrying a current is
wound around it.3) A wire free to move in a magnetic field moves when a current passes through it.4) The transformer is not suitable to convert DC voltage.5) The electric motor rotates with uniform velocity. 6) The induced current dies out in a straight wire faster than in a coil with air core, and
in a coil with air core faster than in a coil wound around an iron core.7) The metallic cylinder used to obtain a unidirectional current in the dynamo is split into
two halves completely insulated from each other.
V) Drills 1) A coil of 80 turns, and cross sectional area 0.2 m2 is suspended in a perpendicular
position to a uniform magnetic field. The average induced emf is 2 V when it rotates1/4 revolution through 0.5 s. Find the magnetic flux density.
(0.0625T)2- If the magnetic flux density between the two poles of the magnet of a dynamo is 0.7
Tesla, and the length of its coil is 0.4 m, find the velocity of motion in such a field toobtain an induced emf in the wire equal to 1V. (3.57m/s)
3) A coil of a dynamo consists of 800 turns each of face area 0.25 m2. It rotates at a rateof 600 revolutions per minute, in a field of magnetic flux density 0.3 Tesla. Calculatethe induced emf when the angle made between the normal to the coil and the magneticflux is 30˚.
4) A rod of copper of length 30 cm moves with at velocity 0.5 m/s in a perpendicular directionto a magnetic field of density 0.8 Tesla. Calculate the emf induced in such a rod.
(1885v)
(0.12v)
270
Uni
t 4 :
Elec
trici
ty D
ynam
ic E
lictri
city
and
Ele
ctro
mag
netis
m
Cha
pter
11:
Ele
ctro
mag
netic
Indu
ctio
n III) Essay questions: 1) What are the factors on which the emf induced in a conductor depends ? Mention the
relation between the emf. and such factors. 2) State Faraday's law of the emf induced in a coil, then show how to verify this practically?3) What is meant by mutual induction between two coils? and what is meant by the
coefficient of mutual induction? How - using the mutual induction - one can verifyLenz's rule?
4) If a current passes through a coil, deduce an equation relating the induced emf in the coiland the rate of change of the current in the coil. From this, deduce a definition for thecoefficent of self induction and the Henry.
5) When does the emf induced in a coil become maximum ? and when does it become zero?6) Explain an experiment to show the conversion of the mechanical energy into electrical
energy, and another experiment to show the opposite conversion. Then, state the ruleused to define the direction of the current in the first case and the direction of motion inthe second case.
7) Deduce the relation by which one can evaluate the instantaneous emf induced in an ACgenerator.
8) What are the modifications introduced to the AC generator to render it a unidrectionalgenerator ?
9) Describe the structure of the electric transformer ? then explain the principle of itsopertion. What is meant by saying that the efficiency of the transformer is 80%?.
10) What is meant by the efficiency of the transformer? What are the factors which lowersuch an efficiency and how to deal with them?
11) Draw a labelled diagram showing the structure of the motor and explain its operation.
272
Uni
t 4 :
Elec
trici
ty D
ynam
ic E
lictri
city
and
Ele
ctro
mag
netis
m
Cha
pter
11:
Ele
ctro
mag
netic
Indu
ctio
n 5) An antenna of length one meter fixed in a motor car, which moves at velocity80km/hour in a direction perpendicular to the horizontal component of the Earth’smagnetic field. An emf of 4 x 10-4 V is induced in the antenna. In such a case,calculate the magnetic flux density of the considerd horizontal field.
(18 x 10-6T)6) Calculate the coefficient of self-induction for a coil in which an emf of 10 V is
induced if the passing current changes at a rate of 40 A/s (0.25 Henry)
7) The mutual induction between two faces of opposite coils is 0.1 Henry and theintensity of current in one of them is 4 A. If this intensity drops to zero in 0.01s, findthe emf induced in the other coil.
(40V)8) A rectangular coil of dimensions 0.4m x 0.2m and of 100 turns rotates with a uniform
velocity 500 revolutions per minute in a uniform field of magnetic flux density 0.1Tesla. The axis of rotation in the plane of the coil is perpendicular to the field.Calculate the emf induced in the coil.
(41.89 V.)9) A step-down transformer of efficiency 90% has a primary coil voltage of 200 V and
that of the secondary is 9 V. If the intensity of the electric current in the primary is 0.5A, and the number of turns of the secondary is 90 turns, what is the intensity of thecurrent of the secondary coil, and what is the number of turns of the primary?
(10 A, 1800 turns )10) A step-down transformer connected to an AC power source of 2500 V gives a
current of 80 A.The ratio between the number of turns of the primary and thesecondary coils is 20:1 Assuming that its efficiency is 80%, find the emf inducedacross the two terminals of the secondary, and find also the current in the primary coil.
(100V,4A)
275
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
OverviewAll what we have studied so far can be lumped under the title of classical physics. By
classical, we do not mean outdated or obsolete. In fact, classical physics explainseverything in our daily life and our common experiences. The present unit, however,entails some of the basic concepts of modern physics and a general view of a quantumphysics. This branch of physics (modern or quantum) deals with a great collection ofscientific phenomena which might not be directly observed in our daily life, but treat anumber of situations in the universe which classical physics cannot explain, especiallywhen we deal with atomic and subatomic systems, i.e, down to the subatomic scale .
Also, this kind of physics explains all phenomena involved in electronics which isthe basis for all modern electronic and communication systems. It also explainschemical reactions on the level of the molecule. Some of such reactions werephotographed by Ahmed Zewail using a high speed laser camera. Such work entitledhim to earn the Noble Prize in chemistry in1999.
Blackbody Radiation:We are content so far to regard light as waves. Waves have common features, i.e,
UV visible light
wavelength (m)
Fig (12-1)Electromagnetic spectrum
Chapter 12 Wave Particle Duality
276
Un
it 5
:
Intr
od
uct
ion
to
Mo
de
rn P
hys
ics
C
ha
pte
r 1
2:
W
ave
Pa
rtic
le D
ua
lity
reflection, refraction, interference and diffraction. We know also that visible light is but a
small portion of the electromagnetic (em) spectrum (Fig 12-1). Electromagnetic waves may
differ in frequency, and hence, in wavelength, but they propagate in free space at a constant
speed c = 3 x 108 m/s. Electromagnetic waves do not need necessarily a medium to
propagate in. We all observe that hot bodies emit light and heat. An example is the Sun (Fig
Fig (12-2)The Sun as a source
of em radiationFig (12-3)
A burnig charcoal emitsem radiation
Fig (12-4a)A glowing incandescentlamp emits em radiation
Fig (12-4b)A lamp emitting less
em radiation
277
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
wave length (nm)nanometer)
Fig(12-5)The wavelength at the peak is
inversely proportional to temperature
UV visable light IR
12-2) and other stars, a burning charcoal (Fig 12-3) and a glowing incandescent lamp (Fig
12-4). We also note that the dominant color of light emitted from these sources varies.
Hence,an em source does not emit all wavelengths equally, but the intensity of radiation
varies with wavelength. The distribution of the radiation intensity with wavelength is called
Planck’s distribution (Fig 12-5). It was also found that the wavelength λm at which the peak
of the curve occurs is inversely proportional to temperature. This is known as Wien’s law.
Therefore, the higher the temperature,
the smaller the wavelength of the peak .
We also note that as the wavelength
tends to infinity (very large)or to zero
(very small ) the intensity of radiation
tends to zero. For example, the
temperature at the surface of the Sun is
6000˚K. Hence, the wavelength at the
peak is 5000˚A ( 0.5 µm). This is within
the visible range. Thus, almost 40% of
the total energy emitted by the Sun is in
the visible range and almost 50% is heat
(infrared radiation), while the rest is distributed over the remaining spectrum. We
practically obtain the same shape of radiation intensity distribution for a glowing
incandescent lamp, except that the temperature is now 3000˚K which puts the wavelength
at the peak at 1000 nm = 10-6 m = 10000˚A = 1 Micron. From such lamps we get nearly 20
% as visible light and most of the rest as heat.We cannot explain these observations using classical physics. It can be argued from
Rad
iati
on i
nten
sity
276
Un
it 5
:
Intr
od
uct
ion
to
Mo
de
rn P
hys
ics
C
ha
pte
r 1
2:
W
ave
Pa
rtic
le D
ua
lity
reflection, refraction, interference and diffraction. We know also that visible light is but a
small portion of the electromagnetic (em) spectrum (Fig 12-1). Electromagnetic waves may
differ in frequency, and hence, in wavelength, but they propagate in free space at a constant
speed c = 3 x 108 m/s. Electromagnetic waves do not need necessarily a medium to
propagate in. We all observe that hot bodies emit light and heat. An example is the Sun (Fig
Fig (12-2)The Sun as a source
of em radiationFig (12-3)
A burnig charcoal emitsem radiation
Fig (12-4a)A glowing incandescentlamp emits em radiation
Fig (12-4b)A lamp emitting less
em radiation
277
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
wave length (nm)nanometer)
Fig(12-5)The wavelength at the peak is
inversely proportional to temperature
UV visable light IR
12-2) and other stars, a burning charcoal (Fig 12-3) and a glowing incandescent lamp (Fig
12-4). We also note that the dominant color of light emitted from these sources varies.
Hence,an em source does not emit all wavelengths equally, but the intensity of radiation
varies with wavelength. The distribution of the radiation intensity with wavelength is called
Planck’s distribution (Fig 12-5). It was also found that the wavelength λm at which the peak
of the curve occurs is inversely proportional to temperature. This is known as Wien’s law.
Therefore, the higher the temperature,
the smaller the wavelength of the peak .
We also note that as the wavelength
tends to infinity (very large)or to zero
(very small ) the intensity of radiation
tends to zero. For example, the
temperature at the surface of the Sun is
6000˚K. Hence, the wavelength at the
peak is 5000˚A ( 0.5 µm). This is within
the visible range. Thus, almost 40% of
the total energy emitted by the Sun is in
the visible range and almost 50% is heat
(infrared radiation), while the rest is distributed over the remaining spectrum. We
practically obtain the same shape of radiation intensity distribution for a glowing
incandescent lamp, except that the temperature is now 3000˚K which puts the wavelength
at the peak at 1000 nm = 10-6 m = 10000˚A = 1 Micron. From such lamps we get nearly 20
% as visible light and most of the rest as heat.We cannot explain these observations using classical physics. It can be argued from
Rad
iati
on i
nten
sity
279
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
tomography (tumor detection) (Fig 12-11), inembryology and in criminology, since the heat radiatedfrom a person lingers for a while even after the personhas left. All these applications are called remotesensing. Egypt has been a pioneer in this field. Howcan we explain the bell shape of radiation? Planck in1900 came up with the answer.
Planck called this phenomenon black body radiation.
The reason for naming it so is that a black body absorbs
all radiation falling on it, regardless of the wavelength.
It is, thus, a perfect absorber. It then re-emits this
radiation wholly. It is therefore a perfect emitter.
If we imagine an enclosed cavity with a small hole,
the inside of the cavity appears black because all of the
radiation within the cavity remains trapped due to multiple reflections. Only a small part
of it leaks out, which is called blackbody radiation (Fig 12-12).
Planck managed to explain this blackbody phenomenon with an interpretation that
Fig (12-10)An image taken by a night vision system
Fig (12-9)A night vision system
Fig (12-8)An image of southern Sinaitaken by Land sat satellites
278
Un
it 5
:
Intr
od
uct
ion
to
Mo
de
rn P
hys
ics
C
ha
pte
r 1
2:
W
ave
Pa
rtic
le D
ua
lity classical physics that since the radiation is an em wave, the intensity of radiation increases
with frequency. Why then should the intensity of radiation go down at the high frequencyend, (Fig 12-6)? This curve is repeated forall hot bodies which emit continuousradiation not only the Sun but also theEarth, and all bodies even living creatures.But the Earth- being a non glowing body -it absorbs the radiation from the Sun andreemits it. But its temperature is far lessthan that of the Sun. Therefore, we find thewavelength at the peak to be nearly 10Micron ,which is within the infrared region(Fig 12-7).There are satellites, and airborneas well as terrestrial equipment which mapand photograph the surface of the Earth,using different regions of the spectrumincluding the infrared radiation emitted bythe surface of the Earth ,in addition to thereflected visible light (Fig 12-8). Also,microwaves are used for the same purposein radars. Scientists analyze such images todetermine possible natural Earth resources.This technique is also used for military purposes such as night vision systems, whichdetect and image moving objects in the dark due to the heat radiation which these objectsre- emit (Figs 12-9,12-10). Thermal imaging is also used in medicine, particularly in
λ µm
Fig (12-6)Radiation decreases with increasing
frequency in disagreement withclassical expectations
rad
iati
on i
nte
nsi
tyPlanck's
distribution
classicalexpectation
Fig (12-7)Radiation from the Earth
and from the Sun
em radiationfrom the Sun
em radiationfrom the Earthra
dia
tion
in
ten
sity
279
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
tomography (tumor detection) (Fig 12-11), inembryology and in criminology, since the heat radiatedfrom a person lingers for a while even after the personhas left. All these applications are called remotesensing. Egypt has been a pioneer in this field. Howcan we explain the bell shape of radiation? Planck in1900 came up with the answer.
Planck called this phenomenon black body radiation.
The reason for naming it so is that a black body absorbs
all radiation falling on it, regardless of the wavelength.
It is, thus, a perfect absorber. It then re-emits this
radiation wholly. It is therefore a perfect emitter.
If we imagine an enclosed cavity with a small hole,
the inside of the cavity appears black because all of the
radiation within the cavity remains trapped due to multiple reflections. Only a small part
of it leaks out, which is called blackbody radiation (Fig 12-12).
Planck managed to explain this blackbody phenomenon with an interpretation that
Fig (12-10)An image taken by a night vision system
Fig (12-9)A night vision system
Fig (12-8)An image of southern Sinaitaken by Land sat satellites
281
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
(a)
(d)
(b)
(e)
(c)
(f)Fig (12-13)
An image where each shot has a differentnumber of photons in increasing order from(a) to(f)
image taken for an object for different numbers of photons. It is worth mentioning though
that the human eye is so sensitive that it can detect as little as one photon falling on it.
Photoelectric Effect:A metal contains positive ions and free electrons which can move around inside the
metal but cannot leave it, due to the attractive forces of the surface which may berepresented by a surface potential barrier. But some of these electrons can escape ifgiven enough energy in the form of heat or light (Fig 12-14).
This is the idea behind the cathode ray tube (CRT), which is used in TV andcomputer monitors (Fig 12 -15).
This tube consists of metal surface called the cathode, which is heated by a filament.Electrons are, thus, emitted by the so called electron gun (E-gun). Due to heat, someelectrons may overcome the forces of attraction at the surface. These electrons are thenfreed (liberated) from the metal and are then picked up by the screen, which is connected
Photoelectric Effect and thermoionic effect:
280
Un
it 5
:
Intr
od
uct
ion
to
Mo
de
rn P
hys
ics
C
ha
pte
r 1
2:
W
ave
Pa
rtic
le D
ua
lity sounded weird at the time. He proposed that radiation was
made up of small units (or packets) of energy, each he
called quantum (or photon). Therefore, we may consider
radiation from a glowing object as a flux of emitted
photons. The photons’ energy increases with frequency,
but their number decreases with increasing energy.
The photons emanate from the vibrations of atoms. The
energy of these vibrating atoms is not continuous but
quantized (discrete or discontinuous) into levels. These
energy levels take values E=nhν, where h is Planck's
constant h=6.625x10-34 Js, and ν is the frequency (Hertz
- Hz). The atom does not radiate as long as it remains in
one energy level. But if the vibrating atom shifts from a
high energy level to a lower energy level, it emits a
photon whose energy E = hν. Thus, photons with high
frequency have high energy and those with low frequency
have low energy. Radiation consists
of billions upon billions of these
photons. We do not see separate
photons, but we observe the features
of the stream of photons as a whole.
These features express in, the stream
of photons represent the classical
properties of radiation Fig (12-13) shows an
Fig (12-11)A thermal image for the
face and neck
Fig (12-12a)Radiation inside the cavity istrapped so it appears black
Fig (12-12b)A small part of energy leaks out of the
hole which is called blackbody radiation
281
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
(a)
(d)
(b)
(e)
(c)
(f)Fig (12-13)
An image where each shot has a differentnumber of photons in increasing order from(a) to(f)
image taken for an object for different numbers of photons. It is worth mentioning though
that the human eye is so sensitive that it can detect as little as one photon falling on it.
Photoelectric Effect:A metal contains positive ions and free electrons which can move around inside the
metal but cannot leave it, due to the attractive forces of the surface which may berepresented by a surface potential barrier. But some of these electrons can escape ifgiven enough energy in the form of heat or light (Fig 12-14).
This is the idea behind the cathode ray tube (CRT), which is used in TV andcomputer monitors (Fig 12 -15).
This tube consists of metal surface called the cathode, which is heated by a filament.Electrons are, thus, emitted by the so called electron gun (E-gun). Due to heat, someelectrons may overcome the forces of attraction at the surface. These electrons are thenfreed (liberated) from the metal and are then picked up by the screen, which is connected
283
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
Fig (12-14c)A more tightly bound electronneeds higher energy to escape
fluorescent screen
a photon a freedelectron
energy
Fig (12-15)Light spot on a fluorescent screen(emits photons when struck by
electrons)
E-beam
E-gun
cathode grid anode plate Xplate Y lightspot
filamentheater
to a positive pole called the anode, thus causing current in the external circuit. When theelectrons hit the screen, they emit light which varies in intensity from point to point on thefluorescent screen, depending on the intensity of the electrical signal transmitted. Such asignal controls the intensity of the electron beam emitted from the E -gun through a negativegrid in its way.
The E -beam can be controlled by electric or magnetic fields to sweep the screen point by
282
Un
it 5
:
Intr
od
uct
ion
to
Mo
de
rn P
hys
ics
C
ha
pte
r 1
2:
W
ave
Pa
rtic
le D
ua
lity
Fig (12-14a)Electrons may be freed from a metal
if given sufficient energy
Fig (12-14b)Minimum energy needed to free an
electron is called work function
uv radiation
zinc plate
freedelectrons
a photonAn electron barely
escaping
energy
283
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
Fig (12-14c)A more tightly bound electronneeds higher energy to escape
fluorescent screen
a photon a freedelectron
energy
Fig (12-15)Light spot on a fluorescent screen(emits photons when struck by
electrons)
E-beam
E-gun
cathode grid anode plate Xplate Y lightspot
filamentheater
to a positive pole called the anode, thus causing current in the external circuit. When theelectrons hit the screen, they emit light which varies in intensity from point to point on thefluorescent screen, depending on the intensity of the electrical signal transmitted. Such asignal controls the intensity of the electron beam emitted from the E -gun through a negativegrid in its way.
The E -beam can be controlled by electric or magnetic fields to sweep the screen point by
285
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
Einstein put forth an interpretation for all this, which led him to Nobel prize in 1921. Heproposed that a photon with ν > νc falling on a metallic surface , has energy hν , while theenergy needed to free an electron ( called the work function) is Ew = hνc ( Fig 12-14).
Thus, the photon is barely able to free an electron, if it has energy hν=hνc= Ew. If thephoton energy exceeds this limit, the electron is freed and the energy difference
hν -Ew is carried by the electrons as kinetic energy, i.e., it moves faster as hν increases.Whereas if hν<Ew, the electron would not be emitted at all, no matter how intense thelight might be. Also, the emission is instantaneous. There is no need for time to collectenergy. The emission takes place instantly, once hν > hνc .
It is to be noted that νc and Ew vary for different materials, and do not depend on thelight intensity, the exposure time or the voltage difference between the anode and the cathode.
Fig (12-17a)Photocurrent versus light
intensity for ν < νc
ph
otoc
urr
ent
light intensity
ph
otoc
urr
ent
light intensity
Fig (12-17b)Photocurrent versus light
intensity for ν > νc
284
Un
it 5
:
Intr
od
uct
ion
to
Mo
de
rn P
hys
ics
C
ha
pte
r 1
2:
W
ave
Pa
rtic
le D
ua
lity point generating the picture,so called raster until the frame
is completed (Fig 12-15).When light falls on a cold cathode instead of heating a
filament, a current flows in the circuit too. This meanselectrons have been freed due to light. The emission ofsuch electrons due to light falling on a metallic surface is aphenomenon called photoelectric effect (Fig 12-16). Thisphenomenon cannot be explained by the classical theory oflight. Considering light as a wave, part of the light fallingon the surface of the metal is absorbed, giving someelectrons enough energy to escape. We are then up aganistcertain difficulties with the classical theory. If we attemptusing the classical model, the current intensity or theemission of such electrons (called photoelectrons) shoulddepend on the intensity of the incident wave, regardless ofits frequency. Also, the kinetic energy (or velocity) of theemitted electrons should increase with increasing intensityof the incident radiation. Even in the case of low lightintensity, giving sufficient time should give some electronsenough energy to be freed, regardless of the frequency of the incident light. But thepractical observations are contrary to these classical expectations. It has been observedthat the emission of electrons depends primarily on the frequency of the incident light noton its intensity. Such electrons are not emitted if the frequency is under a threshold(critical value) νc no matter how intense light may be. If the frequency exceeds νc,photocurrent increases with the intensity of light (Fig 12-17). Also, the kinetic energy (orvelocity) of the emitted electrons depends on the frequency of the incident wave not itsintensity.In addition, the emission of electrons occurs instantly as long as ν > νc. Theelectrons do not need time to collect energy if the light intensity is low, provided ν > νc .
Fig (12-16)Photoelectric current
achieved by absorbingphotons on a metal surface
incident light
emittedphotoelectrons
vacuum
ammeter
voltmeter
battery
285
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
Einstein put forth an interpretation for all this, which led him to Nobel prize in 1921. Heproposed that a photon with ν > νc falling on a metallic surface , has energy hν , while theenergy needed to free an electron ( called the work function) is Ew = hνc ( Fig 12-14).
Thus, the photon is barely able to free an electron, if it has energy hν=hνc= Ew. If thephoton energy exceeds this limit, the electron is freed and the energy difference
hν -Ew is carried by the electrons as kinetic energy, i.e., it moves faster as hν increases.Whereas if hν<Ew, the electron would not be emitted at all, no matter how intense thelight might be. Also, the emission is instantaneous. There is no need for time to collectenergy. The emission takes place instantly, once hν > hνc .
It is to be noted that νc and Ew vary for different materials, and do not depend on thelight intensity, the exposure time or the voltage difference between the anode and the cathode.
Fig (12-17a)Photocurrent versus light
intensity for ν < νc
ph
otoc
urr
ent
light intensity
ph
otoc
urr
ent
light intensity
Fig (12-17b)Photocurrent versus light
intensity for ν > νc
)Photo electeric cell(
287
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
Thus, Vs serves as a measure for KEmax. If the
photon energy is hν, we have:
hν = Ew + KEmax
KE max = hν - Ew
Thus, KEmax is directly proportional to hν,
regardness of light intensity φL (light flux is the
number of photons/s). If ν becomes νc , we have
hν=Ew, then KEmax = 0 and Vs = 0, i.e., no stopping
voltage is needed to stop the current (Fig 12-19).
constant lightintensity
constantcurrent
Fig (12-19a)Measuring KEmax for different
frequencies at constant photon flux
Fig (12-19b)A linear relation between
KEmax and ν
286
Un
it 5
:
Intr
od
uct
ion
to
Mo
de
rn P
hys
ics
C
ha
pte
r 1
2:
W
ave
Pa
rtic
le D
ua
lity
1e (12-1)
Learn at Leisure
Interpretation of the photo electric effect:
To measure the velocity of photoelectrons(and their KE), we apply a negativeretarding voltage between the anode and thecathode. The magnitude of the voltage.which causes the photocurrent to cease iscalled the stopping voltage Vs. At thisvoltage, electrons barely make it to theanode.Vs is the lowest voltage that does that.The kinetic energy of electrons at the anodebecome zero. The kinetic energy at thecathode which would enable electrons tohardly reach the anode is the maximumkinetic energy at the cathode (most energeticelectrons), and hence, is called the maximumkinetic energy KEmax where e is the electroncharge since the total energy at the cathodeequals the total energy at the anode.
-eVs + KEmax
= 0
Vs = KEmax
ν > νc
constant light of
constantcurrent
stopping voltage
constant light of
Fig (12-18a)A circuit for measuring
photocurrent versus voltage
anodecathode
light
φL2
φL1constantcurrent
Fig (12-18b)Photocurrent versus voltage
for different light intensity and constantfrequency ν > νc
287
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
Thus, Vs serves as a measure for KEmax. If the
photon energy is hν, we have:
hν = Ew + KEmax
KE max = hν - Ew
Thus, KEmax is directly proportional to hν,
regardness of light intensity φL (light flux is the
number of photons/s). If ν becomes νc , we have
hν=Ew, then KEmax = 0 and Vs = 0, i.e., no stopping
voltage is needed to stop the current (Fig 12-19).
constant lightintensity
constantcurrent
Fig (12-19a)Measuring KEmax for different
frequencies at constant photon flux
Fig (12-19b)A linear relation between
KEmax and ν
289
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
Compton Effect
It was observed that when a photon ( X or γ rays ) collided with a free electron, thephoton frequency decreased and changed its direction. Also, the electron velocity increasedand it changed its direction (Fig 12-21). This observation could not be explained by thewave (classical) theory of light. It can be argued based on Planck,s hypothesis thatelectromagnetic radiation consists of photons which can collide with electrons as billiardballs collide. In this collision, linear momentum must be conserved (law of conservation oflinear momentum), i.e., the linear momentum before collision must equal the linearmomentum after collision. Also, the law of conservation of energy must apply, i.e., the thesum of the energy of the photon and the electron after collision must equal the sum of theenergy of the photon and the electron before collision . We must, therefore, consider aphoton as a particle with a linear momentum, i.e., it has mass and velocity as much as theelectron is a particle which has mass and velocity, and hence a linear momentum.
Photon Properties A photon is a concentrated packet of energy which has mass, velocity and linear
momentum. Its energy E = hν, it always moves at the speed of light c regardless of itsfrequency. Einstein showed that mass and energy were equivalent E = mc2.A loss of mass
Fig (12-21)
Compton effect
incident photon
scattered electron electron
scattered photon
288
Un
it 5
:
Intr
od
uct
ion
to
Mo
de
rn P
hys
ics
C
ha
pte
r 1
2:
W
ave
Pa
rtic
le D
ua
lity
Fig (12-20a)Relation of photocurrent
with voltage for material A
Fig (12-20b)Relation of photocurrent with
voltage for material B
high light intensity
low light intensity
low light intensity
high light intensity
Fig (12-20c)Relation between KEmax with ν
for different materials
ν ν ν
289
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
Compton Effect
It was observed that when a photon ( X or γ rays ) collided with a free electron, thephoton frequency decreased and changed its direction. Also, the electron velocity increasedand it changed its direction (Fig 12-21). This observation could not be explained by thewave (classical) theory of light. It can be argued based on Planck,s hypothesis thatelectromagnetic radiation consists of photons which can collide with electrons as billiardballs collide. In this collision, linear momentum must be conserved (law of conservation oflinear momentum), i.e., the linear momentum before collision must equal the linearmomentum after collision. Also, the law of conservation of energy must apply, i.e., the thesum of the energy of the photon and the electron after collision must equal the sum of theenergy of the photon and the electron before collision . We must, therefore, consider aphoton as a particle with a linear momentum, i.e., it has mass and velocity as much as theelectron is a particle which has mass and velocity, and hence a linear momentum.
Photon Properties A photon is a concentrated packet of energy which has mass, velocity and linear
momentum. Its energy E = hν, it always moves at the speed of light c regardless of itsfrequency. Einstein showed that mass and energy were equivalent E = mc2.A loss of mass
Fig (12-21)
Compton effect
incident photon
scattered electron electron
scattered photon
290
Un
it 5
:
Intr
od
uct
ion
to
Mo
de
rn P
hys
ics
C
ha
pte
r 1
2:
W
ave
Pa
rtic
le D
ua
lity
(12-2)2pcPw
c2 ( ) φLhν
=
is translated to released energy as in the atomicbomb (Fig 12-22). Nuclear fission is associatedwith a small loss of mass which is converted tolarge amount of energy due to the c2(c2 = 9 x1016 m2/s2) factor. Therefore, the law ofconservation of mass and the law of theconservation of energy blend into the law ofconservation of mass and energy. Thus, aphoton whose energy is hν has a mass of hν/c2,while in motion . Since it is moving at velocityc, its momentum which is the product of massand velocity becomes hν/c . If a beam ofphotons is incident on a certain surface at the rate of φL (photons/s,), then each photonimpinges on the surface and bounces off, and hence, suffers a change in linear momentum2mc. The force which a beam of photons applies to the surface is the change in linearmomentum per second:
F = 2mcφL
F =
where Pw is the power in watts of the light incident on the surface. This force is toosmall to be noticed. But it is appreciable if it affects a free electron instead, due to its smallmass and size, so it throws it off. This is the explanation of Compton effect. In themicroscopic model, we can image a phaton as a sphere whose radius is roughly equal to λand oscillates at frequency ν. The stream of photons collectively has a magnetic field andan electric field. These two fields are perpendicular to one another and to the direction ofpropagation.Both oscillate at ν. The photon flux (or stream) carries the energy of the wave.The wave properties are ,thus, manifested by the photon stream as a whole. The wave
Fig (12-22)Atomic bomb
292
Un
it 5
:
Intr
od
uct
ion
to
Mo
de
rn P
hys
ics
C
ha
pte
r 1
2:
W
ave
Pa
rtic
le D
ua
lity
eye nasalcavity
ear
earcavityskull
braintissue
Fig (12-23b)A CT scan image of the head
Fig (12-23d)A CT scan image of the
abdomen
Fig (12-23c)An image of blood vessels
using X- rays and a dye
intestines
kidneyspine
lever
295
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
Thus, the wavelength is Planck’s constant divided by the linear momentum PL. It
should be noted that when photons fall on a surface, a comparison is made between λ and
the interatomic distance of the surface. If λ is greater than the interatomic distance, these
photons sense the surface as a continuous one and get reflected from it as in wave theory.
If the interatomic distance is comparable to λ, photons penetrate through the atoms. This
is what happens in the case of X- rays.
ExampleCalculate the photon mass and linear momentum if λ = 380 nm
Solution
= 7.89 x1014 Hz
= 5.81x10-36 kg
= 1.74 x10-27 kgm/s
Wave properties of a particle In the universe, there is a great deal of symmetry. If waves have particle nature, could
it be that particles might have wave properties ? a question posed by De Broglie in 1923led to a hypothesis of wave particle duality applying to particles. The wavelength of aparticle must be in analogy with a photon
λ = h/PL
where pL is the linear momentum of the particle.
(6.625 x 10-34 Js)(380) (1x10-9m)
PL = h/λ =
(3x108 m/s)(380) (1x10-9m)
ν = c/λ =
(6.625 x 10-34 Js) (7.89x1014 s-1)(3x108m/s)2m= E/c2 = hν/c2 =
(12 − 4)
294
Un
it 5
:
Intr
od
uct
ion
to
Mo
de
rn P
hys
ics
C
ha
pte
r 1
2:
W
ave
Pa
rtic
le D
ua
lity MRI
Magnetic Ressonance Imaging (MRI) is another method of tomography. Instead ofusing X- ray with their possibly harmful side effects, radio(RF) waves are used. It has theability to detect tumors, and it depends also on making computerized image slices. Thepatient lies on a moving bed surrounded by a strong (superconductive) magnet. Areceiving detector is used to receiving RF waves from hydrogen nuclei in the body . Thestrong magnet orients the spins of the nuclei, RF waves produced by a source disturb thisspin orientation . Such waves are then stopped. As nuclei relax to their original state, theyemit RF waves which are received by the detector. The computer reconstructs the imagesshowing the concentration of hydrogen, and hence water collections, indicating tumors(Fig 12-24). Superconducting magnets are useful in reducing the heat due to eddy currents.
ExampleCalculate the force applied by a beam of light whose power is 1W on the surface of a wall.
Solution
This force is too diminutive to affect the wall
Relation between photon wavelength and its linear momentumλ = c/ν
Multiplying the numerator by h
λ = =
PL = mc
= c
=
∴ λ =
F = = = 0.67 x 10-8 N2 x 13 x 108
2 Pwc ∴
hchν
hhν/c
hνc2
hνc
(12-3)
hPL
295
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
Thus, the wavelength is Planck’s constant divided by the linear momentum PL. It
should be noted that when photons fall on a surface, a comparison is made between λ and
the interatomic distance of the surface. If λ is greater than the interatomic distance, these
photons sense the surface as a continuous one and get reflected from it as in wave theory.
If the interatomic distance is comparable to λ, photons penetrate through the atoms. This
is what happens in the case of X- rays.
ExampleCalculate the photon mass and linear momentum if λ = 380 nm
Solution
= 7.89 x1014 Hz
= 5.81x10-36 kg
= 1.74 x10-27 kgm/s
Wave properties of a particle In the universe, there is a great deal of symmetry. If waves have particle nature, could
it be that particles might have wave properties ? a question posed by De Broglie in 1923led to a hypothesis of wave particle duality applying to particles. The wavelength of aparticle must be in analogy with a photon
λ = h/PL
where pL is the linear momentum of the particle.
(6.625 x 10-34 Js)(380) (1x10-9m)
PL = h/λ =
(3x108 m/s)(380) (1x10-9m)
ν = c/λ =
(6.625 x 10-34 Js) (7.89x1014 s-1)(3x108m/s)2m= E/c2 = hν/c2 =
(12 − 4)
294
Un
it 5
:
Intr
od
uct
ion
to
Mo
de
rn P
hys
ics
C
ha
pte
r 1
2:
W
ave
Pa
rtic
le D
ua
lity MRI
Magnetic Ressonance Imaging (MRI) is another method of tomography. Instead ofusing X- ray with their possibly harmful side effects, radio(RF) waves are used. It has theability to detect tumors, and it depends also on making computerized image slices. Thepatient lies on a moving bed surrounded by a strong (superconductive) magnet. Areceiving detector is used to receiving RF waves from hydrogen nuclei in the body . Thestrong magnet orients the spins of the nuclei, RF waves produced by a source disturb thisspin orientation . Such waves are then stopped. As nuclei relax to their original state, theyemit RF waves which are received by the detector. The computer reconstructs the imagesshowing the concentration of hydrogen, and hence water collections, indicating tumors(Fig 12-24). Superconducting magnets are useful in reducing the heat due to eddy currents.
ExampleCalculate the force applied by a beam of light whose power is 1W on the surface of a wall.
Solution
This force is too diminutive to affect the wall
Relation between photon wavelength and its linear momentumλ = c/ν
Multiplying the numerator by h
λ = =
PL = mc
= c
=
∴ λ =
F = = = 0.67 x 10-8 N2 x 13 x 108
2 Pwc ∴
hchν
hhν/c
hνc2
hνc
(12-3)
hPL
296
Un
it 5
:
Intr
od
uct
ion
to
Mo
de
rn P
hys
ics
C
ha
pte
r 1
2:
W
ave
Pa
rtic
le D
ua
lity
source ofelectrons
screenFig (12-25b)Calculation of the path difference between two
e-rays through a double slit
double slitsource ofelectrons
screen
intensity distribution
Fig (12-25a)Diffraction of electrons through a double slit
But what does this mean? Weconsider light as a huge stream ofphotons. Photons collectively have awave property accompanying them,thus, manifesting reflection,refraction, interference anddiffraction.The wave intensitydescribes the photon concentractionas if a photon carries the genes oflight, regarding frequency, speed andwavelength. By the same token, an electron ray (e-beam) is a huge stream of electrons.Collectively, there must be a wave accompanying them. An electron carries the genes ofthe stream, regarding charge, spin and linear momentum. The accompanying wave haswavelength λ The intensity of the accompanying wave describes the electronconcentration. Such a wave can disperse, reflect, refract, interfere and diffract, just as lightdoes (Fig 12-25). But does this mean we can use an electron ray as much as we use a lightray in a microscope? The answer is yes. This answer is verified by the discovery of theelectron microscope.
299
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
path which has an integral multiple of λ , or else the
path over which a standing wave is formed (Fig
12-27), or the path over which ψ2 is maximum (Fig
12-26 b).
Electron Microscope
Electron microscope is an important labinstrument which depends in its operation onthe wave nature of electrons. It resembles anoptical microscope in many ways. Theimportant difference is in the resolvingpower. The e-microscope has a highresolving power, because the electrons cancarry a high kinetic energy, and hence veryshort λ (equation 12-3). Thus, itsmagnification is so high that it can detectvery small objects, so small that an ordinaryoptical microscope fails to observe(Fig 12-28).
The optical microscope uses light, whilethe e-microscope uses an electron beam. Theelectron beam might have a wavelength 1000times or more shorter than visible light.
Therefore, the electron-microscope candistinguish fine details. The lenses used ine-microscope are usually magnetic. Theselenses focus the e-beam. Their design fallsunder the topic of electron optics.
Fig (12-27c)An orbit as a standing wave
formed over a string fixed at both ends
opticalmicroscope
electronmicroscope
source
lens
object
objectiveimage
image
projectionlens
screen orphotographic
plate
Fig (12-28a)Electron microscope
298
Un
it 5
:
Intr
od
uct
ion
to
Mo
de
rn P
hys
ics
C
ha
pte
r 1
2:
W
ave
Pa
rtic
le D
ua
lity
with the nucleus is zero (so the electron does not fall on the nucleus or else the universewould vanish). Also, the probability for an electron to exist infinitely away from the nucleusis zero, or else the atom is ionized by itself. Ionization needs external energy. We concludethat the energy of the electron when trapped in an atom is less than that of a free electron bythe magnitude of the ionization energy. Therefore, an electron remains trapped in the atom,unless acted upon by an external stimulus. Heisenberg showed also that we could notdetermine an exact path (orbit) for an electron in an atom. But we can say that an orbit is the
Fig (12-27a)An orbit is an integer of λ as a traveling wave ina closed path whose end coincides with its start
Fig (12-27b)The order of the orbit is
determined by the integer value
Neuclus
299
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
path which has an integral multiple of λ , or else the
path over which a standing wave is formed (Fig
12-27), or the path over which ψ2 is maximum (Fig
12-26 b).
Electron Microscope
Electron microscope is an important labinstrument which depends in its operation onthe wave nature of electrons. It resembles anoptical microscope in many ways. Theimportant difference is in the resolvingpower. The e-microscope has a highresolving power, because the electrons cancarry a high kinetic energy, and hence veryshort λ (equation 12-3). Thus, itsmagnification is so high that it can detectvery small objects, so small that an ordinaryoptical microscope fails to observe(Fig 12-28).
The optical microscope uses light, whilethe e-microscope uses an electron beam. Theelectron beam might have a wavelength 1000times or more shorter than visible light.
Therefore, the electron-microscope candistinguish fine details. The lenses used ine-microscope are usually magnetic. Theselenses focus the e-beam. Their design fallsunder the topic of electron optics.
Fig (12-27c)An orbit as a standing wave
formed over a string fixed at both ends
opticalmicroscope
electronmicroscope
source
lens
object
objectiveimage
image
projectionlens
screen orphotographic
plate
Fig (12-28a)Electron microscope
301
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
Quantum Mechanics
The interpretation of previous observations has paved the way to a new set of laws of
mechanics, namely, Quantum Mechanics. This branch of science is based on the following
assumptions formed by Schrodinger:
1) An electron in an atom has energy values which belong to a set of allowable values
called energy levels. The atom does not emit energy unless it falls from a high level to a
lower level.
2) Such emission is in the form of a photon whose energy hν is equal to the difference
between the two energy levels. This process is called relaxation.
3) The absorption of energy by an atom occurs if the photon energy is exactly equal to the
energy difference between two levels . In this case, the atom is excited by having its
electron move up in energy to the higher level. This process is called excitation .
4) If the photon energy is greater than the ionization energy of the atom, an electron is
totally freed from the parent atom, and the atom becomes ionized (ion).
5) Relaxation and excitation are simultaneous processes. At thermal equilibrium, the atom
is stable due to the balance and simultaneity of these processes.
6) There is a function which is always positive that describes the electron in the atom. This
functions tends to zero at the nucleus and at infinity (at the border of the atom).
Therefore, the electron remains trapped within the atom due to nuclear attraction
without falling onto the nucleus. To explain this, we may say that as the electron draws
near the nucleus, its velocity increases so much that it flies away (back off). It was also
found that the assumptions of quantum mechanics agree with experimental observations
in all cases when electrons are tightly bound in a limited size. However, if the size is
extensive, then classical mechanics may be used.
303
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
In a Nutshell
• Classical physics cannot explain many phenomena, particularly those in which light ( orem radiation) interacts with electrons or atoms.
• Light or any em radiation consists of a huge collection of photons, each photon havingenergy hν, where h is Planck’s constant and ν is the frequency.
• An evidence for photons is the photoelectric effect, where photocurrent depends on theintensity of incident light as long as the frequency is greater than a critical value νc. Butif the frequency is less than νc , no photocurrent flows. The kinetic energy of theelectron freed by the photoelectric effect depends on the frequency not on the lightintensity.
• A photon has a mass, a linear momentum and a constant speed which is the speed oflight. It has a size denoted by the wavelength. If a photon falls on a wall, it applies asmall force on it, but if it falls on an electron, the electron will be thrown off due to itssmall mass and size.
• Compton effect proves the particle nature of photons, where a photon has mass, speedand linear momentum.
• A wave describes the collective behavior of photons.• The wavelength of a photon is Planck’s constant divided by the linear momentum. The
same relation applies to a free particle, where the wavelength describes the wave natureof the particle ,i.e., the wave accompanying the particle.
• The electron microscope proves de Broglie relation for particles. It is used to detectdiminutive particles.
• Quantum mechanics is based on assumptions which agree with experimentalobservations, for the cases when an electron is trapped within a limited confinement.While classical physics applies when the electron is free to move or when the continingsize is extensive.
300
Un
it 5
:
Intr
od
uct
ion
to
Mo
de
rn P
hys
ics
C
ha
pte
r 1
2:
W
ave
Pa
rtic
le D
ua
lity
Fig (12-28b) Head of a fly as seen by
an e-microscopeFig (12-28c)
Uranium atoms as seen by a specialtype of e-microscopes
Fig (12-28d)Rubella virus as seen by an
e-microscope (white spots are on thesurface of the infected cells)
301
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
Quantum Mechanics
The interpretation of previous observations has paved the way to a new set of laws of
mechanics, namely, Quantum Mechanics. This branch of science is based on the following
assumptions formed by Schrodinger:
1) An electron in an atom has energy values which belong to a set of allowable values
called energy levels. The atom does not emit energy unless it falls from a high level to a
lower level.
2) Such emission is in the form of a photon whose energy hν is equal to the difference
between the two energy levels. This process is called relaxation.
3) The absorption of energy by an atom occurs if the photon energy is exactly equal to the
energy difference between two levels . In this case, the atom is excited by having its
electron move up in energy to the higher level. This process is called excitation .
4) If the photon energy is greater than the ionization energy of the atom, an electron is
totally freed from the parent atom, and the atom becomes ionized (ion).
5) Relaxation and excitation are simultaneous processes. At thermal equilibrium, the atom
is stable due to the balance and simultaneity of these processes.
6) There is a function which is always positive that describes the electron in the atom. This
functions tends to zero at the nucleus and at infinity (at the border of the atom).
Therefore, the electron remains trapped within the atom due to nuclear attraction
without falling onto the nucleus. To explain this, we may say that as the electron draws
near the nucleus, its velocity increases so much that it flies away (back off). It was also
found that the assumptions of quantum mechanics agree with experimental observations
in all cases when electrons are tightly bound in a limited size. However, if the size is
extensive, then classical mechanics may be used.
303
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
In a Nutshell
• Classical physics cannot explain many phenomena, particularly those in which light ( orem radiation) interacts with electrons or atoms.
• Light or any em radiation consists of a huge collection of photons, each photon havingenergy hν, where h is Planck’s constant and ν is the frequency.
• An evidence for photons is the photoelectric effect, where photocurrent depends on theintensity of incident light as long as the frequency is greater than a critical value νc. Butif the frequency is less than νc , no photocurrent flows. The kinetic energy of theelectron freed by the photoelectric effect depends on the frequency not on the lightintensity.
• A photon has a mass, a linear momentum and a constant speed which is the speed oflight. It has a size denoted by the wavelength. If a photon falls on a wall, it applies asmall force on it, but if it falls on an electron, the electron will be thrown off due to itssmall mass and size.
• Compton effect proves the particle nature of photons, where a photon has mass, speedand linear momentum.
• A wave describes the collective behavior of photons.• The wavelength of a photon is Planck’s constant divided by the linear momentum. The
same relation applies to a free particle, where the wavelength describes the wave natureof the particle ,i.e., the wave accompanying the particle.
• The electron microscope proves de Broglie relation for particles. It is used to detectdiminutive particles.
• Quantum mechanics is based on assumptions which agree with experimentalobservations, for the cases when an electron is trapped within a limited confinement.While classical physics applies when the electron is free to move or when the continingsize is extensive.
303
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
In a Nutshell
• Classical physics cannot explain many phenomena, particularly those in which light ( orem radiation) interacts with electrons or atoms.
• Light or any em radiation consists of a huge collection of photons, each photon havingenergy hν, where h is Planck’s constant and ν is the frequency.
• An evidence for photons is the photoelectric effect, where photocurrent depends on theintensity of incident light as long as the frequency is greater than a critical value νc. Butif the frequency is less than νc , no photocurrent flows. The kinetic energy of theelectron freed by the photoelectric effect depends on the frequency not on the lightintensity.
• A photon has a mass, a linear momentum and a constant speed which is the speed oflight. It has a size denoted by the wavelength. If a photon falls on a wall, it applies asmall force on it, but if it falls on an electron, the electron will be thrown off due to itssmall mass and size.
• Compton effect proves the particle nature of photons, where a photon has mass, speedand linear momentum.
• A wave describes the collective behavior of photons.• The wavelength of a photon is Planck’s constant divided by the linear momentum. The
same relation applies to a free particle, where the wavelength describes the wave natureof the particle ,i.e., the wave accompanying the particle.
• The electron microscope proves de Broglie relation for particles. It is used to detectdiminutive particles.
• Quantum mechanics is based on assumptions which agree with experimentalobservations, for the cases when an electron is trapped within a limited confinement.While classical physics applies when the electron is free to move or when the continingsize is extensive.
305
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
whose mass is 10 kg , what happens if the object is an electron and why ?
II) Essay questions
1) Show why the wave theory failed to explain the photoelectric effect, and how Einstein
managed to interpret the experimental results of this phenomenon.
2) Show how to verify the particle nature of light from the blackbody radiation .
3) Explain the Compton effect and show how it proves the particle nature of light ?
0.67x10-3N
302
Un
it 5
:
Intr
od
uct
ion
to
Mo
de
rn P
hys
ics
C
ha
pte
r 1
2:
W
ave
Pa
rtic
le D
ua
lity Learn at Leisure
Can you identify the contribution of each of the following scientists to modern physics ?
Schrodinger
Planck Einstein
Heisenberg
de Broglie
Compton
Bohr
Thompson
303
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
In a Nutshell
• Classical physics cannot explain many phenomena, particularly those in which light ( orem radiation) interacts with electrons or atoms.
• Light or any em radiation consists of a huge collection of photons, each photon havingenergy hν, where h is Planck’s constant and ν is the frequency.
• An evidence for photons is the photoelectric effect, where photocurrent depends on theintensity of incident light as long as the frequency is greater than a critical value νc. Butif the frequency is less than νc , no photocurrent flows. The kinetic energy of theelectron freed by the photoelectric effect depends on the frequency not on the lightintensity.
• A photon has a mass, a linear momentum and a constant speed which is the speed oflight. It has a size denoted by the wavelength. If a photon falls on a wall, it applies asmall force on it, but if it falls on an electron, the electron will be thrown off due to itssmall mass and size.
• Compton effect proves the particle nature of photons, where a photon has mass, speedand linear momentum.
• A wave describes the collective behavior of photons.• The wavelength of a photon is Planck’s constant divided by the linear momentum. The
same relation applies to a free particle, where the wavelength describes the wave natureof the particle ,i.e., the wave accompanying the particle.
• The electron microscope proves de Broglie relation for particles. It is used to detectdiminutive particles.
• Quantum mechanics is based on assumptions which agree with experimentalobservations, for the cases when an electron is trapped within a limited confinement.While classical physics applies when the electron is free to move or when the continingsize is extensive.
304
Un
it 5
:
Intr
od
uct
ion
to
Mo
de
rn P
hys
ics
C
ha
pte
r 1
2:
W
ave
Pa
rtic
le D
ua
lity Questions and Drills
I)Drills
1) Calculate the energy of a photon whose wavelength is 770 nm and find its mass and
linear momentum?
(2.58x10-19J , 0.29x10-35kg , 0.86x10-27kgm/s)
2) Calculate the mass of an X- ray photon and a γ ray photon if the wavelength of X-ray is
100 nm , and that of γ-ray is 0.05 nm
(mX=2.2 X10-35kg , mγ=4.4x10-32kg)
3) Calculate the wavelength of a ball whose mass is 140 kg which moves at velocity 40
m/s. Also, calculate the wavelength of an electron if it has the same velocity.
(λ=1.18x10-37m , λe=1.8 x10-5m)
4) A radio station emits a wave whose frequency is 92.4 MHz. Calculate the energy of
each photon emitted from this station. Also, calculate the rate of photons φL if the
power of the station is 100 kW.
(E=612.15x10-28J , φL=16.3 x1029 s-1)
5) An electron is under a potential difference 20 kV. Calculate its velocity upon collision
with the anode from the law of conservation of energy. The electron charge is 1.6 x 10-19C,
its mass is 9.1 x 10-31 kg. Then calculate λ and PL.
(v=0.838x108m/s , λ=0.868x10-11m , PL=7.625x10-23kgm/s)
6) If the least distance detected with an electron microscope is 1nm, calculate the velocity
of the electrons and the potential of the anode.
(velocity=0.725x106m/s , V=1.5Volt)
7) Calculate the force by which an e-beam whose power is 100 kW affects an object
305
Unit 5
: Intro
ductio
n to
Modern
Physics C
hapte
r 12: W
ave
Particle
Duality
whose mass is 10 kg , what happens if the object is an electron and why ?
II) Essay questions
1) Show why the wave theory failed to explain the photoelectric effect, and how Einstein
managed to interpret the experimental results of this phenomenon.
2) Show how to verify the particle nature of light from the blackbody radiation .
3) Explain the Compton effect and show how it proves the particle nature of light ?
0.67x10-3N
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
1
304311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
305311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
Ene
rgy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell E
nerg
y
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels 311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
307
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
OverviewThe word atom goes back to a Greek origin, meaning the indivisible. Different models forthe atom have been put forth since then by many great scientists based on manyexperimental evidences.
a) Thompson’s Atom (1898) 1 – After Thompson conducted several experiments leading to the discovery of theelectron and the determination of the ratio for the electron, he put forth a model of asolid positively charged substance in which negative electrons were immersed (Fig13–1).2 – Since the atom is neutral, the sum of the negative charge is equal to the sum of thepositive charge.
b) Rutherford’s Atom (1911)Rutherford performed his well known experiment based on which he formulated a modelfor the structure of the atom and showed that it was not solid.In his experiment, Rutherford bombarded a thin gold plate (10-4cm) with a beam of alphaparticles (4He2) (Fig 13 – 2 a, b).
Atomic SpectraChapter 13
Fig (13-1)Thompson’s
atomic model Rutherford
eme
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell E
ner
gy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
305
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
307313
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Emission of Light from Bohr’s Atom.
1) When hydrogen atoms are stimulated (given energy) not all of them are excited the sameway. Thus, electrons in different atoms move from the first level K (n = 1) to differenthigher levels (n = 2, 3, 4..)
2) Electrons remain in excited levels (or states) only for a short period of time ,calledlifetime (nearly 10-8s), then they revert to the lowest level (ground state).
3) In going down from level E2 to level E1 , the electron emits a photon whose energyhν = E2 - E1 : where ν is the frequency of the photon and its wavelength is :
4) The line spectrum of hydrogen consists of a particular energy value, and hence aparticular frequency.
Fig (13 – 7)Standing waves
Fig (13 – 6)Electron transitions
between atomic levels
cνλ =
313
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Emission of Light from Bohr’s Atom.
1) When hydrogen atoms are stimulated (given energy) not all of them are excited the sameway. Thus, electrons in different atoms move from the first level K (n = 1) to differenthigher levels (n = 2, 3, 4..)
2) Electrons remain in excited levels (or states) only for a short period of time ,calledlifetime (nearly 10-8s), then they revert to the lowest level (ground state).
3) In going down from level E2 to level E1 , the electron emits a photon whose energyhν = E2 - E1 : where ν is the frequency of the photon and its wavelength is :
4) The line spectrum of hydrogen consists of a particular energy value, and hence aparticular frequency.
Fig (13 – 7)Standing waves
Fig (13 – 6)Electron transitions
between atomic levels
cνλ =
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
306312
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
He then added three more postulates:1) If an electron moves from an outer shell of energy E2 to an inner shell of energy E1 (E2 > E1 ), an amount of energy E2 – E1 is released in the form of a photon, whoseenergy hν = E2 – E1, where ν is the frequency of the emitted photon (Fig 13 – 6).
2) The electric (Coulomb’s) forces and mechanical (Newton’s) forces are at work in theatom.3) We can estimate the radius of the shell by considering that the wave accompanying theelectron forms a standing wave (Fig 13 – 7 ).(calculate the orbit radius for n = 1, 2, 3,)
Fig (13-5c)Absorption of photons
ener
gy
emittedphoton
absorbedphoton unabsorbed
photon
Fig (13 - 5d)Emissin of photons
ener
gy
312
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
He then added three more postulates:1) If an electron moves from an outer shell of energy E2 to an inner shell of energy E1 (E2 > E1 ), an amount of energy E2 – E1 is released in the form of a photon, whoseenergy hν = E2 – E1, where ν is the frequency of the emitted photon (Fig 13 – 6).
2) The electric (Coulomb’s) forces and mechanical (Newton’s) forces are at work in theatom.3) We can estimate the radius of the shell by considering that the wave accompanying theelectron forms a standing wave (Fig 13 – 7 ).(calculate the orbit radius for n = 1, 2, 3,)
Fig (13-5c)Absorption of photons
ener
gy
emittedphoton
absorbedphoton unabsorbed
photon
Fig (13 - 5d)Emissin of photons
ener
gy
312
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
He then added three more postulates:1) If an electron moves from an outer shell of energy E2 to an inner shell of energy E1 (E2 > E1 ), an amount of energy E2 – E1 is released in the form of a photon, whoseenergy hν = E2 – E1, where ν is the frequency of the emitted photon (Fig 13 – 6).
2) The electric (Coulomb’s) forces and mechanical (Newton’s) forces are at work in theatom.3) We can estimate the radius of the shell by considering that the wave accompanying theelectron forms a standing wave (Fig 13 – 7 ).(calculate the orbit radius for n = 1, 2, 3,)
Fig (13-5c)Absorption of photons
ener
gy
emittedphoton
absorbedphoton unabsorbed
photon
Fig (13 - 5d)Emissin of photons
ener
gy
313
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Emission of Light from Bohr’s Atom.
1) When hydrogen atoms are stimulated (given energy) not all of them are excited the sameway. Thus, electrons in different atoms move from the first level K (n = 1) to differenthigher levels (n = 2, 3, 4..)
2) Electrons remain in excited levels (or states) only for a short period of time ,calledlifetime (nearly 10-8s), then they revert to the lowest level (ground state).
3) In going down from level E2 to level E1 , the electron emits a photon whose energyhν = E2 - E1 : where ν is the frequency of the photon and its wavelength is :
4) The line spectrum of hydrogen consists of a particular energy value, and hence aparticular frequency.
Fig (13 – 7)Standing waves
Fig (13 – 6)Electron transitions
between atomic levels
cνλ =
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
309314
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
Different series of atomic spectral lines for hydrogen are produced, and are arranged asfollows (Fig 13 – 8):
1) Leyman’s series: where the electron moves down tolevel K (n = 1) from higher levels. This series liesin the ultraviolet range (short wavelengths andhigh frequencies).
2) Balmer’s series: where the electron moves down to
Fig (13 – 8 b)Atomic model for
hydrogen spectrum spect
Fig (13 – 8 a)Atomic spectral series for hydrogen
Pfund’sseries
Brackett’sseries
Paschen’sseries
Balmer’sseries
Lyman’s series
Leyman’sseries
Balmer’sseries
K
ultraviolet
IR
paschen’s
series
P
Q
N
M
L
paschen’sseries
n= n=6
315
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Fig (13 – 9b)Spectrometer schematic
level L (n = 2) from higher levels. This series lies in the visible range. 3) Paschen’s series: where the electron moves down to level M (n = 3) from higher
levels. This series lies in the infrared (IR) range.4) Bracket’s series: where the electron moves down to level N (n = 4) from higher
levels. This series lies in the IR range.5) Pfund’s series: where the electron moves down to level O (n = 5) from higher levels.
This series lies in the far IR and is the longest wavelengths ( the lowest frequencies)among the line spectrum of hydrogen.
SpectrometerTo obtain a pure spectrum, a spectrometer is
used (Fig 13 – 9). It consists of 3 parts :1) a source of rays : a light source in front of
which there is a slit whose width can beadjusted by a screw. This slit is at the focalpoint of a convex lens.
sourceslit
prism
violet
red
yellow
Fig (13 – 9a)Spectrometer
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
308314
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
Different series of atomic spectral lines for hydrogen are produced, and are arranged asfollows (Fig 13 – 8):
1) Leyman’s series: where the electron moves down tolevel K (n = 1) from higher levels. This series liesin the ultraviolet range (short wavelengths andhigh frequencies).
2) Balmer’s series: where the electron moves down to
Fig (13 – 8 b)Atomic model for
hydrogen spectrum spect
Fig (13 – 8 a)Atomic spectral series for hydrogen
Pfund’sseries
Brackett’sseries
Paschen’sseries
Balmer’sseries
Lyman’s series
Leyman’sseries
Balmer’sseries
K
ultraviolet
IR
paschen’s
series
P
Q
N
M
L
paschen’sseries
n= n=6
315
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Fig (13 – 9b)Spectrometer schematic
level L (n = 2) from higher levels. This series lies in the visible range. 3) Paschen’s series: where the electron moves down to level M (n = 3) from higher
levels. This series lies in the infrared (IR) range.4) Bracket’s series: where the electron moves down to level N (n = 4) from higher
levels. This series lies in the IR range.5) Pfund’s series: where the electron moves down to level O (n = 5) from higher levels.
This series lies in the far IR and is the longest wavelengths ( the lowest frequencies)among the line spectrum of hydrogen.
SpectrometerTo obtain a pure spectrum, a spectrometer is
used (Fig 13 – 9). It consists of 3 parts :1) a source of rays : a light source in front of
which there is a slit whose width can beadjusted by a screw. This slit is at the focalpoint of a convex lens.
sourceslit
prism
violet
red
yellow
Fig (13 – 9a)Spectrometer
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
311316
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
2) a turntable on which a prism is placed.3) a telescope consisting of two convex lenses (objective and eye
piece).To use the spectrometer for obtaining a pure spectrum, the slit is lit
with bright light falling from the slit onto the prism at the minimumangle of deviation. The telescope is directed to receive the lightpassing through the telescope.The objective focuses the raysbelonging to the same color at the focal plane of the lens . That ishow we obtain a sharp (pure) spectrum.
From studying the line spectra of different elements whose atoms are excited , wenotice different types of spectra (continuous and line) :- the spectrum consisting of all wavelengths in a continuous manner is called the
continuous spectrum.- the spectrum occurring at specified frequencies and not continuously distributed is called
the line spectrum. Alternatively, they may be divided as emission and absorption spectra :- the spectrum resulting from the transfer of excited atoms from a high level to lower level
Fraunhofer
Fig (13-9c)Use of a spectrometer to measure the temperature
of the stars and their gases
317
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
is called the emission spectrum.- It was found experimentally that when white light passes through a certain gas, some
wavelengths in the continuous spectrum are missing. These wavelengths are the same asthose which appear in the emission spectrum of the gas (Fig 13 – 9) . This type ofspectrum is called the absorption spectrum. Fraunhofer lines in the solar spectrum areexamples of the absorption spectrum of the elements in the Sun, basically helium andhydrogen.
X-raysWhat are X-rays ?
They are invisible electromagnetic waves of short wavelength (10-13 – 10-8m) betweenuv and gamma rays. They were first discovered by Rontgen. He called it so (the unknownrays) because he did not know what they were .
Properties of X-rays:- They can penetrate media easily .- They can ionize gases .- They diffract in crystals . - They affect sensitive photographic plates .
Fig (13 – 10)Emission line spectra for some elements
atomichydrogen
barium
mercury
sodium
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
310316
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
2) a turntable on which a prism is placed.3) a telescope consisting of two convex lenses (objective and eye
piece).To use the spectrometer for obtaining a pure spectrum, the slit is lit
with bright light falling from the slit onto the prism at the minimumangle of deviation. The telescope is directed to receive the lightpassing through the telescope.The objective focuses the raysbelonging to the same color at the focal plane of the lens . That ishow we obtain a sharp (pure) spectrum.
From studying the line spectra of different elements whose atoms are excited , wenotice different types of spectra (continuous and line) :- the spectrum consisting of all wavelengths in a continuous manner is called the
continuous spectrum.- the spectrum occurring at specified frequencies and not continuously distributed is called
the line spectrum. Alternatively, they may be divided as emission and absorption spectra :- the spectrum resulting from the transfer of excited atoms from a high level to lower level
Fraunhofer
Fig (13-9c)Use of a spectrometer to measure the temperature
of the stars and their gases
317
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
is called the emission spectrum.- It was found experimentally that when white light passes through a certain gas, some
wavelengths in the continuous spectrum are missing. These wavelengths are the same asthose which appear in the emission spectrum of the gas (Fig 13 – 9) . This type ofspectrum is called the absorption spectrum. Fraunhofer lines in the solar spectrum areexamples of the absorption spectrum of the elements in the Sun, basically helium andhydrogen.
X-raysWhat are X-rays ?
They are invisible electromagnetic waves of short wavelength (10-13 – 10-8m) betweenuv and gamma rays. They were first discovered by Rontgen. He called it so (the unknownrays) because he did not know what they were .
Properties of X-rays:- They can penetrate media easily .- They can ionize gases .- They diffract in crystals . - They affect sensitive photographic plates .
Fig (13 – 10)Emission line spectra for some elements
atomichydrogen
barium
mercury
sodium
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
313318
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
Coolidge TubeThis is used to produce X-rays. When the filament is
heated, electrons are produced and directed at the targetunder the influence of the electric field,which gives themhigh energy, depending on the voltage difference betweenthe target and the hot filament. When an electron collideswith the tungsten target, part- if not all- of its energy isconverted to X-rays (Fig 13 – 11).
Spectrum of X-raysAnalyzing a beam of X-rays generated from a target to
components of different wavelengths, we find that the
spectrum consists of two parts :
a) the continuous spectrum of all wavelengths (within a
certain range) regardless of the target material.
b) the line spectrum corresponding to certain wavelengths
characteristic of the target material, called the
characteristic X-ray radiation.
Interpretation of X-ray generation
a) characteristic radiation
The line spectrum is generated when an electron collides with an electron close to thenucleus of the target material atom. If the latter electron receives sufficient energy, itjumps to a higher level, or leaves the atom altogether, and is replaced by another electron
Fig(13 –11)Coolidge tube
cooling fins
copper rod
vacuum tube
high dcvoltage
X rays hotfilament
heating source
targe
t
target
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
312318
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
Coolidge TubeThis is used to produce X-rays. When the filament is
heated, electrons are produced and directed at the targetunder the influence of the electric field,which gives themhigh energy, depending on the voltage difference betweenthe target and the hot filament. When an electron collideswith the tungsten target, part- if not all- of its energy isconverted to X-rays (Fig 13 – 11).
Spectrum of X-raysAnalyzing a beam of X-rays generated from a target to
components of different wavelengths, we find that the
spectrum consists of two parts :
a) the continuous spectrum of all wavelengths (within a
certain range) regardless of the target material.
b) the line spectrum corresponding to certain wavelengths
characteristic of the target material, called the
characteristic X-ray radiation.
Interpretation of X-ray generation
a) characteristic radiation
The line spectrum is generated when an electron collides with an electron close to thenucleus of the target material atom. If the latter electron receives sufficient energy, itjumps to a higher level, or leaves the atom altogether, and is replaced by another electron
Fig(13 –11)Coolidge tube
cooling fins
copper rod
vacuum tube
high dcvoltage
X rays hotfilament
heating source
targe
t
target
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
315320
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
Fig (13 – 13)Use of X-rays in studying crystals
energy continually due to the braking effect of the surrounding electrons, giving rise to
electromagnetic radiation covering all different possible wavelengths, since the electron
loses energy gradually. This is the origin of the continuous radiation of X-rays.
Important Applications of X-rays1) One of the important features of X-rays is diffraction, as they penetrate materials. That
is why X-rays are used in studying the crystalline structure of materials (Fig 13 – 13).The atoms in the crystal act as a diffraction grating (which is a generalization of
diffraction from a double slit). Bright and dark fringes form, depending on the
difference in the optical path.
X-ray tube anode
high DCvoltage
X-rays
crystal
vacuum
aperture
filamentheatingsource
321
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
2) X- rays have a great penetrating power.
This is why they are used to detect defects in metallic
structures.3) X- rays are used in imaging bones and fractures and
some other medical diagnosis (Fig 13 – 14).
Fig (13 – 14) An X-ray image for the chest
rib spinal cord
chest heart lung
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
314320
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
Fig (13 – 13)Use of X-rays in studying crystals
energy continually due to the braking effect of the surrounding electrons, giving rise to
electromagnetic radiation covering all different possible wavelengths, since the electron
loses energy gradually. This is the origin of the continuous radiation of X-rays.
Important Applications of X-rays1) One of the important features of X-rays is diffraction, as they penetrate materials. That
is why X-rays are used in studying the crystalline structure of materials (Fig 13 – 13).The atoms in the crystal act as a diffraction grating (which is a generalization of
diffraction from a double slit). Bright and dark fringes form, depending on the
difference in the optical path.
X-ray tube anode
high DCvoltage
X-rays
crystal
vacuum
aperture
filamentheatingsource
321
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
2) X- rays have a great penetrating power.
This is why they are used to detect defects in metallic
structures.3) X- rays are used in imaging bones and fractures and
some other medical diagnosis (Fig 13 – 14).
Fig (13 – 14) An X-ray image for the chest
rib spinal cord
chest heart lung
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
317
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
316322
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
In a Nutshell
·• Bohr’s postulates and model of the hydrogen atom : When an electron jumps from a high level to a lower level, it produces radiation in the
form of a photon of frequency ν and energy hν, which is equal to the differencebetween the two levels
hν = E2 – E1, E2 > E1 .
·• The line spectrum of hydrogen consists of 5 series. Each line corresponds to a definiteenergy difference, frequency and wavelength
Lyman uvBalmer visiblePaschen IR (infrared)Brackett IRPfund far IR
·• The spectrometer is an apparatus used to decompose light to its components (visibleand invisible)
·• X-rays are an invisible radiation of short wavelengths, first discovered by Rontgen(1895).He called them the unknown (X) rays
·• X-ray diffraction is used in studying the crystalline structure, and also in the industrialand medical applications.
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
1
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
319325
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
Fig (14-1)Spontaneous emission
Excitation by absorption ofenergy from an external source
Relaxation to a lower level after alifetime and release of excitation
energy
Chapter 14 Laser
OverviewRarely has any discovery left an impact on applied science as the discovery of laser has
done. Soon after its discovery, laser has been introduced into optics, biology, chemistry,medicine and engineering especially communications.
The word laser is an acronym for Light Amplification by Stimulated Emission ofRadiation. In 1960, Maiman built the first laser out of chromium-doped Ruby. Later,He-Ne laser was manufactured along with other types of lasers.
Spontaneous Emission and Stimulated EmissionThe atom has energy levels, the lowest of which is called ground state(E1) in which the
atom initially exists. The atom may be excited to one of higher states E2, E3 etc.
If we shine a photon with energy hν = E2– E1 on the atom, the atom absorbs this photon
and gets excited to E2. Soon enough after a lifetime (nearly 10-8 s), the atom gets rid of this
excitation energy in the form of a photon and goes back to its original state (Fig 14 -1).
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
319
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
321326
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
This type of radiation is called spontaneous radiation. It is the type of radiation common
in ordinary light sources. The emitted photon has the same frequency and energy as the
photon that caused the excitation . But the phase and direction are arbitrary. In 1917,
Einstein showed that in addition to spontaneous radiation, there is another type of
radiation, called stimulated emission (the dominant emission in lasers). If a photon of
energy E2-E1 falls on an excited atom at level E2 before the lifetime is over, this photon
pushes the atom back to the ground state, and hence, the atom radiates the excitation
energy in the form of a photon of the same frequency, phase and direction of the falling
photon. (Fig 14–2).
Thus, throughout stimulated radiation, there are two types of photons; the stimulatingand the stimulated photons moving together at the same frequency, phase and direction.
The emission of photons from the atoms of the material in this way renders thesephotons coherent and collimated for long distances. They are highly concentrated, and
Fig (14-2)Stimulated emission
Relaxation to a lower leveldue to an external photonbefore its lifetime is over
A photon passes by an
excited atom
incident photon
327
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
- occurs where an external photonstimulates excited atoms to emitthe energy difference is the formof a photon before the lifetimeinterval is over.
- The emitted photons are monochromatic(single) wavelength.
- The emitted photons are coherentand propagate in one direction as acollimated parallel beam.
- The intensity remains constant over longdistances contrary to the inversesquare law. It has been possible tosend a laser beam to the Moon andreceive it back, without much loseses,despite the long distance involved.Spreading effect is nil and limitedscattering takes place.
- This is the dominant radiation inlaser sources.
- occurs when the atom relaxes froman excited state to a lower state,emitting spontaneously the energydifference in the form of a photonwithout the effect of an externalphoton. It occurs after the lifetimeinterval is over.
- The emitted photons have awide range of wavelengths.
- The emitted photons propagaterandomly
- The intensity of photonsdecreases according to theinverse square law. This iscalled spreading. Whilecollisions with particles iscalled scattering. In ordinarylight sources both spreadingand scattering occur.
- This is the dominant radiationin ordinary light sources.
1
2
3
4
5
Spontaneous emission Stimulated emissions
remain unspread and unscattered, unlike photons emitted spontaneously.The following table gives a comparison between spontaneous and stimulated emissions :
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
320326
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
This type of radiation is called spontaneous radiation. It is the type of radiation common
in ordinary light sources. The emitted photon has the same frequency and energy as the
photon that caused the excitation . But the phase and direction are arbitrary. In 1917,
Einstein showed that in addition to spontaneous radiation, there is another type of
radiation, called stimulated emission (the dominant emission in lasers). If a photon of
energy E2-E1 falls on an excited atom at level E2 before the lifetime is over, this photon
pushes the atom back to the ground state, and hence, the atom radiates the excitation
energy in the form of a photon of the same frequency, phase and direction of the falling
photon. (Fig 14–2).
Thus, throughout stimulated radiation, there are two types of photons; the stimulatingand the stimulated photons moving together at the same frequency, phase and direction.
The emission of photons from the atoms of the material in this way renders thesephotons coherent and collimated for long distances. They are highly concentrated, and
Fig (14-2)Stimulated emission
Relaxation to a lower leveldue to an external photonbefore its lifetime is over
A photon passes by an
excited atom
incident photon
327
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
- occurs where an external photonstimulates excited atoms to emitthe energy difference is the formof a photon before the lifetimeinterval is over.
- The emitted photons are monochromatic(single) wavelength.
- The emitted photons are coherentand propagate in one direction as acollimated parallel beam.
- The intensity remains constant over longdistances contrary to the inversesquare law. It has been possible tosend a laser beam to the Moon andreceive it back, without much loseses,despite the long distance involved.Spreading effect is nil and limitedscattering takes place.
- This is the dominant radiation inlaser sources.
- occurs when the atom relaxes froman excited state to a lower state,emitting spontaneously the energydifference in the form of a photonwithout the effect of an externalphoton. It occurs after the lifetimeinterval is over.
- The emitted photons have awide range of wavelengths.
- The emitted photons propagaterandomly
- The intensity of photonsdecreases according to theinverse square law. This iscalled spreading. Whilecollisions with particles iscalled scattering. In ordinarylight sources both spreadingand scattering occur.
- This is the dominant radiationin ordinary light sources.
1
2
3
4
5
Spontaneous emission Stimulated emissions
remain unspread and unscattered, unlike photons emitted spontaneously.The following table gives a comparison between spontaneous and stimulated emissions :
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
323329
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
an ordinary light source isscattered during propagation
laser light travels in parallel rays for longdistances without much scattering
ordinary light source
Fig (14-4a)Scattering of an ordinary light source and a laser
Fig (14-4b)Launching a laser beam from the Earth to a reflector on the
surface of the Moon, 380000 km away
laser source
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
322328
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
Properties of a laser beam
1) Monochromaticity: Each line in the visible spectrum in ordinary light sources includesa band of wavelengths (this is why the ordinary color appears to have different shades tothe naked eye). The intensity of each wavelength in this band width is shown inFig (14–3a). A laser source emits one spectral line with a very limited bandwidth andthe intensity is concentrated at the wavelength of that spectral line Fig (14–3b), hence itis called monochromatic.
Fig (14-3a)Spectral width for an ordinary
monochromatic light source Fig (14-3b)Spectral width for
a laser source
ligh
t in
tens
ity
ligh
t in
tens
ity
(φL)max
φLφL
λ
329
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
an ordinary light source isscattered during propagation
laser light travels in parallel rays for longdistances without much scattering
ordinary light source
Fig (14-4a)Scattering of an ordinary light source and a laser
Fig (14-4b)Launching a laser beam from the Earth to a reflector on the
surface of the Moon, 380000 km away
laser source
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
325330
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
2) Collimation : In ordinary light sources, the diameter of the emitted light beam increaseswith distance , where in lasers, the diameter stays constant for long distances withoutmuch unscattering. Thus ,energy is transmitted without much losses .
3) Coherence: Photons of ordinary light sources propagate randomly or incoherently.They emanate at different instants of time, and have inconsistent and varying phase. Inlasers, however, photons emanate coherently both in time and place, since they comeout together at the same time sequence, and maintains the same phase differencethroughout, during propagation over long distances. This makes radiation intense andfocused.
4) Intensity: Light produced by ordinary sources is subject to the inverse square law, since theintensity of radiation falling on unit area decreases, the further away from the light source,due to spreading (Fig 14-4a). The laser rays falling on a unit surface are unspread. Theymaintain a constant intensity and are not subject to the inverse squarelaw.
Fig (14-4c)Measuring astronomical
distances by a laser beam
Fig (14-4d)Measuring the distance between the Moonand the Earth by the reflection of a laser
beam from a reflector on the lunar surface
331
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
Theory of the Laser Action:
Laser action depends on driving the atoms or molecules of the active medium into a
state of population inversion, while maintaining a form of dynamic equilibrium. In this
state, the number of atoms in the excited state exceeds the number of atoms in the lower
The intensity of ordinary lightdecreases with distance from the
source due to the inverse square law
incoherent light
coherent light
Fig (14-5)Coherence
Fig (14-6)Laser light maintains the same
intensity as it propagates
low lightintensityhigh light
intensity
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
324330
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
2) Collimation : In ordinary light sources, the diameter of the emitted light beam increaseswith distance , where in lasers, the diameter stays constant for long distances withoutmuch unscattering. Thus ,energy is transmitted without much losses .
3) Coherence: Photons of ordinary light sources propagate randomly or incoherently.They emanate at different instants of time, and have inconsistent and varying phase. Inlasers, however, photons emanate coherently both in time and place, since they comeout together at the same time sequence, and maintains the same phase differencethroughout, during propagation over long distances. This makes radiation intense andfocused.
4) Intensity: Light produced by ordinary sources is subject to the inverse square law, since theintensity of radiation falling on unit area decreases, the further away from the light source,due to spreading (Fig 14-4a). The laser rays falling on a unit surface are unspread. Theymaintain a constant intensity and are not subject to the inverse squarelaw.
Fig (14-4c)Measuring astronomical
distances by a laser beam
Fig (14-4d)Measuring the distance between the Moonand the Earth by the reflection of a laser
beam from a reflector on the lunar surface
331
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
Theory of the Laser Action:
Laser action depends on driving the atoms or molecules of the active medium into a
state of population inversion, while maintaining a form of dynamic equilibrium. In this
state, the number of atoms in the excited state exceeds the number of atoms in the lower
The intensity of ordinary lightdecreases with distance from the
source due to the inverse square law
incoherent light
coherent light
Fig (14-5)Coherence
Fig (14-6)Laser light maintains the same
intensity as it propagates
low lightintensityhigh light
intensity
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
327332
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
state. Thus, when stimulated emission occurs, it will be amplified as photons are increased
in number going back and forth in the active medium, due to multiple reflections between
two enclosing mirrors. In so doing, more and more excited atoms are poised to generate
stimulated emission, which is further amplified and so on. This is the origin of
amplification of the laser (Fig 14–7), called laser action .
Main Components of a Laser
Despite the variations in size, type and frequency, three common elements must exist in
any laser:
1) Active medium: This can be a crystalline solid (e.g. ruby), semiconductor (chapter15) a
liquid dye, gas atoms (e.g. He – Ne laser), ionized gases (e.g. Argon laser), or
molecular gases (e.g. CO2 laser).
2) Sources of energy responsible for exciting the active medium as follows :
(a) excitation by electrical energy, either by using radio frequency (RF) waves or by
using electric discharge under high DC voltage gas lasers:( HeNe – Ar – CO2).
(b) excitation by optical energy, also known as optical pumping, which can be done
either by flash lamps (e.g. in ruby laser) or using a laser beam as a source of energy
(liquid dye laser).
(c) thermal excitation, by using the thermal effects resulting from the kinetic energy of
gases to excite the active material (e.g. in He-Ne laser).
(d) excitation by chemical energy as chemical reactions between giving gases energy to
stimulate atoms toward lasing (e.g. the reaction between hydrogen and fluorine or
the reaction between Deuterium fluoride and CO2 ) .
333
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
3) Resonant cavity is the container and the activating catalyst for amplification . It can be
one of two types :
(a) external resonant cavity in the form of two parallel mirrors enclosing the active
medium permitting multiple reflections leading to amplification as in gas lasers
(Fig 14 – 7a).
b) internal resonant cavity where the ends of the active material are polished so as to act
as mirrors as in ruby laser (Fig 14 – 7b). One of the two mirrors is semitransparent
to allow some of the laser radiation to leak out (Fig 14 – 8).
two reflectingmirrors
Fig (14-7a)External resonant cavity
the two polished ends ofthe active medium act as
mirrors
Fig (14-7b)Internal resonant cavity
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
326332
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
state. Thus, when stimulated emission occurs, it will be amplified as photons are increased
in number going back and forth in the active medium, due to multiple reflections between
two enclosing mirrors. In so doing, more and more excited atoms are poised to generate
stimulated emission, which is further amplified and so on. This is the origin of
amplification of the laser (Fig 14–7), called laser action .
Main Components of a Laser
Despite the variations in size, type and frequency, three common elements must exist in
any laser:
1) Active medium: This can be a crystalline solid (e.g. ruby), semiconductor (chapter15) a
liquid dye, gas atoms (e.g. He – Ne laser), ionized gases (e.g. Argon laser), or
molecular gases (e.g. CO2 laser).
2) Sources of energy responsible for exciting the active medium as follows :
(a) excitation by electrical energy, either by using radio frequency (RF) waves or by
using electric discharge under high DC voltage gas lasers:( HeNe – Ar – CO2).
(b) excitation by optical energy, also known as optical pumping, which can be done
either by flash lamps (e.g. in ruby laser) or using a laser beam as a source of energy
(liquid dye laser).
(c) thermal excitation, by using the thermal effects resulting from the kinetic energy of
gases to excite the active material (e.g. in He-Ne laser).
(d) excitation by chemical energy as chemical reactions between giving gases energy to
stimulate atoms toward lasing (e.g. the reaction between hydrogen and fluorine or
the reaction between Deuterium fluoride and CO2 ) .
333
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
3) Resonant cavity is the container and the activating catalyst for amplification . It can be
one of two types :
(a) external resonant cavity in the form of two parallel mirrors enclosing the active
medium permitting multiple reflections leading to amplification as in gas lasers
(Fig 14 – 7a).
b) internal resonant cavity where the ends of the active material are polished so as to act
as mirrors as in ruby laser (Fig 14 – 7b). One of the two mirrors is semitransparent
to allow some of the laser radiation to leak out (Fig 14 – 8).
two reflectingmirrors
Fig (14-7a)External resonant cavity
the two polished ends ofthe active medium act as
mirrors
Fig (14-7b)Internal resonant cavity
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
329334
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
a
b
c
d
e
Fig (14-8a)Stimulated emission by an external photon
relaxing electron
emitted photon
incident photon
after excitation
incident photon
excitedelectron
before excitation
higher level
ground statenormal condition
inverted population
a photon approacheswhich causes excitation
stimulated emission is generated
recurrence of stimulated emission
Fig (14-8b)Laser action
335
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
unexcited condition
excited condition
metastable state
incident photon hν =E2 - E1
incident photon
emitted photon
Fig (14-8c)Population inversion through a third
metastable state
a
b
c
d
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
328334
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
a
b
c
d
e
Fig (14-8a)Stimulated emission by an external photon
relaxing electron
emitted photon
incident photon
after excitation
incident photon
excitedelectron
before excitation
higher level
ground statenormal condition
inverted population
a photon approacheswhich causes excitation
stimulated emission is generated
recurrence of stimulated emission
Fig (14-8b)Laser action
335
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
unexcited condition
excited condition
metastable state
incident photon hν =E2 - E1
incident photon
emitted photon
Fig (14-8c)Population inversion through a third
metastable state
a
b
c
d
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
331336
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser mirror semitransparent mirror
mirror glass tube excited atomsemitransparent
mirror
Fig (14-8e)Amplification by
multiple reflections
Fig (14-8f)Output radiation from the semitransparent mirror
Fig (14-8d)Multiple reflections between the two mirrors
337
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
Helium – Neon (He – Ne) laserThese two elements have been selected due to the
near equality of the values of the same metastable
excited energy levels in these two elements.
(a) Construction of He-Ne laser :1) A quartz tube including a mixture of helium and
neon in the ratio 10 : 1 at a low pressure of nearly 0.6
mm Hg (Fig 14 – 9).
2) At both ends of the tube there are two plane or concave
parallel mirrors which are perpendicular to the tube axis.
One has a reflection coefficient of nearly 99.5%, while
the other mirror is semitransparent with a reflection
coefficient of 98%.
3) High frequency electric field feeding the tube from the
outside to excite the helium and neon atoms, or a high
DC voltage difference inside the tube causing electric
discharge.
(b) Operation :1) The voltage difference inside the tube leads to the excitation of the helium atoms to
higher levels (Fig 14 – 10).
2) The excited helium atoms collide with the unexcited neon atoms inelastic collisions.
Thus, energy is transferred from the excited helium atoms to the neon atoms due to the
near equality of the excited levels in both atoms. Neon atoms are, thus, excited.
Fig (14-9a)He – Ne laser schematic
mirror laser beamvacuum tube
Fig (14-9b)He – Ne laser
window
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
330336
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser mirror semitransparent mirror
mirror glass tube excited atomsemitransparent
mirror
Fig (14-8e)Amplification by
multiple reflections
Fig (14-8f)Output radiation from the semitransparent mirror
Fig (14-8d)Multiple reflections between the two mirrors
337
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
Helium – Neon (He – Ne) laserThese two elements have been selected due to the
near equality of the values of the same metastable
excited energy levels in these two elements.
(a) Construction of He-Ne laser :1) A quartz tube including a mixture of helium and
neon in the ratio 10 : 1 at a low pressure of nearly 0.6
mm Hg (Fig 14 – 9).
2) At both ends of the tube there are two plane or concave
parallel mirrors which are perpendicular to the tube axis.
One has a reflection coefficient of nearly 99.5%, while
the other mirror is semitransparent with a reflection
coefficient of 98%.
3) High frequency electric field feeding the tube from the
outside to excite the helium and neon atoms, or a high
DC voltage difference inside the tube causing electric
discharge.
(b) Operation :1) The voltage difference inside the tube leads to the excitation of the helium atoms to
higher levels (Fig 14 – 10).
2) The excited helium atoms collide with the unexcited neon atoms inelastic collisions.
Thus, energy is transferred from the excited helium atoms to the neon atoms due to the
near equality of the excited levels in both atoms. Neon atoms are, thus, excited.
Fig (14-9a)He – Ne laser schematic
mirror laser beamvacuum tube
Fig (14-9b)He – Ne laser
window
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
333338
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
3) An accumulation of excited neon atoms
ensues. The excited level of a neon atom has arelatively long lifetime (nearly 10-3s). Such alevel is called metastable state. Hence,population inversion occurs in neon atoms.
4) A group of neon atoms that are excited relaxto a lower excited state. In so doing, they emitspontaneous photons, which have energyequal to the difference in energy levels. Then,photons propagate randomly in all directionsinside the tube.
5) Photons which propagate along the axis of thetube are reflected back by one of the two
He Ne
collisionsof atoms
rapiddecay rapid
decay
collisions with thewalls of the container
excitation ofelectrons
ener
gy
Fig (14-10b)Transitions between energy levels in He-Ne laser
ground state
Fig (14-10a)He – Ne laser energy levels
metastable state
laserbeam
rapiddecay
excitationby
collisions
energytransfer
bycollisionsbetweenHe and
Ne atoms
groundstateNeHe
ener
gy
energy
339
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
mirrors in its way, they bounce off inside the tube and cannot get out.
6) During the propagation of these photons inside the tube between the two mirrors, they
may well collide with some neon atoms in the excited metastable state, well before
lifetime is over. Thus, they stimulate the neon atoms to emit photons of the same energy
and direction as the colliding photon. Thus, the number of photons moving inside the
tube multiplies.
7) The new stream of photons repeat the excursion, and thus, they remultiply by the lasing
action. This is how amplification takes place .
8) When radiation inside the tube reaches a certain level, we let it out partially through the
semitransparent mirror, while the rest of the radiation remains trapped inside the tube.
The stimulated emission and the lasing action go on.
9) As to the neon atoms which have relaxed to a lower level, soon enough they lose further
whatever left of their energy in different forms, and finally go back to the ground state.
Helium atoms collide again with neon atoms, and the cycle repeats.
10) As to the helium atoms which have lost their energy by collision, they regain energy
through the electric discharge and so on.
Laser applicationsToday there are different types and sizes of lasers. Laser light covers different regions of
the electromagnetic spectrum from visible to uv and IR. Some laser systems can focus a
laser beam in a small spot, where energy might get so high as to melt - and even evaporate
- iron, or pierce diamond. There are lasers which may have enough energy to destroy
missiles and planes in what is termed Star War. Some applications of lasers also include
holography (Fig 14 - 11) and medical applications.
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
332338
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
3) An accumulation of excited neon atoms
ensues. The excited level of a neon atom has arelatively long lifetime (nearly 10-3s). Such alevel is called metastable state. Hence,population inversion occurs in neon atoms.
4) A group of neon atoms that are excited relaxto a lower excited state. In so doing, they emitspontaneous photons, which have energyequal to the difference in energy levels. Then,photons propagate randomly in all directionsinside the tube.
5) Photons which propagate along the axis of thetube are reflected back by one of the two
He Ne
collisionsof atoms
rapiddecay rapid
decay
collisions with thewalls of the container
excitation ofelectrons
ener
gy
Fig (14-10b)Transitions between energy levels in He-Ne laser
ground state
Fig (14-10a)He – Ne laser energy levels
metastable state
laserbeam
rapiddecay
excitationby
collisions
energytransfer
bycollisionsbetweenHe and
Ne atoms
groundstateNeHe
ener
gy
energy
339
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
mirrors in its way, they bounce off inside the tube and cannot get out.
6) During the propagation of these photons inside the tube between the two mirrors, they
may well collide with some neon atoms in the excited metastable state, well before
lifetime is over. Thus, they stimulate the neon atoms to emit photons of the same energy
and direction as the colliding photon. Thus, the number of photons moving inside the
tube multiplies.
7) The new stream of photons repeat the excursion, and thus, they remultiply by the lasing
action. This is how amplification takes place .
8) When radiation inside the tube reaches a certain level, we let it out partially through the
semitransparent mirror, while the rest of the radiation remains trapped inside the tube.
The stimulated emission and the lasing action go on.
9) As to the neon atoms which have relaxed to a lower level, soon enough they lose further
whatever left of their energy in different forms, and finally go back to the ground state.
Helium atoms collide again with neon atoms, and the cycle repeats.
10) As to the helium atoms which have lost their energy by collision, they regain energy
through the electric discharge and so on.
Laser applicationsToday there are different types and sizes of lasers. Laser light covers different regions of
the electromagnetic spectrum from visible to uv and IR. Some laser systems can focus a
laser beam in a small spot, where energy might get so high as to melt - and even evaporate
- iron, or pierce diamond. There are lasers which may have enough energy to destroy
missiles and planes in what is termed Star War. Some applications of lasers also include
holography (Fig 14 - 11) and medical applications.
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
335340
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
a) Holography:Images of objects are formed by collecting rays reflected from them. An image
represents variations in the intensity of light from point to point. But is light intensity all
there is to it in the information about an image? If we have two rays leaving off an
illuminated object at two points on it, there is a difference in intensity alright (proportional
to the square of the amplitude of the wave). But in addition, there is a path difference
between the two lit points and the corresponding points on the photographic plate where
the image is recorded due to the topology of the object. Thus, the waves leaving off the
object carry information in both amplitude and phase (phase difference = x path
difference). The photographic plate records only the intensity (square of the amplitude) and
does not record the phase. That is why a 2D image does not carry the 3D detail. In other
words, a plane image has only half the truth (only the intensity). In 1948, a Hungarian
scientist Gabor (Nobel prize laurette) proposed a method to obtain the component that is
missing from the information in the image and retrieve it from the beam, using another
beam of the same wavelength called the reference beam. A laser beam is split into two
beams. One is used to illuminate the object, and the other is used as the reference beam.
The reflected beam and the reference beam meet at the photographic plate, and interference
takes place. After the photographic plate is developed, resulting interference fringes appear
coded, and we call such an image a hologram. Illuminating a hologram with a laser of the
same wavelength and looking through it with the naked eye, we see an identical 3D image
of the object without using any lenses. The full information (intensity and phase) is now
retrieved due to the coherence nature of the laser. Fig (14 – 12) shows the optical system
used to obtain a hologram using a laser beam. Tens of photos may be stored in one
hologram. We may also obtain 3D images in holograms of moving objects.
2
341
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
Learn at Leisure
Types of hologramsA hologram is a kind of diffraction grating which is a
generalization of a double slit, where interference occurs
between the penetrating waves. To make a hologram, the
object is illuminated with a laser light. The reflected rays
are recorded on the holographic plate, while at the same
time the reference beam is recorded. Interference pattern is
formed and a hologram is developed the same way as a
photographic plate. To read a hologram, a laser beam is
used in the same direction as the laser beam during
recording. The light rays read out the patterns formed on the
hologram, giving an imge as if the object is seen from that
angle, Looking through the hologram in the direction
opposite to the reference beam, a virtual image is seen as if the object lay behind the
hologram, i.e., on the beam side. We can also see a real image in the direction opposite to
the reference beam, .i.e.,on the same side of the observer, if a screen or smoke is used,
where a 3D image may be formed in space (Fig 14–11). What we described above is the
transmission hologram, which is lit from behind (Fig 14–12). It may be lit also by an
ordinary light source, but the image will have many colors. There are yet other types of
holograms such as the reflection hologram, which is lit up front, in which ordinary light may also
be used. There is also embossed hologram. Like the reflection hologram, it may be lit up front,
i.e., on the observer side, and uses ordinary light. It may be considered as a transmission hologram
Fig (14-11)Hologram generates
a 3D image
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
334340
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
a) Holography:Images of objects are formed by collecting rays reflected from them. An image
represents variations in the intensity of light from point to point. But is light intensity all
there is to it in the information about an image? If we have two rays leaving off an
illuminated object at two points on it, there is a difference in intensity alright (proportional
to the square of the amplitude of the wave). But in addition, there is a path difference
between the two lit points and the corresponding points on the photographic plate where
the image is recorded due to the topology of the object. Thus, the waves leaving off the
object carry information in both amplitude and phase (phase difference = x path
difference). The photographic plate records only the intensity (square of the amplitude) and
does not record the phase. That is why a 2D image does not carry the 3D detail. In other
words, a plane image has only half the truth (only the intensity). In 1948, a Hungarian
scientist Gabor (Nobel prize laurette) proposed a method to obtain the component that is
missing from the information in the image and retrieve it from the beam, using another
beam of the same wavelength called the reference beam. A laser beam is split into two
beams. One is used to illuminate the object, and the other is used as the reference beam.
The reflected beam and the reference beam meet at the photographic plate, and interference
takes place. After the photographic plate is developed, resulting interference fringes appear
coded, and we call such an image a hologram. Illuminating a hologram with a laser of the
same wavelength and looking through it with the naked eye, we see an identical 3D image
of the object without using any lenses. The full information (intensity and phase) is now
retrieved due to the coherence nature of the laser. Fig (14 – 12) shows the optical system
used to obtain a hologram using a laser beam. Tens of photos may be stored in one
hologram. We may also obtain 3D images in holograms of moving objects.
2
341
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
Learn at Leisure
Types of hologramsA hologram is a kind of diffraction grating which is a
generalization of a double slit, where interference occurs
between the penetrating waves. To make a hologram, the
object is illuminated with a laser light. The reflected rays
are recorded on the holographic plate, while at the same
time the reference beam is recorded. Interference pattern is
formed and a hologram is developed the same way as a
photographic plate. To read a hologram, a laser beam is
used in the same direction as the laser beam during
recording. The light rays read out the patterns formed on the
hologram, giving an imge as if the object is seen from that
angle, Looking through the hologram in the direction
opposite to the reference beam, a virtual image is seen as if the object lay behind the
hologram, i.e., on the beam side. We can also see a real image in the direction opposite to
the reference beam, .i.e.,on the same side of the observer, if a screen or smoke is used,
where a 3D image may be formed in space (Fig 14–11). What we described above is the
transmission hologram, which is lit from behind (Fig 14–12). It may be lit also by an
ordinary light source, but the image will have many colors. There are yet other types of
holograms such as the reflection hologram, which is lit up front, in which ordinary light may also
be used. There is also embossed hologram. Like the reflection hologram, it may be lit up front,
i.e., on the observer side, and uses ordinary light. It may be considered as a transmission hologram
Fig (14-11)Hologram generates
a 3D image
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
337342
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
with a mirror behind. It is a cheap alternative to a hologram. There is also pulse hologram,which uses powerful laser pulses. Holograms can be made of people and moving objects atsuccessive times, which may lead to the future 3D movies (Fig 14 – 13).
Fig (14-12a)Hologram formation
Fig (14-12b)Hologram as a grating
referencerays
referencerays
mirror
object
hologramreal image virtual image
hologram
Fig (14-13)Successive stationary shots giving
an illusion of motion
343
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
b) Lasers in medicine:The retina contains light sensitive cells. In case of
retinal detachment, part of the retinal loses its function.
Unless quickly treated, the eye may lose sight
completely. In early stages, the eye may be treated by
reconnecting the detached part with the layer underneath.
This used to be a strenuous and delicate operation.
Nowadays, lasers are used for that purpose Fig (14 – 14).
The operation takes less time and efford than before. The
thermal heat from the laser cauterizes the points of
detachment (endothermy). Lasers are also used to treat
cases of far and near sightedness, so the patient can
dispose with glasses Fig (14-15). Other applications of
lasers in medicine include endoscopy, where lasers with
optical fibers are used for diagnosis and even operative
surgery (Fig 14-16).
Other applications of laserc) communications , where optical fibers carry information - loaded laser beam instead of
a wire carring electrical signals.
d) industry, particularly fine industries.
e) military applications include precision guidance , smart bombs and laser radar
(LADAR).
f) CD recording (Fig 14 – 17).
Fig (14-14) Use of a laser beam in
treating retinal detachment
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
336342
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
with a mirror behind. It is a cheap alternative to a hologram. There is also pulse hologram,which uses powerful laser pulses. Holograms can be made of people and moving objects atsuccessive times, which may lead to the future 3D movies (Fig 14 – 13).
Fig (14-12a)Hologram formation
Fig (14-12b)Hologram as a grating
referencerays
referencerays
mirror
object
hologramreal image virtual image
hologram
Fig (14-13)Successive stationary shots giving
an illusion of motion
343
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
b) Lasers in medicine:The retina contains light sensitive cells. In case of
retinal detachment, part of the retinal loses its function.
Unless quickly treated, the eye may lose sight
completely. In early stages, the eye may be treated by
reconnecting the detached part with the layer underneath.
This used to be a strenuous and delicate operation.
Nowadays, lasers are used for that purpose Fig (14 – 14).
The operation takes less time and efford than before. The
thermal heat from the laser cauterizes the points of
detachment (endothermy). Lasers are also used to treat
cases of far and near sightedness, so the patient can
dispose with glasses Fig (14-15). Other applications of
lasers in medicine include endoscopy, where lasers with
optical fibers are used for diagnosis and even operative
surgery (Fig 14-16).
Other applications of laserc) communications , where optical fibers carry information - loaded laser beam instead of
a wire carring electrical signals.
d) industry, particularly fine industries.
e) military applications include precision guidance , smart bombs and laser radar
(LADAR).
f) CD recording (Fig 14 – 17).
Fig (14-14) Use of a laser beam in
treating retinal detachment
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
339344
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
g) laser printing where a laser beam is used to carry information
from the computer to a drum coated by a photosensitive
material. A toner is used to print off from the drum onto paper
(Fig 14-18).
h) arts and laser shows (Fig 14-19).
i) surveying (determining dimensions and areas).
j) space research.
Fig (14-15)Cornea treatment by a
laser
345
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
Fig (14-16)Use of lasers in endoscopy
Fig (14-17a)Use of a laser in writing on CDs
pit
pitless arealaser
lensback side
mirror
lensdetector
glass plate
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
338344
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
g) laser printing where a laser beam is used to carry information
from the computer to a drum coated by a photosensitive
material. A toner is used to print off from the drum onto paper
(Fig 14-18).
h) arts and laser shows (Fig 14-19).
i) surveying (determining dimensions and areas).
j) space research.
Fig (14-15)Cornea treatment by a
laser
345
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
Fig (14-16)Use of lasers in endoscopy
Fig (14-17a)Use of a laser in writing on CDs
pit
pitless arealaser
lensback side
mirror
lensdetector
glass plate
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
341346
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
Fig (14-17b)CDs
Fig (14-18)Use of a laser in printing
(Fig (14-19)Laser show
scanning with laser drum covered with alight sensitive material
informationstart
laser
light intensitycontroller
347
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
In a Nutshell• Spontaneous emission:
It is the emission from one excited atom as it relaxes from a high energy level to a low
energy level after its lifetime interval is over and under no external stimulus.
• Stimulated emission:
It is the emission from one excited atom as a result of a collision with an external
photon, which has the same energy as the one that caused it to be excited. Photons at the
end , come out in coherence ,i.e.,having the same phase, (direction and frequency).
• Properties of a laser beam :
1 ) spectral purity (monochromatic).
2 ) collimation (parallel rays).
3 ) coherence (same phase and direction).
4 ) concentration (high intensity and small diameter).
• Laser action :
1) the active medium must be in the state of population inversion .
2) emission of radiation for the excited atom through the stimulated emission.
3) amplification of stimulated emission through the resonant cavity
• Basic elements of a laser :
1 ) an active medium.
2 ) a source of energy (pumping).
3 ) a resonant cavity.
• He - Ne laser is a gas laser:
in which the active medium is a mixture of helium and neon in the ratio 10 : 1
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
340346
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
Fig (14-17b)CDs
Fig (14-18)Use of a laser in printing
(Fig (14-19)Laser show
scanning with laser drum covered with alight sensitive material
informationstart
laser
light intensitycontroller
347
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
In a Nutshell• Spontaneous emission:
It is the emission from one excited atom as it relaxes from a high energy level to a low
energy level after its lifetime interval is over and under no external stimulus.
• Stimulated emission:
It is the emission from one excited atom as a result of a collision with an external
photon, which has the same energy as the one that caused it to be excited. Photons at the
end , come out in coherence ,i.e.,having the same phase, (direction and frequency).
• Properties of a laser beam :
1 ) spectral purity (monochromatic).
2 ) collimation (parallel rays).
3 ) coherence (same phase and direction).
4 ) concentration (high intensity and small diameter).
• Laser action :
1) the active medium must be in the state of population inversion .
2) emission of radiation for the excited atom through the stimulated emission.
3) amplification of stimulated emission through the resonant cavity
• Basic elements of a laser :
1 ) an active medium.
2 ) a source of energy (pumping).
3 ) a resonant cavity.
• He - Ne laser is a gas laser:
in which the active medium is a mixture of helium and neon in the ratio 10 : 1
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
343348
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
• Laser applications:
1 ) 3D photography (holography).2 ) medicine (e.g. treating retinal detachment).3 ) communications.4 ) industry.5 ) military applications.6) CD recording7) printing8) arts and shows9) surveying10) space research
349
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
Questions and Drills
1- What is meant by laser ?2- Compare between spontaneous emission and stimulated emission operation - wise and
feature - wise. 3- Laser light has special characteristics which distinguish it from ordinary light .
Discuss this statement . 4- Discuss clearly the laser action .5- What is meant by optical pumping and population inversion?6- What is the role of the resonant cavity in laser operation ?7- Lasers have 3 main components, what are they ?8- On what basis have helium and neon been chosen as an active medium in He - Ne
laser ?9- What is the role of helium in He - Ne laser ? 10- Explain clearly how a laser beam is generated in He - Ne laser .11- Explain how holography works using lasers . 12- Lasers are used extensively in medicine. Discuss one of its applications . 13- Lasers play an important role in missile guidance in modern warfare. Why is laser
used as such?
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
342348
Unit 5:
m
od
ern
ph
ysic
s C
hapte
r 14:
Laser
• Laser applications:
1 ) 3D photography (holography).2 ) medicine (e.g. treating retinal detachment).3 ) communications.4 ) industry.5 ) military applications.6) CD recording7) printing8) arts and shows9) surveying10) space research
349
Un
it 5
: m
odern
physic
s C
ha
pte
r 1
4:
L
ase
r
Questions and Drills
1- What is meant by laser ?2- Compare between spontaneous emission and stimulated emission operation - wise and
feature - wise. 3- Laser light has special characteristics which distinguish it from ordinary light .
Discuss this statement . 4- Discuss clearly the laser action .5- What is meant by optical pumping and population inversion?6- What is the role of the resonant cavity in laser operation ?7- Lasers have 3 main components, what are they ?8- On what basis have helium and neon been chosen as an active medium in He - Ne
laser ?9- What is the role of helium in He - Ne laser ? 10- Explain clearly how a laser beam is generated in He - Ne laser .11- Explain how holography works using lasers . 12- Lasers are used extensively in medicine. Discuss one of its applications . 13- Lasers play an important role in missile guidance in modern warfare. Why is laser
used as such?
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
1
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
345351
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Overview:The world witnesses a tremendous mushrooming in the field of electronics and
communication to the point where they have become an insignia for this era. Electronics
and communication are now indispensible in our life. TV, cellular (mobile) phone ,
computer, satellites and other systems are evidences for the vast progress in the
applications of electronics and communications, whether in business, e-government,
information technology(IT), entertainment or culture. They have become also an essential
ingredient in modern warfare. Weapons do not fare from the point of view of fire power
only, but guidance, surveillance, monitoring jamming and deception ,called electronic
counter measures (or ECM) play an important role in combat. Also, in medicine whether in
diagnois, prognosis, or operations, electronics plays a key role. In short, there is no single
field in all walks of life where electronics has no part, starting from e –games to e– warfare.
Therefore, you must attain a certain level of awareness about electronics – simplified as it
may be, yet essential regardless of the prospective career you might end up with.
Origin of electronics:The word electronics stems from the electron. Electronics describes the behavior of electrons.
There are two states for an electron: a free electron and a bound electron. The free electron – as
in the case of CRT- is subject to classical physics. A bound electron – however – is subject to
quantum physics. The binding of an electron might be within an atom, a molecule or the bulk of
matter. Matter has different forms : gas, liquid, solid or plasma (when the gases are ionized as
in the fluorescent lamp). Matter in each form consists of molecules. What distinguishes states of
Modern ElectronicsChapter 15
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
345
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
347353
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
level and the excited level. If the electron goes back to a lower level, it emits energy in the
form of a photon. The probability of finding an electron in a particular excited level
decreases as the energy of that level increases. There is a balance between the process of
excitation and the process of relaxation, noting that the electron tends to go back to the
ground state.
Pure Semiconductors:There are three types of materials from the point of view of
electrical conductivity. Conductors conduct electricity and heat
easily (as in metals). Insulators do not conduct electricity and heat
(as in wood and plastics). Semiconductors are in between. At
absolute zero, they act as insulators, whereas as temperature
increases ,their conductivity increases (as in silicon).
Silicon is one of the important and common elements in the
universe. It exists in sand (SiO2) and rocks of the Earth’s crust. But
crystals of pure silicon consist of silicon atoms bound together in covalent bonds. A crystal is a
regular arrangement of atoms in the solid state. A silicon atom has four electrons in the
outermost shell (Fig 15 – 1). Therefore, each silicon atom shares 4 electrons with 4 neighboring
atoms, so that the outer shell of each is complete on sharing basis to contain 8 electrons each
(Fig 15 – 2 a,b). We must distinguish here between two types of electrons in silicon. The first
type is the innermost (tightly bound) electrons, which are strongly attracted to their parents
atoms. The second type is the valence electrons, which have more freedom to move across
interatomic distances. They exist in the outermost shell. At low temperatures (Fig 15 - 2c), all
bonds in the crystal are intact (unbroken).
In this case – unlike metals – there are no free electrons. But as temperature increases,
Fig (15-1)A silicon atom
Core
354
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Fig (15-3b)As temperature increasesmore bonds are broken
Fig (15-3a)Breaking a bond requires
energy
generation ofan electronhole pair
recombinationof an electron
hole pair
thermal energy
freeelectron
thermalenergy
Fig (15-2a)Each atom shareselectrons with its
neighbors
Fig (15-2b)Covalent bonds. We may
represent a Si atom (-14 e)around (+14 e) nucleus as a core
(+4e) and (-4e ) in the outer shell
+4e Core
Fig (15-2c)Silicon crystal at T=0˚K
all bonds are intact
some bonds, are broken and electrons are freed. Such an electron leaves behind a vacancy
in the broken bond.This vacancy is called a hole (Fig 15 – 3). Because the atom is neutral,
then the absence of an electron entails the appearance of a positive charge. We, thus, say
that the hole has a positive charge. We do not call a silicon atom which loses an electron
from its bond an ion, because soon enough, this atom may capture a free electron or an
electron from another bond to fill its own vacancy. Then, the atom returns neutral, and the
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
346352
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
matter apart is the intermolecular distance. In the case of a solid, this distance is very small. In
the case of a gas, this distance is large. In the case of a liquid, it is somewhere in between. If we
consider the solid state, the atoms or molecules of matter get close enough to each other within a
certain distance due to the forces of attraction between them. If we imagine that they are made to
get close, then the forces of repulsion act in to prevent further proximity. Thus, the interatomic
distance represents a point of equilibrium (or balance) between the forces of attraction and the
forces of repulsion among the atoms. It is to be noted that these atoms oscillate around their
equilibrium positions due to heat. But they are separated by space. We cannot see this space by
our naked eye, because the interatomic distance is much smaller than the wavelength of the
photons of visible light to which our eyes are sensitive.
An electron in an atom:An electron in an atom is considered a bound electron. It cannot depart on its own. It
needs energy to do that. This energy is called the ionization energy, i.e., the energy of an
electron in bondage is less than its energy when free by this amount. That is why the
electron remains in bondage in the first place. This energy is called the binding energy. It is
the cause of keeping the atom stable. The electron in the atom has a set of discrete energy
levels according to Bohr’s model. It occupies one of the allowable levels and cannot have
an energy value in between. The electron in the atom is governed by the laws of quantum
mechanics. That is why the probability of having an electron fall onto the nucleus, or
having the electron outside the atom (without external help) is zero. What binds the electron
to the nucleus is the electric force of attraction.
As long as the electron remains in one energy level, it does not gain or lose energy. But if
the electron acquires energy by absorption, it is excited to a higher energy level, provided
the energy absorbed is exactly equal to the energy difference between the original (ground)
353
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
level and the excited level. If the electron goes back to a lower level, it emits energy in the
form of a photon. The probability of finding an electron in a particular excited level
decreases as the energy of that level increases. There is a balance between the process of
excitation and the process of relaxation, noting that the electron tends to go back to the
ground state.
Pure Semiconductors:There are three types of materials from the point of view of
electrical conductivity. Conductors conduct electricity and heat
easily (as in metals). Insulators do not conduct electricity and heat
(as in wood and plastics). Semiconductors are in between. At
absolute zero, they act as insulators, whereas as temperature
increases ,their conductivity increases (as in silicon).
Silicon is one of the important and common elements in the
universe. It exists in sand (SiO2) and rocks of the Earth’s crust. But
crystals of pure silicon consist of silicon atoms bound together in covalent bonds. A crystal is a
regular arrangement of atoms in the solid state. A silicon atom has four electrons in the
outermost shell (Fig 15 – 1). Therefore, each silicon atom shares 4 electrons with 4 neighboring
atoms, so that the outer shell of each is complete on sharing basis to contain 8 electrons each
(Fig 15 – 2 a,b). We must distinguish here between two types of electrons in silicon. The first
type is the innermost (tightly bound) electrons, which are strongly attracted to their parents
atoms. The second type is the valence electrons, which have more freedom to move across
interatomic distances. They exist in the outermost shell. At low temperatures (Fig 15 - 2c), all
bonds in the crystal are intact (unbroken).
In this case – unlike metals – there are no free electrons. But as temperature increases,
Fig (15-1)A silicon atom
Core
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
349355
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
broken bond is mended, and the hole shifts somewhere else, and so on.
As the temperature increases, the number of free electrons and holes increases, noting
that the number of free electrons equals the number of free holes in a pure semiconductor.
But a state of dynamic equilibrium is reached (called thermal equilibrium) at which only a
small percentage of bonds are broken. The number of bonds broken per second will be
equal to the number of bonds mended per second, so that a fixed number of free electrons
and free holes remains constant at every temperature. But not the same electrons and same
holes remain free.They reshuffle ,but their number stays constant.
Free electrons (a class of valance electrons) represent a third type of electrons in silicon.
Such electrons in fact are still confined, but they are confined to the full size of the crystal
itself, i.e., are limited by the so called surface of the crystal. Breaking a bond requires a
minimum energy (thermal or optical). In the case of mending a bond (called
recombination), energy is released (thermal or optical).
As the electrons move in a random motion ,so do the holes, since electrons in the bond move
around randomly to fill in vacancies (voids) within the broken bonds (Fig 15 – 4 ).
Fig (15-4a)Holes move randomly between bonds
356
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Fig (15-4b)Motion of holes is equivalent to motion of
electrons within bonds (in the opposite direction)Fig (15-4c)
At a certain temperature, thenumber of free electrons and
holes is constant
Doping:Semiconductors are known to be sensitive to impurities and to temperature. Since silicon
is tetravalent, the addition of an element as phosphorus (P) or antimony (Sb) or any other
pentavalent element will cause such an impurity atom to replace a silicon atom in the
crystal (Fig 15 – 5 a). Then, the phosphorus atom will try to do the same bonding with the
neighbors as the silicon atom would do.
Fig (15-5a)An antimony atom (pentavalent)
replaces a silicon atom
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
348354
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Fig (15-3b)As temperature increasesmore bonds are broken
Fig (15-3a)Breaking a bond requires
energy
generation ofan electronhole pair
recombinationof an electron
hole pair
thermal energy
freeelectron
thermalenergy
Fig (15-2a)Each atom shareselectrons with its
neighbors
Fig (15-2b)Covalent bonds. We may
represent a Si atom (-14 e)around (+14 e) nucleus as a core
(+4e) and (-4e ) in the outer shell
+4e Core
Fig (15-2c)Silicon crystal at T=0˚K
all bonds are intact
some bonds, are broken and electrons are freed. Such an electron leaves behind a vacancy
in the broken bond.This vacancy is called a hole (Fig 15 – 3). Because the atom is neutral,
then the absence of an electron entails the appearance of a positive charge. We, thus, say
that the hole has a positive charge. We do not call a silicon atom which loses an electron
from its bond an ion, because soon enough, this atom may capture a free electron or an
electron from another bond to fill its own vacancy. Then, the atom returns neutral, and the
355
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
broken bond is mended, and the hole shifts somewhere else, and so on.
As the temperature increases, the number of free electrons and holes increases, noting
that the number of free electrons equals the number of free holes in a pure semiconductor.
But a state of dynamic equilibrium is reached (called thermal equilibrium) at which only a
small percentage of bonds are broken. The number of bonds broken per second will be
equal to the number of bonds mended per second, so that a fixed number of free electrons
and free holes remains constant at every temperature. But not the same electrons and same
holes remain free.They reshuffle ,but their number stays constant.
Free electrons (a class of valance electrons) represent a third type of electrons in silicon.
Such electrons in fact are still confined, but they are confined to the full size of the crystal
itself, i.e., are limited by the so called surface of the crystal. Breaking a bond requires a
minimum energy (thermal or optical). In the case of mending a bond (called
recombination), energy is released (thermal or optical).
As the electrons move in a random motion ,so do the holes, since electrons in the bond move
around randomly to fill in vacancies (voids) within the broken bonds (Fig 15 – 4 ).
Fig (15-4a)Holes move randomly between bonds
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
351357
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Because the impurity atom has 5 electrons, four of them willtake part in the bonding scheme, sparing one valence extra(excess) electron. The force of attraction on the excess electronwhich is left out is weak. Hence, it can easily be detached fromits parent atom, which becomes a positive ion.This extra electronjoins the stock of the free electrons in the crystal . In otherwords, the crystal has an added source of free electrons besidesbroken bonds, namely, impurity atoms. Such impurity atoms arecalled donors (givers). At thermal equilibrium, the sum of thepositive charge equals the sum of the negative charge.
n = p + ND+
where ND+ is the positive donor ion concentration, n is the free electron density and p is
the hole density. In this case, n > p and the material is called n–type. Conversely, if we add
aluminum (Al) or boron (B) or any trivalent element, to
pure silicon, the impurity atom replaces a silicon atom.
Since the impurity atom now has 3 electrons in the
outershell, it detaches an electron from a neighboring
bond to complete its own bond creating an extra hole,
becoming a negative ion. At thermal equilibrium ,
p= NA- + n
where NA- is the negative impurity concentration.
Thus, p>n. Such an atom is called acceptor (taker). In all
cases, we have np = ni
2
(Fig 15-5b)Doping with a pentavalent atomprovides an extra free electron.A pentavalent atom has a core
(+5 e) and 5 electrons
Fig (15-6a)A boron atom replaces
a silicon atom
(15 - 2)
(15 - 1)
(15 - 3)
an excess electron
357
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Because the impurity atom has 5 electrons, four of them willtake part in the bonding scheme, sparing one valence extra(excess) electron. The force of attraction on the excess electronwhich is left out is weak. Hence, it can easily be detached fromits parent atom, which becomes a positive ion.This extra electronjoins the stock of the free electrons in the crystal . In otherwords, the crystal has an added source of free electrons besidesbroken bonds, namely, impurity atoms. Such impurity atoms arecalled donors (givers). At thermal equilibrium, the sum of thepositive charge equals the sum of the negative charge.
n = p + ND+
where ND+ is the positive donor ion concentration, n is the free electron density and p is
the hole density. In this case, n > p and the material is called n–type. Conversely, if we add
aluminum (Al) or boron (B) or any trivalent element, to
pure silicon, the impurity atom replaces a silicon atom.
Since the impurity atom now has 3 electrons in the
outershell, it detaches an electron from a neighboring
bond to complete its own bond creating an extra hole,
becoming a negative ion. At thermal equilibrium ,
p= NA- + n
where NA- is the negative impurity concentration.
Thus, p>n. Such an atom is called acceptor (taker). In all
cases, we have np = ni
2
(Fig 15-5b)Doping with a pentavalent atomprovides an extra free electron.A pentavalent atom has a core
(+5 e) and 5 electrons
Fig (15-6a)A boron atom replaces
a silicon atom
(15 - 2)
(15 - 1)
(15 - 3)
an excess electron
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
350356
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Fig (15-4b)Motion of holes is equivalent to motion of
electrons within bonds (in the opposite direction)Fig (15-4c)
At a certain temperature, thenumber of free electrons and
holes is constant
Doping:Semiconductors are known to be sensitive to impurities and to temperature. Since silicon
is tetravalent, the addition of an element as phosphorus (P) or antimony (Sb) or any other
pentavalent element will cause such an impurity atom to replace a silicon atom in the
crystal (Fig 15 – 5 a). Then, the phosphorus atom will try to do the same bonding with the
neighbors as the silicon atom would do.
Fig (15-5a)An antimony atom (pentavalent)
replaces a silicon atom
357
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Because the impurity atom has 5 electrons, four of them willtake part in the bonding scheme, sparing one valence extra(excess) electron. The force of attraction on the excess electronwhich is left out is weak. Hence, it can easily be detached fromits parent atom, which becomes a positive ion.This extra electronjoins the stock of the free electrons in the crystal . In otherwords, the crystal has an added source of free electrons besidesbroken bonds, namely, impurity atoms. Such impurity atoms arecalled donors (givers). At thermal equilibrium, the sum of thepositive charge equals the sum of the negative charge.
n = p + ND+
where ND+ is the positive donor ion concentration, n is the free electron density and p is
the hole density. In this case, n > p and the material is called n–type. Conversely, if we add
aluminum (Al) or boron (B) or any trivalent element, to
pure silicon, the impurity atom replaces a silicon atom.
Since the impurity atom now has 3 electrons in the
outershell, it detaches an electron from a neighboring
bond to complete its own bond creating an extra hole,
becoming a negative ion. At thermal equilibrium ,
p= NA- + n
where NA- is the negative impurity concentration.
Thus, p>n. Such an atom is called acceptor (taker). In all
cases, we have np = ni
2
(Fig 15-5b)Doping with a pentavalent atomprovides an extra free electron.A pentavalent atom has a core
(+5 e) and 5 electrons
Fig (15-6a)A boron atom replaces
a silicon atom
(15 - 2)
(15 - 1)
(15 - 3)
an excess electron
357
Unit 5: m
odern physics Chapter 15: M
odern Electronics
Because the impurity atom has 5 electrons, four of them willtake part in the bonding scheme, sparing one valence extra(excess) electron. The force of attraction on the excess electronwhich is left out is weak. Hence, it can easily be detached fromits parent atom, which becomes a positive ion.This extra electronjoins the stock of the free electrons in the crystal . In otherwords, the crystal has an added source of free electrons besidesbroken bonds, namely, impurity atoms. Such impurity atoms arecalled donors (givers). At thermal equilibrium, the sum of thepositive charge equals the sum of the negative charge.
n = p + ND+
where ND+ is the positive donor ion concentration, n is the free electron density and p is
the hole density. In this case, n > p and the material is called n–type. Conversely, if we add
aluminum (Al) or boron (B) or any trivalent element, to
pure silicon, the impurity atom replaces a silicon atom.
Since the impurity atom now has 3 electrons in the
outershell, it detaches an electron from a neighboring
bond to complete its own bond creating an extra hole,
becoming a negative ion. At thermal equilibrium ,
p= NA- + n
where NA- is the negative impurity concentration.
Thus, p>n. Such an atom is called acceptor (taker). In all
cases, we have np = ni
2
(Fig 15-5b)Doping with a pentavalent atomprovides an extra free electron.A pentavalent atom has a core
(+5 e) and 5 electrons
Fig (15-6a)A boron atom replaces
a silicon atom
(15 - 2)
(15 - 1)
(15 - 3)
an excess electron
357
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Because the impurity atom has 5 electrons, four of them willtake part in the bonding scheme, sparing one valence extra(excess) electron. The force of attraction on the excess electronwhich is left out is weak. Hence, it can easily be detached fromits parent atom, which becomes a positive ion.This extra electronjoins the stock of the free electrons in the crystal . In otherwords, the crystal has an added source of free electrons besidesbroken bonds, namely, impurity atoms. Such impurity atoms arecalled donors (givers). At thermal equilibrium, the sum of thepositive charge equals the sum of the negative charge.
n = p + ND+
where ND+ is the positive donor ion concentration, n is the free electron density and p is
the hole density. In this case, n > p and the material is called n–type. Conversely, if we add
aluminum (Al) or boron (B) or any trivalent element, to
pure silicon, the impurity atom replaces a silicon atom.
Since the impurity atom now has 3 electrons in the
outershell, it detaches an electron from a neighboring
bond to complete its own bond creating an extra hole,
becoming a negative ion. At thermal equilibrium ,
p= NA- + n
where NA- is the negative impurity concentration.
Thus, p>n. Such an atom is called acceptor (taker). In all
cases, we have np = ni
2
(Fig 15-5b)Doping with a pentavalent atomprovides an extra free electron.A pentavalent atom has a core
(+5 e) and 5 electrons
Fig (15-6a)A boron atom replaces
a silicon atom
(15 - 2)
(15 - 1)
(15 - 3)
an excess electron
357
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Because the impurity atom has 5 electrons, four of them willtake part in the bonding scheme, sparing one valence extra(excess) electron. The force of attraction on the excess electronwhich is left out is weak. Hence, it can easily be detached fromits parent atom, which becomes a positive ion.This extra electronjoins the stock of the free electrons in the crystal . In otherwords, the crystal has an added source of free electrons besidesbroken bonds, namely, impurity atoms. Such impurity atoms arecalled donors (givers). At thermal equilibrium, the sum of thepositive charge equals the sum of the negative charge.
n = p + ND+
where ND+ is the positive donor ion concentration, n is the free electron density and p is
the hole density. In this case, n > p and the material is called n–type. Conversely, if we add
aluminum (Al) or boron (B) or any trivalent element, to
pure silicon, the impurity atom replaces a silicon atom.
Since the impurity atom now has 3 electrons in the
outershell, it detaches an electron from a neighboring
bond to complete its own bond creating an extra hole,
becoming a negative ion. At thermal equilibrium ,
p= NA- + n
where NA- is the negative impurity concentration.
Thus, p>n. Such an atom is called acceptor (taker). In all
cases, we have np = ni
2
(Fig 15-5b)Doping with a pentavalent atomprovides an extra free electron.A pentavalent atom has a core
(+5 e) and 5 electrons
Fig (15-6a)A boron atom replaces
a silicon atom
(15 - 2)
(15 - 1)
(15 - 3)
an excess electron
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
353358
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
where ni is the electron or hole concentration in pure silicon,i.e., if n increases, p decreases and vice versa. This is called lawof mass action. As an approximation, we may say :
in case of n-type
n = ND+
p = ni2/ND
+
In case of p-type,
p = NA-
n = ni2/NA
-
Electronic Components and DevicesElectronic components and devices are the building blocks for all electronic systems (Fig
15 – 7). Some of these components are simple, e.g., resistor (R), inductor (L), capacitor
(C). Some are more complex, such as pn junction (diode), transistor. There are also other
specialized devices, such as optoelectronic and control devices. Semiconductors from
which most of these devices are made are known to be sensitive to environmental
conditions, such as light, heat, pressure, radiation and chemical pollution. That is why they
are used as sensors or means for measuring external stimulii. Using these sensors, we can
measure the intensity of incident light, temperature, pressure, humidity, pollution,
radiation,etc.
Fig (15-6b)Doping with a trivalent atom
provides an extra hole. Atrivalent atom has a core (+3e)
and 3 electrons
(15 - 4)
(15 - 5)
(15 - 6)
(15 - 7)
359
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Fig (15-7a)Resistors
Fig (15-7b)Diodes and transistors
Fig (15-7c)Inductors
Fig (15-7d)Capacitors
Fig (15-7e)Transformers
Fig (15-7f)Switches
Fig (15-7g)A different set of components and devices
(Can you recognize some?)
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
352358
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
where ni is the electron or hole concentration in pure silicon,i.e., if n increases, p decreases and vice versa. This is called lawof mass action. As an approximation, we may say :
in case of n-type
n = ND+
p = ni2/ND
+
In case of p-type,
p = NA-
n = ni2/NA
-
Electronic Components and DevicesElectronic components and devices are the building blocks for all electronic systems (Fig
15 – 7). Some of these components are simple, e.g., resistor (R), inductor (L), capacitor
(C). Some are more complex, such as pn junction (diode), transistor. There are also other
specialized devices, such as optoelectronic and control devices. Semiconductors from
which most of these devices are made are known to be sensitive to environmental
conditions, such as light, heat, pressure, radiation and chemical pollution. That is why they
are used as sensors or means for measuring external stimulii. Using these sensors, we can
measure the intensity of incident light, temperature, pressure, humidity, pollution,
radiation,etc.
Fig (15-6b)Doping with a trivalent atom
provides an extra hole. Atrivalent atom has a core (+3e)
and 3 electrons
(15 - 4)
(15 - 5)
(15 - 6)
(15 - 7)
359
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Fig (15-7a)Resistors
Fig (15-7b)Diodes and transistors
Fig (15-7c)Inductors
Fig (15-7d)Capacitors
Fig (15-7e)Transformers
Fig (15-7f)Switches
Fig (15-7g)A different set of components and devices
(Can you recognize some?)
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
355360
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
pn junction:A pn junction (Fig 15 – 8) consists of an n–type
region and a p-type region. The name pn stands for p-region and n-region not positive and negative. Alsop,n regions are not two regions glued together but ann-material is converted in part to p-material or viceversa. Holes in the p–type region have high concentration, while holes in the n–type regionhave low concentration. Therefore, some holes diffuse from the p-type region to the n–typeregion. Also, some electrons diffuse from the n–type region (high concentration forelectrons) to the p–type region (low concentration forelectrons). Since each region is neutral (the sum ofpositive charge equals the sum of negative charge), thetransfer of some electrons from the n–type regionuncovers an equal number of positive donor ions, andthe transfer of some holes from the p–type regionuncovers an equal number of negative acceptor ions.This results in a middle region composed of positive ionson one side, and negative ions on the other, while noelectrons or holes exist in this region. This region iscalled transition (depletion) region. In such a region, anelectric field is set up, directed from the positive ions tothe negative ions. This electric field causes a driftcurrent to flow in a direction opposite to the diffusioncurrent. At equilibrium, the forward current is balancedwith a reverse current, so that the net current is zero (Fig 15 – 9).If we apply an external voltage such that the p-typeregion is connected to the positive terminal of the battery
Fig (15-8)A pn Junction
Fig (15-9a)Electrons diffuse from
n to p and holes from p to n
Fig (15-9b)Transition region depleted from
electrons and holes, only ions exist
Transition†(depletion) re-
361
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
and the n-type region to the negative terminal of the battery, the field due to the battery isopposite to the internal field.in the transition region, and therefore, it weakens it. If wereverse the battery, then the two fields will aid each other. In the first case (forward bias), anet current will flow, and in the second case (reverse bias) current is almost zero (Fig15-12). The action of the pn junction is like a switch, which is closed in the forwarddirection (conducting) and open (non conducting) in the reverse direction (Fig 15-13). Wecan make sure that the pn diode is functioning by using an ohmmeter, since the diodeshould have a small resistance in the forward direction and a large resistance in the reverse
-region -region
external potential difference
-region -regiontransition region
barriar voltage
-region
Fig (15-10a)Forward Bios
Fig (15-10b)Motion of electrons and holes
due to forward bias
-region
Fig (15-11a)Diode in reverse bias
Fig (15-11b)Motion of electrons and holes
due to reverse bias
Fig (15-12)I - V characteristic in a pn diode
reverse voltageforwardvoltage
curr
ent
-regiontransition regionp-region
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
354360
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
pn junction:A pn junction (Fig 15 – 8) consists of an n–type
region and a p-type region. The name pn stands for p-region and n-region not positive and negative. Alsop,n regions are not two regions glued together but ann-material is converted in part to p-material or viceversa. Holes in the p–type region have high concentration, while holes in the n–type regionhave low concentration. Therefore, some holes diffuse from the p-type region to the n–typeregion. Also, some electrons diffuse from the n–type region (high concentration forelectrons) to the p–type region (low concentration forelectrons). Since each region is neutral (the sum ofpositive charge equals the sum of negative charge), thetransfer of some electrons from the n–type regionuncovers an equal number of positive donor ions, andthe transfer of some holes from the p–type regionuncovers an equal number of negative acceptor ions.This results in a middle region composed of positive ionson one side, and negative ions on the other, while noelectrons or holes exist in this region. This region iscalled transition (depletion) region. In such a region, anelectric field is set up, directed from the positive ions tothe negative ions. This electric field causes a driftcurrent to flow in a direction opposite to the diffusioncurrent. At equilibrium, the forward current is balancedwith a reverse current, so that the net current is zero (Fig 15 – 9).If we apply an external voltage such that the p-typeregion is connected to the positive terminal of the battery
Fig (15-8)A pn Junction
Fig (15-9a)Electrons diffuse from
n to p and holes from p to n
Fig (15-9b)Transition region depleted from
electrons and holes, only ions exist
Transition†(depletion) re-
361
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
and the n-type region to the negative terminal of the battery, the field due to the battery isopposite to the internal field.in the transition region, and therefore, it weakens it. If wereverse the battery, then the two fields will aid each other. In the first case (forward bias), anet current will flow, and in the second case (reverse bias) current is almost zero (Fig15-12). The action of the pn junction is like a switch, which is closed in the forwarddirection (conducting) and open (non conducting) in the reverse direction (Fig 15-13). Wecan make sure that the pn diode is functioning by using an ohmmeter, since the diodeshould have a small resistance in the forward direction and a large resistance in the reverse
-region -region
external potential difference
-region -regiontransition region
barriar voltage
-region
Fig (15-10a)Forward Bios
Fig (15-10b)Motion of electrons and holes
due to forward bias
-region
Fig (15-11a)Diode in reverse bias
Fig (15-11b)Motion of electrons and holes
due to reverse bias
Fig (15-12)I - V characteristic in a pn diode
reverse voltageforwardvoltage
curr
ent
-regiontransition regionp-region
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
357362
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
reverse bias forward bias
cathode
anode
diode
diode
Fig (15-13b)Ideal I-V characteristic
Fig (15-13a)Diode symbol
Fig (15-13c)In forward bias the diode is like a closed switch
Fig (15-13d)In reverse bias the diode is like an open switch
363
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
V
direction. This is in contrast with a linear resistor, where the magnitude of the current is thesame, whether or not the voltage polarity is reversed (symmetrical characteristic).
A pn diode is important in rectification. It is used in charging car batteries, and mobilebatteries, where AC is converted to DC (Fig 15- 14).
Learn at Leisure
How to convert AC to DCTo convert AC to DC several steps may be followed. First a diode may be used as a half wave rectifier (HWR) (Fig 15 - 14a), using a resistor
and a diode (Fig 15-14b). Four diodes may be used in a bridge (Fig 15- 14 c,d) for a full
diode
Ou
tpu
t vo
ltag
e
Fig (15-14a)Waveform of a rectified half wave
Fig (15-14b)A simple half wave rectifier
Fig (15-14d)A full wave rectifier inthe negative half cycle
Fig (15-14c)A full wave rectifier inthe positive half cycle
V
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
356362
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
reverse bias forward bias
cathode
anode
diode
diode
Fig (15-13b)Ideal I-V characteristic
Fig (15-13a)Diode symbol
Fig (15-13c)In forward bias the diode is like a closed switch
Fig (15-13d)In reverse bias the diode is like an open switch
363
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
V
direction. This is in contrast with a linear resistor, where the magnitude of the current is thesame, whether or not the voltage polarity is reversed (symmetrical characteristic).
A pn diode is important in rectification. It is used in charging car batteries, and mobilebatteries, where AC is converted to DC (Fig 15- 14).
Learn at Leisure
How to convert AC to DCTo convert AC to DC several steps may be followed. First a diode may be used as a half wave rectifier (HWR) (Fig 15 - 14a), using a resistor
and a diode (Fig 15-14b). Four diodes may be used in a bridge (Fig 15- 14 c,d) for a full
diode
Ou
tpu
t vo
ltag
e
Fig (15-14a)Waveform of a rectified half wave
Fig (15-14b)A simple half wave rectifier
Fig (15-14d)A full wave rectifier inthe negative half cycle
Fig (15-14c)A full wave rectifier inthe positive half cycle
V
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
359364
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Fig (15-14e)Waveform of a rectified full wave
Fig (15-14f)Waveform of a capacitor input filter
Fig (15-14g)Capacitor input filter
input Output
wave rectifier (FWR) (Fig 15- 14e). Also, we may obtain a nearly constant current (Fig15-14f) by using a capacitor input filter (Fig 15- 14g).
365
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Learn at Leisure
Electronic tuningTo tune up a TV or radio onto a certain station, we need to
adjust the value of a capacitor to set the frequency of thereceiver to the frequency of the selected broadcast station. Thiscondition is called resonance. In modern receivers, the capacitoris replaced by a reverse biased pn diode . The width of thetransition region increases with increasing reverse bias (Fig 15 -15). The increase of the width of the transition region entails anincrease of the fixed ionic charge on both sides of the transitionregion with reverse voltage. This is tantamount to capacitoraction. Thus, we can change the value of the capacitor bycontrolling the revese voltage. This is called electronic tuning(and the device is called a varactor).
Transistor:The transistor was cenceived in 1955 by Bardeen,
Schockley and Brattain. There are many types of transistors,
but we focus here on bipolar junction transistor (BJT), i.e.,
pnp or npn. Such a transistor consists of a p-region followed
by an n-region then a p-region (pnp), or an n-region followed
by a p-region then an n-region (npn) (Fig 15- 16). The three
regions are called emitter (E) -base (B) and collector (C).
Consider an npn transistor. The first junction (np) is forward
biased. The second (pn) junction is reverse biased. In this case, electrons are emitted from the
Bardeen, Schochley and Brattain
Fig (15-15)The width of the transition region
increases with increasing reverse bias
forward bias
reverse bias
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
358364
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Fig (15-14e)Waveform of a rectified full wave
Fig (15-14f)Waveform of a capacitor input filter
Fig (15-14g)Capacitor input filter
input Output
wave rectifier (FWR) (Fig 15- 14e). Also, we may obtain a nearly constant current (Fig15-14f) by using a capacitor input filter (Fig 15- 14g).
365
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Learn at Leisure
Electronic tuningTo tune up a TV or radio onto a certain station, we need to
adjust the value of a capacitor to set the frequency of thereceiver to the frequency of the selected broadcast station. Thiscondition is called resonance. In modern receivers, the capacitoris replaced by a reverse biased pn diode . The width of thetransition region increases with increasing reverse bias (Fig 15 -15). The increase of the width of the transition region entails anincrease of the fixed ionic charge on both sides of the transitionregion with reverse voltage. This is tantamount to capacitoraction. Thus, we can change the value of the capacitor bycontrolling the revese voltage. This is called electronic tuning(and the device is called a varactor).
Transistor:The transistor was cenceived in 1955 by Bardeen,
Schockley and Brattain. There are many types of transistors,
but we focus here on bipolar junction transistor (BJT), i.e.,
pnp or npn. Such a transistor consists of a p-region followed
by an n-region then a p-region (pnp), or an n-region followed
by a p-region then an n-region (npn) (Fig 15- 16). The three
regions are called emitter (E) -base (B) and collector (C).
Consider an npn transistor. The first junction (np) is forward
biased. The second (pn) junction is reverse biased. In this case, electrons are emitted from the
Bardeen, Schochley and Brattain
Fig (15-15)The width of the transition region
increases with increasing reverse bias
forward bias
reverse bias
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
361
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
360366
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
negative emitter (n) diffusing to the base (p), where they wander around in the base untilpicked up by the positive collector (n).A portion of electrons gets recombined with holes. Ifthe emitted electorn current is IE and the portion that reaches the collector Ic is Ic = αe IE,then the protion lost in the base by recombination with holes is IB = (1-αe)IE. This must bethe base current supplying holes to the base to make up for the losses due to therecombination process. Therefore, the ratio of the collector current to the base current is:
β αα
ααe
C
B
e E
e E
e
e
II
I(1- I
= = =−) 1
transistor symbolpnp
Fig (15-16a) A pnp transistor
Fig (15-16b) A pnp transistor
Fig (15-16c)An npn transistor
Fig (15-16d)An npn transistor
(15 - 8)
npn Transistorsymbol
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
363368
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Transistor as a switch:Considering the collector circuit, we have
VCC = VCE + ICRC
where VCC is the collector battery, and vCE is
the voltage difference between the collecttor and
the emitter, IC is the collector current and RC is
the collector resistance. As IC increases, VCE
decreases, until it reaches a value as low as 0.2V
for a high base current. Considering the base as
the input ,the collector as the output and the
emitter as common (ground), we note that as the
input increases, (or positive) the transistor is ON,
and the output decreases and vice versa. The
circuit behaves as an inverter, for positive
voltage in the base (high), current flows in the
collector, and the output voltage is very small
(low). If the base voltage is small (or negative) or
(low). The transistor is OFF and the current in the
collector ceases, and the output voltage on the
collector increases (high). The transistor as such
operates as a switch (Fig 15-18).
Fig (15-18a)Transistor as a switch (ON condition)
Fig (15-18b)Transistor as a switch (OFF condition)
Fig (15-18c)Inverter characteristic
(15-9)
369
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Digital Electronics:All electronic systems deal with natural quantities and convert them to electrical signals.
As an example, a microphone converts sound to an electrical signal. A video camera
converts an image to electrical signals. In TV, the image (video) and sound (audio) are
transformed into electrical signals, then into electromagnetic waves. All this occurs at the
transmitter. At the receiver, the em signal is transformed back into electrical (video and
audio) signals. The electronics which deals with natural quantities is called analog
electronics. A new branch of electronics has developed, namely, digital electronics. In this
case, the electrical signal is not transmitted continuously (all values are allowed), but is
coded, such that the signal is in terms of one of two possible values representing two states
0 or 1. So, if we want to represent 3, it can be written as 112, where subscript 2 denotes the
binary system (not eleven).
3 = 1x20+1x21
as we may express 17 in decimal system as
17 = 7x100+1x101
similarly in the binary system, we use the weights of 20, 21, 22 … instead of 100, 101, 102,… .Thus, each numeral, symbol and alphabet is coded with a binary code. Analog quantities
may be encoded by an analog – digital converter (ADC). At the reciever, digital quantities are
decoded into analog quantities using a digital to analog converter (DAC). Why do all this? In
nature, there are unwanted spurious signals, called electrical noise. Noise is caused by the
random motion of electrons. Electrons are charged particles. As they move randomly, they
cause minute randomly varying currents. These currents interfere with and disturb the
information - bearing signals. We notice that in weak radio stations, noise appears as a hiss,
and in weak TV stations (or with a bad antenna an aerial) noise appears as spots (salt and
pepper). Electrical noise marrs the useful signals, and is difficult to get rid of. In case of digital
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
362368
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Transistor as a switch:Considering the collector circuit, we have
VCC = VCE + ICRC
where VCC is the collector battery, and vCE is
the voltage difference between the collecttor and
the emitter, IC is the collector current and RC is
the collector resistance. As IC increases, VCE
decreases, until it reaches a value as low as 0.2V
for a high base current. Considering the base as
the input ,the collector as the output and the
emitter as common (ground), we note that as the
input increases, (or positive) the transistor is ON,
and the output decreases and vice versa. The
circuit behaves as an inverter, for positive
voltage in the base (high), current flows in the
collector, and the output voltage is very small
(low). If the base voltage is small (or negative) or
(low). The transistor is OFF and the current in the
collector ceases, and the output voltage on the
collector increases (high). The transistor as such
operates as a switch (Fig 15-18).
Fig (15-18a)Transistor as a switch (ON condition)
Fig (15-18b)Transistor as a switch (OFF condition)
Fig (15-18c)Inverter characteristic
(15-9)
369
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Digital Electronics:All electronic systems deal with natural quantities and convert them to electrical signals.
As an example, a microphone converts sound to an electrical signal. A video camera
converts an image to electrical signals. In TV, the image (video) and sound (audio) are
transformed into electrical signals, then into electromagnetic waves. All this occurs at the
transmitter. At the receiver, the em signal is transformed back into electrical (video and
audio) signals. The electronics which deals with natural quantities is called analog
electronics. A new branch of electronics has developed, namely, digital electronics. In this
case, the electrical signal is not transmitted continuously (all values are allowed), but is
coded, such that the signal is in terms of one of two possible values representing two states
0 or 1. So, if we want to represent 3, it can be written as 112, where subscript 2 denotes the
binary system (not eleven).
3 = 1x20+1x21
as we may express 17 in decimal system as
17 = 7x100+1x101
similarly in the binary system, we use the weights of 20, 21, 22 … instead of 100, 101, 102,… .Thus, each numeral, symbol and alphabet is coded with a binary code. Analog quantities
may be encoded by an analog – digital converter (ADC). At the reciever, digital quantities are
decoded into analog quantities using a digital to analog converter (DAC). Why do all this? In
nature, there are unwanted spurious signals, called electrical noise. Noise is caused by the
random motion of electrons. Electrons are charged particles. As they move randomly, they
cause minute randomly varying currents. These currents interfere with and disturb the
information - bearing signals. We notice that in weak radio stations, noise appears as a hiss,
and in weak TV stations (or with a bad antenna an aerial) noise appears as spots (salt and
pepper). Electrical noise marrs the useful signals, and is difficult to get rid of. In case of digital
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
365370
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
electronics, the information does not lie in the absolute value of the signal (which might be
contaminated by noise), but lies in the code in terms of 0 or 1. It does not matter if the valuecorresponding to 0 or to 1 has some noise superimposed on it. What matters is the state (0 or 1).This is the main advantage of digital electronics. For this reason, it has permeated our modernlife extensively, as in cellular (mobile) telephony, digital satellite TV, and CDs. What hasincreased the importance of digital electronics is the advent of the computer. Everythingthat is entered into the computer-whether numbers or letters-must be transformed into abinary code. Even images are divided into small elements, each called a pixel (pictureelement). These too must be encoded. The computer performs all arithmetic and logicoperations using binary (Boolean) algebra. It also stores information in the binary codetemporarily in the RAM (Random Access Memory) or permanently in the hard disk, bymagnetizing in one direction for 0 and in the opposite direction for 1.
Logic Gates:Modern applications of electronics, such as computer circuits and modern communication
systems depend on digital circuits, called logic gates. These are the circuits that perform logic
Fig (15-19a)Not gate symbol
Fig (15-19b)States of a NOT gate
Fig (15-19a)An equivalent drawing for a NOT
gate. When the switch is closed (ON)the lamp is (OFF) and vice versa
input output
371
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
operations, such as inversion (NOT), simultaneity or coincidence (AND) and optionality (OR) asfollows :1) Inverter (NOT Gate) has one input and one output, and has the following truth table:
input output1 00 1
2) AND Gate: has two inputs or more and one output and has the following truth table: input output
00 001 010 011 1
input Ainput B
output
Fig (15-20a)AND gate symbol
Fig (15-20c)An equivalent drawing for an AND gate. The lamp
does not glow until both switches are closed
lamp
Fig (15-20b)States of an AND gate
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
364370
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
electronics, the information does not lie in the absolute value of the signal (which might be
contaminated by noise), but lies in the code in terms of 0 or 1. It does not matter if the valuecorresponding to 0 or to 1 has some noise superimposed on it. What matters is the state (0 or 1).This is the main advantage of digital electronics. For this reason, it has permeated our modernlife extensively, as in cellular (mobile) telephony, digital satellite TV, and CDs. What hasincreased the importance of digital electronics is the advent of the computer. Everythingthat is entered into the computer-whether numbers or letters-must be transformed into abinary code. Even images are divided into small elements, each called a pixel (pictureelement). These too must be encoded. The computer performs all arithmetic and logicoperations using binary (Boolean) algebra. It also stores information in the binary codetemporarily in the RAM (Random Access Memory) or permanently in the hard disk, bymagnetizing in one direction for 0 and in the opposite direction for 1.
Logic Gates:Modern applications of electronics, such as computer circuits and modern communication
systems depend on digital circuits, called logic gates. These are the circuits that perform logic
Fig (15-19a)Not gate symbol
Fig (15-19b)States of a NOT gate
Fig (15-19a)An equivalent drawing for a NOT
gate. When the switch is closed (ON)the lamp is (OFF) and vice versa
input output
371
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
operations, such as inversion (NOT), simultaneity or coincidence (AND) and optionality (OR) asfollows :1) Inverter (NOT Gate) has one input and one output, and has the following truth table:
input output1 00 1
2) AND Gate: has two inputs or more and one output and has the following truth table: input output
00 001 010 011 1
input Ainput B
output
Fig (15-20a)AND gate symbol
Fig (15-20c)An equivalent drawing for an AND gate. The lamp
does not glow until both switches are closed
lamp
Fig (15-20b)States of an AND gate
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
367372
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Fig (15-21c)An equivalent drawing for an OR gate.
One switch need be closed for the lamp to glow
Fig (15-21b)States of OR gate
Fig (15-21a)OR gate symbol
Thus, there is no output (1) unless both inputs are (1) each, i.e., two conditions or moreare met to satisfy an output (1). It can be represented by two switches in series. They bothhave to be closed at the same time for current to flow and the lamp to glow.3) OR Gate has two inputs or more and one output (Fig 15–21). One condition (1) may
suffice to have an output (1) . input Output0 0 001 110 111 1
input Ainput B
output
Lamp
373
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
This can be represented by two switches inparallel, one of them only need be closed to passcurrent.
All operations performed by the computer arebased on these gates and others.
These gates can be implemented by transistors.In this case, the transistor may not be lookedupon as an amplifier but as a switch .
Thus, we can use the transistor as an inverter(NOT gate).
A transistor with more than one emitter maybe used as an AND gate, so that the transistorwill not pass current unless each emitter has positive voltage (1).
Also, we may envision the transistor as an OR gate in the form of a pair of paralleltransistors. If (1) exists at either one of the inputs, one of the transistors conducts and (1)appears at the output .
Transistors are also used in memory circuits, where data ( 0 or 1 ) is retained temporarilyin the RAM and permanently in the hard disk or CD. In a CD, a laser beam engraves a bit ina plastic disk for 1 and no bit for 0. This is the Write process. In a CD drive, a laser beam isused for the Read process (Fig 15 – 22a). A DVD is a modified version of a CD with higherstorage capacity. There are also digital cameras, which convert images to electrical signalspixel by pixel, and store them on a magnetic tape, or download them onto a PC (Fig 15 –23). These cameras use a new technique for handling and transferring electrical charges,namely, charge coupled devices (CCD). This makes the cameras light weighted (portable)and inexpensive. This is the basis for the camcorder, Fax machines and mobile camera.
Fig (15-22a)A CD - drive
Plasticdisk
Collimatingcoil
prism
laser
pits
lens
sensitivelight
detector
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
366372
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Fig (15-21c)An equivalent drawing for an OR gate.
One switch need be closed for the lamp to glow
Fig (15-21b)States of OR gate
Fig (15-21a)OR gate symbol
Thus, there is no output (1) unless both inputs are (1) each, i.e., two conditions or moreare met to satisfy an output (1). It can be represented by two switches in series. They bothhave to be closed at the same time for current to flow and the lamp to glow.3) OR Gate has two inputs or more and one output (Fig 15–21). One condition (1) may
suffice to have an output (1) . input Output0 0 001 110 111 1
input Ainput B
output
Lamp
373
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
This can be represented by two switches inparallel, one of them only need be closed to passcurrent.
All operations performed by the computer arebased on these gates and others.
These gates can be implemented by transistors.In this case, the transistor may not be lookedupon as an amplifier but as a switch .
Thus, we can use the transistor as an inverter(NOT gate).
A transistor with more than one emitter maybe used as an AND gate, so that the transistorwill not pass current unless each emitter has positive voltage (1).
Also, we may envision the transistor as an OR gate in the form of a pair of paralleltransistors. If (1) exists at either one of the inputs, one of the transistors conducts and (1)appears at the output .
Transistors are also used in memory circuits, where data ( 0 or 1 ) is retained temporarilyin the RAM and permanently in the hard disk or CD. In a CD, a laser beam engraves a bit ina plastic disk for 1 and no bit for 0. This is the Write process. In a CD drive, a laser beam isused for the Read process (Fig 15 – 22a). A DVD is a modified version of a CD with higherstorage capacity. There are also digital cameras, which convert images to electrical signalspixel by pixel, and store them on a magnetic tape, or download them onto a PC (Fig 15 –23). These cameras use a new technique for handling and transferring electrical charges,namely, charge coupled devices (CCD). This makes the cameras light weighted (portable)and inexpensive. This is the basis for the camcorder, Fax machines and mobile camera.
Fig (15-22a)A CD - drive
Plasticdisk
Collimatingcoil
prism
laser
pits
lens
sensitivelight
detector
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
369374
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Images may be transferred via the internet using a newfeature called Bluetooth.
Learn at Leisure
LEDWhen current passes through a pn junction, electrons and
holes transporting from one side to the other, are annihiltedin a recombination process. This process is accompanied bythe emission of light in the form of photons. A solid statelamp can, thus, be made out of a pn junction. This is calledlight emitting diode (LED) (Fig 15 – 23 a). It is used in
Fig (15-22b)Digital camera
Fig (15-22c)A CCD element
Fig (15-22d)Storing date on a magnetic tape in the digital
cameraFig (15-22e)
Downloading onto a computer
Fig (15-23a)LED
axis light
transparentdome
terminal
375
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Fig (15-24b)Different versions of transistors and diodes
electronic bulletins, watches and measuring equipment. Iflight from a forward biased heavily doped pn junction isconcentrated by laser action, we may have a junction (solid
state) laser (Fig 15 – 23 b), which is used in surgery,
communication through fiber optics and in modern warfare,
such as missile guidance and radar (laser radar is called
LADAR).
Eelctronic Circuits:Any analog or digital electronic system is composed of
electronic components connected together in a closed path called
circuit. The components may be passive as resistors, inductors,
capacitors, or diodes (Fig 15 – 24). Active components include
transistors in all types.
Circuits formed from separate components and soldered Fig (15-24a)Resistors
Fig (15-23b)A junction laser
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
368374
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Images may be transferred via the internet using a newfeature called Bluetooth.
Learn at Leisure
LEDWhen current passes through a pn junction, electrons and
holes transporting from one side to the other, are annihiltedin a recombination process. This process is accompanied bythe emission of light in the form of photons. A solid statelamp can, thus, be made out of a pn junction. This is calledlight emitting diode (LED) (Fig 15 – 23 a). It is used in
Fig (15-22b)Digital camera
Fig (15-22c)A CCD element
Fig (15-22d)Storing date on a magnetic tape in the digital
cameraFig (15-22e)
Downloading onto a computer
Fig (15-23a)LED
axis light
transparentdome
terminal
375
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Fig (15-24b)Different versions of transistors and diodes
electronic bulletins, watches and measuring equipment. Iflight from a forward biased heavily doped pn junction isconcentrated by laser action, we may have a junction (solid
state) laser (Fig 15 – 23 b), which is used in surgery,
communication through fiber optics and in modern warfare,
such as missile guidance and radar (laser radar is called
LADAR).
Eelctronic Circuits:Any analog or digital electronic system is composed of
electronic components connected together in a closed path called
circuit. The components may be passive as resistors, inductors,
capacitors, or diodes (Fig 15 – 24). Active components include
transistors in all types.
Circuits formed from separate components and soldered Fig (15-24a)Resistors
Fig (15-23b)A junction laser
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
371376
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Fig (15-24c)Tansistors as pairs and quadruples
together are called discrete circuits (Fig 15 – 24). A new era of
integrated circuits(ICs) started in the 1960’s at the peak of space
research. The goal then was to develop electronic circuits with a
new technology ,which would put light weight, compactness,
effectiveness and reliability at prime interest. The answer was
ICs or microchips (Fig 15– 25). The basic idea is to cram all the
needed components onto a silicon wafer, where different regions
are assigned to needed functions without treating them as
separate components. If we want to make a diode, for example,
then starting with an n-type wafer, we allow p atoms to diffuse
in defined regions in the wafer. This is called selective (planar)
diffusion (Fig 15–26). The way this is done is a complicated
chemical process in which a mask is made and light (recently
laser) is used. The process is similar to film developing in
photography, and is called photolithography, which means
carving on stone. If we now want to make an npn transistor, we
open up a window in the p-region, and let n atoms diffuse
selectively there. Interestingly, all these operations are
Fig (15-25a)IC’s in different forms
Fig (15-25b)IC uncovered
Fig (15-25c)Microprocessor in comparison
with a match head
377
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
repeatedly made on a thin wafer of silicon thousands of
times simultaneously. Thus, the wafer is cut up to
thousands of slices, each called a chip, all carrying
exactly the same layout and same specifications. This
technique yields low cost electronic circuits due to the
mass production involved. The burden is really the initial
investment of setting up the foundries, with all the
sophisticated equipment including robots, testing systems,
and in the design, artwork i.e., the brainwork involved in
the programming, particularly when such circuits are
custom made. The commonplace ICs are, however,
inexpensive, since millions are made at a time for the
same design and artwork. This is what has made ICs
popular in both analog and digital electronic systems. In
fact, instead of designing highly complicated and costly
ICs existing or available (off the shelf) ICs are sought
first. Thus, design has shifted toward system engineering,
namely putting the right things together at the lowest
possible cost and highest possible efficiency.
A collection of components including ICs are often
mounted on a board called a printed circuit board (PCB).
An example is the motherboard of a PC (Fig 15 – 27). It
includes the processor, RAM, Arithmetic Logic Unit
(ALU), control circuits etc.
Fig (15-25d)Pentium IC
Fig (15-26)Selective diffusion
Fig (15-27)Motherboard
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
370376
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Fig (15-24c)Tansistors as pairs and quadruples
together are called discrete circuits (Fig 15 – 24). A new era of
integrated circuits(ICs) started in the 1960’s at the peak of space
research. The goal then was to develop electronic circuits with a
new technology ,which would put light weight, compactness,
effectiveness and reliability at prime interest. The answer was
ICs or microchips (Fig 15– 25). The basic idea is to cram all the
needed components onto a silicon wafer, where different regions
are assigned to needed functions without treating them as
separate components. If we want to make a diode, for example,
then starting with an n-type wafer, we allow p atoms to diffuse
in defined regions in the wafer. This is called selective (planar)
diffusion (Fig 15–26). The way this is done is a complicated
chemical process in which a mask is made and light (recently
laser) is used. The process is similar to film developing in
photography, and is called photolithography, which means
carving on stone. If we now want to make an npn transistor, we
open up a window in the p-region, and let n atoms diffuse
selectively there. Interestingly, all these operations are
Fig (15-25a)IC’s in different forms
Fig (15-25b)IC uncovered
Fig (15-25c)Microprocessor in comparison
with a match head
377
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
repeatedly made on a thin wafer of silicon thousands of
times simultaneously. Thus, the wafer is cut up to
thousands of slices, each called a chip, all carrying
exactly the same layout and same specifications. This
technique yields low cost electronic circuits due to the
mass production involved. The burden is really the initial
investment of setting up the foundries, with all the
sophisticated equipment including robots, testing systems,
and in the design, artwork i.e., the brainwork involved in
the programming, particularly when such circuits are
custom made. The commonplace ICs are, however,
inexpensive, since millions are made at a time for the
same design and artwork. This is what has made ICs
popular in both analog and digital electronic systems. In
fact, instead of designing highly complicated and costly
ICs existing or available (off the shelf) ICs are sought
first. Thus, design has shifted toward system engineering,
namely putting the right things together at the lowest
possible cost and highest possible efficiency.
A collection of components including ICs are often
mounted on a board called a printed circuit board (PCB).
An example is the motherboard of a PC (Fig 15 – 27). It
includes the processor, RAM, Arithmetic Logic Unit
(ALU), control circuits etc.
Fig (15-25d)Pentium IC
Fig (15-26)Selective diffusion
Fig (15-27)Motherboard
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
373378
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
ICs have permeated even medical equipment including instrumentation, diagnosis, and
prognosis. One day pacemakers and insulin control circuits using microcapsules involving
microprocessors may be injected into the body to do their work from within.
Miniaturization, where to ?When the first computer was built in the 1950’s, its capabilities were very limited by
today’s standards. It was bulky, about the size of an apartment. It was built from vacuumtube (valves). Then, transistors were used. ICs have led to the development of PCs whichmade computers available to the public. Since the 1970’s PCs are continually beingenhanced. Their capacity and capability to do complicated calculations are on the increase,while calculation time is getting shorter, and size and weight are getting smaller. Also, costis on the decline. These improvements sound contradictory, but they are happening and at ahigh rate, thanks to the understanding and best application of the basic concepts of modernphysics, materials science, chemistry, laser and to the rapid advancement of the technology.There is a common law called Moore’s law, which states that capacity and speed doubleevery 18 months. If a chip (the size of a pin head) contains 100 transistors, this called smallscale integration (SSI). If it contains 1000 transistors, it is called medium scale integration(MSI). If it contains 10000 transistors, it is called large scale integration (LSI). If it contains100000 transistors, it is called very large scale integration (VLSI). If it exceeds that, it iscalled ultra large scale integration (ULSI). Can you imagine 1 million transistors in a pinhead area?
What then if you know that the figure in 2005 has reached 300 millions with prospect ofeven more?
If the miniaturization keeps going at that rate what next ? we shall soon be limited by thediffraction of light as the physical dimension will soon approach λ of the used light. It seems
379
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
that we are headed to reach the size of the atom itself, i.e., 0 and 1 may be stored in the form ofan electron being in either one of two states in the atom, ground state or an excited state.
Alternatively, the two states may be one direction of electron spin, and the other state in theopposite direction. This is called quantum computer. This is the trend of the near future, which isabout to materialize. It is in harmony with future direction trends of science in search for minutedetails of time and space.
This has led to the advent of new technologies such as Nano and Femto technologies.
Learn at Leisure
Selective Diffusion How can we make phosphorus atoms for example diffuse in a very small area and not
through the rest of the chip ? This happens in several steps. First, we cover the silicon waferwith a layer of oxide (Si O2), then we use a mask prepared in a way similar to photographicdevelopment. We cover the oxide layer with a photoresist, which is a light sensitivematerial. We then put a mask with opaque and transparent regions on top of the photoresist.We then expose the surface to ultraviolet (uv) rays. When the photoresist is exposed to uv,it polymerizes (solidifies) in the region where it is exposed, and remains liquid in theunexposed areas. We then lift the mask and use HCl acid, which interacts with SiO2 (aprocess called etching) in the areas where the photoresist is in liquid form (where SiO2 isuncovered with polymerized photoresist). The acid, thus, opens up a hole in the oxide.Then, phosphorus atoms are allowed to diffuse through the opening in the oxide, while theoxide isolates the remaining areas from diffusion. Thus, miniaturization depends on theaccuracy of the mask. Therefore, laser is used in mask making. For more miniaturization,electron beam and molecular beam are used for shorter λ, and hence smaller dimensions (why?)where they directly carve on the chip. But this cannot be used on a large scale, but is used forspecial ICs only.
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
372378
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
ICs have permeated even medical equipment including instrumentation, diagnosis, and
prognosis. One day pacemakers and insulin control circuits using microcapsules involving
microprocessors may be injected into the body to do their work from within.
Miniaturization, where to ?When the first computer was built in the 1950’s, its capabilities were very limited by
today’s standards. It was bulky, about the size of an apartment. It was built from vacuumtube (valves). Then, transistors were used. ICs have led to the development of PCs whichmade computers available to the public. Since the 1970’s PCs are continually beingenhanced. Their capacity and capability to do complicated calculations are on the increase,while calculation time is getting shorter, and size and weight are getting smaller. Also, costis on the decline. These improvements sound contradictory, but they are happening and at ahigh rate, thanks to the understanding and best application of the basic concepts of modernphysics, materials science, chemistry, laser and to the rapid advancement of the technology.There is a common law called Moore’s law, which states that capacity and speed doubleevery 18 months. If a chip (the size of a pin head) contains 100 transistors, this called smallscale integration (SSI). If it contains 1000 transistors, it is called medium scale integration(MSI). If it contains 10000 transistors, it is called large scale integration (LSI). If it contains100000 transistors, it is called very large scale integration (VLSI). If it exceeds that, it iscalled ultra large scale integration (ULSI). Can you imagine 1 million transistors in a pinhead area?
What then if you know that the figure in 2005 has reached 300 millions with prospect ofeven more?
If the miniaturization keeps going at that rate what next ? we shall soon be limited by thediffraction of light as the physical dimension will soon approach λ of the used light. It seems
379
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
that we are headed to reach the size of the atom itself, i.e., 0 and 1 may be stored in the form ofan electron being in either one of two states in the atom, ground state or an excited state.
Alternatively, the two states may be one direction of electron spin, and the other state in theopposite direction. This is called quantum computer. This is the trend of the near future, which isabout to materialize. It is in harmony with future direction trends of science in search for minutedetails of time and space.
This has led to the advent of new technologies such as Nano and Femto technologies.
Learn at Leisure
Selective Diffusion How can we make phosphorus atoms for example diffuse in a very small area and not
through the rest of the chip ? This happens in several steps. First, we cover the silicon waferwith a layer of oxide (Si O2), then we use a mask prepared in a way similar to photographicdevelopment. We cover the oxide layer with a photoresist, which is a light sensitivematerial. We then put a mask with opaque and transparent regions on top of the photoresist.We then expose the surface to ultraviolet (uv) rays. When the photoresist is exposed to uv,it polymerizes (solidifies) in the region where it is exposed, and remains liquid in theunexposed areas. We then lift the mask and use HCl acid, which interacts with SiO2 (aprocess called etching) in the areas where the photoresist is in liquid form (where SiO2 isuncovered with polymerized photoresist). The acid, thus, opens up a hole in the oxide.Then, phosphorus atoms are allowed to diffuse through the opening in the oxide, while theoxide isolates the remaining areas from diffusion. Thus, miniaturization depends on theaccuracy of the mask. Therefore, laser is used in mask making. For more miniaturization,electron beam and molecular beam are used for shorter λ, and hence smaller dimensions (why?)where they directly carve on the chip. But this cannot be used on a large scale, but is used forspecial ICs only.
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
375380
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
In a Nutshell
• A metallic crystal consists of positive ions and a cloud of free electrons roaming around
the crystal in random motion. There is a force of attraction between the ions and the
electron cloud. But the resultant of all forces of attraction on a single free electron is zero
. If an electron tries to escape from the metal, a net force of attraction due to the atom
layer at the surface pulls it in.
• A pure silicon (semiconductor) crystal consists of atoms covalently bonded. At low
temperatures, there are no free electrons. If temperature increases, some bonds are
broken, electrons become free, leaving behind holes. Both electrons and holes move
randomly.
• The number of broken bonds increases with temperature. It may increase also by an
external stimulus, such as light, provided that the photon energy is sufficient to break the
bond.
• The number of free electrons and holes increases by adding impurities (doping). Thus, the
material becomes n-type or p-type.
• The conductivity of a semiconductor depends on the conduction of free electrons and
holes. Thus, a semiconductor has two current carriers: electrons and holes, while in a
metal there is only one current carrier (the electron). Electron concentration in a metal is
constant and does not depend on temperature.
• Semiconductors are environment-sensitive. They can be used as sensors to light, heat,
pressure humidity, chemical pollution, radiation etc.
• A diode (pn junction) consists of a p–type region and an n-type region. If the p-side is
connected to the positive terminal of the battery and the n-side to the negative terminal
381
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
(forward connection or forward bias) current flows. If the battery is reversed no current
flows.
This is why a diode is used in rectification.
• A transistor may be pnp or npn, and can be used as an amplifier, since the ratio of the
collector current to the base current βe is large. Therefore, any small change in the base
current leads to an amplified change in the collector current.
• A transistor may also be used as a switch. It is used in logic gates, such as an inverter
(NOT), AND, OR gates.
• Digital electronics is superceding analog electronics for its ability to overcome electrical
noise . Its basic concept is to code information in binary form (0 , 1).
• ICs have the advantages of small size and weight, increased speeds and capacity, and yet
low cost. This is the reason for the proliferation of PCs.
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
374380
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
In a Nutshell
• A metallic crystal consists of positive ions and a cloud of free electrons roaming around
the crystal in random motion. There is a force of attraction between the ions and the
electron cloud. But the resultant of all forces of attraction on a single free electron is zero
. If an electron tries to escape from the metal, a net force of attraction due to the atom
layer at the surface pulls it in.
• A pure silicon (semiconductor) crystal consists of atoms covalently bonded. At low
temperatures, there are no free electrons. If temperature increases, some bonds are
broken, electrons become free, leaving behind holes. Both electrons and holes move
randomly.
• The number of broken bonds increases with temperature. It may increase also by an
external stimulus, such as light, provided that the photon energy is sufficient to break the
bond.
• The number of free electrons and holes increases by adding impurities (doping). Thus, the
material becomes n-type or p-type.
• The conductivity of a semiconductor depends on the conduction of free electrons and
holes. Thus, a semiconductor has two current carriers: electrons and holes, while in a
metal there is only one current carrier (the electron). Electron concentration in a metal is
constant and does not depend on temperature.
• Semiconductors are environment-sensitive. They can be used as sensors to light, heat,
pressure humidity, chemical pollution, radiation etc.
• A diode (pn junction) consists of a p–type region and an n-type region. If the p-side is
connected to the positive terminal of the battery and the n-side to the negative terminal
381
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
(forward connection or forward bias) current flows. If the battery is reversed no current
flows.
This is why a diode is used in rectification.
• A transistor may be pnp or npn, and can be used as an amplifier, since the ratio of the
collector current to the base current βe is large. Therefore, any small change in the base
current leads to an amplified change in the collector current.
• A transistor may also be used as a switch. It is used in logic gates, such as an inverter
(NOT), AND, OR gates.
• Digital electronics is superceding analog electronics for its ability to overcome electrical
noise . Its basic concept is to code information in binary form (0 , 1).
• ICs have the advantages of small size and weight, increased speeds and capacity, and yet
low cost. This is the reason for the proliferation of PCs.
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
377382
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Questions and Drills
I) Drills:
1) Calculate the number of silicon atoms in 1 cm3, if the density of silicon is 2.33 g/cm3
and its atomic mass is 28
(0.5x1023cm-3)
2) If electron or hole concentration in pure silicon is 1x1010cm-3, phosphorus is added at a
concentration of 1012cm3, calculate the concentrations of electrons and holes in this case.
Is this silicon n-type or p-type? (n=1012cm-3 p=108cm-3 )
(n - type)
3) Calculate the concentration of aluminum to be added so that silicon returns pure .
(NA- = 1012cm-3)
4) A transistor has αe = 0.99 . Calculate βe. Then calculate the collector current if the base
current is 100 µA (βe=99 , Ic = 99x10-4A)
5) The electrical signal in the base of a transistor is 200 µA . The collector current is to be
10 mA. Calculate αe and βe. (αe = 0.98 , βe = 50)
6) A diode can be represented by a forward resistance 100Ω ,while it is infinity in the
reverse direction. We apply +5 V ,and then reverse it to – 5 V. Calculate the current in
both cases. (50 mA, O)
7) If 1 mm2 contains 1 million transistors, calculate the area assigned to each transistor.
(10-12m2)
II) Essay questions:
1) Discuss the importance of digital electronics and mention 5 applications. 2) Deduce the truth table for an AND gate followed by an inverter. 3) Deduce the truth table for an OR gate followed by an inverter.
311
Unit 5
: modern
physic
s C
hapte
r 13: A
tom
ic S
pectra
Bohr’s Model (1913)
Bohr studied the difficulties faced by Rutherford’s model,and proposed a model for the hydrogen atom building onRutherford’s findings :1) At the center of the atom there is a positively charged
nucleus .2) Negatively charged electrons move around the nucleus in
shells. Each shell (loosely often called orbit) has an energyvalue. Electrons do not emit radiation as long as they remainin each shell (Fig 13 – 5).
3) The atom is electrically neutral, since the number ofelectrons around the nucleus equals the number ofpositive charges in the nucleus.
Fig ( 13-5a)Bohr’s Model
first shell
second shell
En
ergy
free electron levelcontinuous levels
Bohr
Fig (13-5b)Energy levels
310
Unit 5
: m
odern
physic
s C
hapte
r 13: A
tom
ic S
pe
ctr
a
spectra of the atoms of all elements would have been continuous. This is contrary to allexperimental observations.The spectra of the elements have a discrete nature, and arecalled line spectra, i.e., occurring at wavelengths characteristic of the element.
Fig (13 – 4a )Apparatus for studying the spectra of the elements
Fig (13 – 4b )Spectra of some elements
gas slit prism screen
potentialdifference
376382
Unit 5
: modern
physic
s C
hapte
r 15: M
odern
Ele
ctro
nic
s
Questions and Drills
I) Drills:
1) Calculate the number of silicon atoms in 1 cm3, if the density of silicon is 2.33 g/cm3
and its atomic mass is 28
(0.5x1023cm-3)
2) If electron or hole concentration in pure silicon is 1x1010cm-3, phosphorus is added at a
concentration of 1012cm3, calculate the concentrations of electrons and holes in this case.
Is this silicon n-type or p-type? (n=1012cm-3 p=108cm-3 )
(n - type)
3) Calculate the concentration of aluminum to be added so that silicon returns pure .
(NA- = 1012cm-3)
4) A transistor has αe = 0.99 . Calculate βe. Then calculate the collector current if the base
current is 100 µA (βe=99 , Ic = 99x10-4A)
5) The electrical signal in the base of a transistor is 200 µA . The collector current is to be
10 mA. Calculate αe and βe. (αe = 0.98 , βe = 50)
6) A diode can be represented by a forward resistance 100Ω ,while it is infinity in the
reverse direction. We apply +5 V ,and then reverse it to – 5 V. Calculate the current in
both cases. (50 mA, O)
7) If 1 mm2 contains 1 million transistors, calculate the area assigned to each transistor.
(10-12m2)
II) Essay questions:
1) Discuss the importance of digital electronics and mention 5 applications. 2) Deduce the truth table for an AND gate followed by an inverter. 3) Deduce the truth table for an OR gate followed by an inverter.
1
383
General Revision
1) A metallic wire is stretched between two vertical fixed pins .Is the velocity ofpropagation of a transverse wave through this wire affected by a change in thetemperature of the surrounding medium? and why ?
2) Two identical strings, one end of each is fixed to the wall, while the other end isstretched by hand . A transverse pulse is sent through one of the wires, and after ashort time another transverse pulse is sent through the other wire. What can be done tomake the second pulse catch up with the first pulse? Give reasons.
3) Give reasons:It is easy to see your reflected image on the window glass of a lit room at night whenit is dark outside the room. But that is difficult when there is light outside the room.
4) Two rays of light converge on a point on a screen. A parallel glass plate is placed inthe path of this light, and the glass plate is parallel to the screen. Will the point ofconvergence remain on the screen or change position ? Give reasons .
5) Explain and give reasons :When a blue light source is placed at the center of a solid glass cube with a whitescreen facing each side, a circular spot of light appears on each screen in front of eachsurface of the cube . When the blue source is replaced by a red color source, the shapeof the spot changes from circular to square .
6) In the following figure, a fiber optic has an external layer from glass. Its refractiveindex is less than that of the glass of the core . If the light beam passes through it asshown in the figure.
377
377383
378385
of distance 11 x 10-4 m apart, and the distance between the double slit and the screenwas 5m. Find the distance between two successive similar fringes .12) A rubber hose is connected to a tap, and the water flows through in a steady flow.Explain why the cross sectional area of the flowing water decreases when the end ofthe rubber hose is directed down, and increases when the end of the rubber hose isdirected up.
13) A balloon filled with air is attached to the bottom of a glass tank. Then the tank isfilled completely with water ,with this balloon completely immersed in water. Supposethat the tank with its contents were transferred from the Earth to the Moon. Discuss andgive reason for all what would happen to the balloon?
14) A hollow copper ball is suspended under the surface of water in a tank. Discussand give reason for all what would happen to the position of the ball in the tank if itwere transferred from the Earth to the Moon?
15) Verify the following statement and correct the mistakes if any:When a person divesin a swimming pool near the bottom , each of the upthrust and pressure exerted overhim increases.
16) An ice cube is placed in a glass beaker, then it is filled completely to the rim withwater . Discuss in the light of Archimedes’ principle what changes may happen whenthe ice melts (fuses) .
17) A glass beaker filled to the rim with water is resting on a scale . A block is placedin water, causing some of it to spill over.The water that is spilled is wiped away ,andthe beaker is still filled to the rim . Compare between the initial and final reading on thescale, if the block is made from:a) woodb) iron
379
379384
a)explain why the direction of the beam does not change at each of S and P.b)explain why there is a total reflection at each of Q and R.c)explain why the double layer fiber optic is preferred to that of a single layer.7- A teacher gave his students the following figure (A) which expresses a path of lightbeam from A to B through a triangular prism made of glass which has a critical angle42˚. He asked the students to draw the path of the beam before reaching A and afterleaving B. The figure (B) expresses the attempt of a student. But the teacher made itclear that the angles of X and Y were not correct. Suggest without calculations thechanges required to correct the angles X and Y, and give reasons for your suggestion.
(Fig A) (Fig B)
8) The tension of a stretched string is changed from 70N to 80N without a change in itslength. Calculate the ratio of the fundamental frequencies as a result .
9) A string of length 0.06 m and mass 2.5 x 10-3 kg is stretched by a force of 400N.Find the frequency of the produced tone if it vibrates in three segments.10) A triangular prism has an angle of 60˚ and refractive index of. 2 . Calculate theminimum angle of deviation and the corresponding angle of incidence.
11) A monochromatic light of 66 x 10-8 m wavelength strikes a double slit
378
380387
24) Show that Van der Waals’ effect can explain the conversion of gases into liquidstate.25) Give reason : Van der Waals’ effect on gases appears clearly at low temperatures.
26) Gas behavior deviates from that of the ideal gas as its density increases. Disuss thisstatement .
27) What is the scientific concept on which the Dewar’s flask is designed?
28) Give reason :Liquid helium is preferred as a cryogenic material.
29) Compare between the characteristics of the adiabatic change and the isothermalchange .
30) What is meant by the transitional (critical) temperature of a metal?
31) Give reason : We use a superconductive coil in manufactering the levitated train.
32) Give reason : A superconductor is used in making satellite's antenna.
33) Give reason: Meissner effect appears only in superconductive materials.
34) Suppose that the atoms of helium gas have the same average velocity as the atomsof argon gas. Which of them has a higher temperature and why?
35) Calculate the average kinetic energy and root mean square of the velocity of a freeelectron at 300˚K , where Boltzmann's constant = 1.38 x 10-23 J/˚K , the mass ofelectron is 9.1 x 10-31kg.
381
381386
where the density of water is 1000 kgm-3, that of wood is 550 kgm-3 and that of iron is7860 kgm-3.18) State the conditions which make the liquid flow in a steady flow and prove that in asteady flow, the velocity of liquid flow at any point is inversely proportional to theacross sectional area of the tube at that point?
19) A major artery of radius 0.5 cm branches out into many capillaries,the radius ofeach is 0.2 cm . The speed of blood in the main artery is 0.4 m/s , and the speed ofblood in each capillary is 0.25 m/s . Find the number of the capillaries ?
20) A cross sectional area of one end of a U- shaped tube is twice the other. When asuitable amount of water is placed in the tube and an amount of oil is poured into thewide end, the surface of water in the tube is lowered by 0.5 cm. Calculate the height ofoil in the tube. Knowing that the density of water equals 1000 kg m-3 and that of oilequals 800 kg m-3.
21) The small and large piston cross sectional areas of a hydraulic press are 4 x 10-4m2
and 20 x 10-4m2, respectively, and a force of 200 N is exerted on the small piston.Calculate the mass required to be placed on the large piston to be in the same level withthe small piston ,g = 10 m/s2.
22) What is the least area for a floating layer of ice of thickness 5 cm above the waterof a river which makes this layer carry a car whose mass is16 x103kg. The density of water is 1000 kg/m3 and that of ice is 920 kg/ m3.
23) A cork ball has a volume of 5 x10-3 m3,placed in water of density 1000 kgm-3.
About 2/5 of its volume was immersed. Calculate the density of the cork and the forceneeded to immerse all the volume of the ball.
380
382389
step - up transformer decreases the current.
44) There are three essential factors that must be considered when designingtransformers to decrease the loss of the electric energy. What are these factors andhow?
45) Give reason: The eddy current is not generated in the metallic blocks unlessa magnetic field of variable intensity exists.
46) Compare between an AC generator and a DC generator.
47) Give reason: To increase the power of a motor ,several coils seperated by smallangles are used.
48) The following table shows values of resistance of wire of cross sectional area 0.1 m2 and different lengths.
Plot the relation between the length ( l ) on the X axis and Resistance (R) on the Y axis. From the plot find:
a) resistance of a part of the wire of length 12 m .b) the resistivity of the material of the wire.c) the conductivity of the material of the wire.
49) A wire 30 cm long and 0.3 cm2 cross sectional area is connected in series with aDC source and an ammeter . The potential difference between the ends of the wire is
Length l m
Resistance R
2
5
4
10
6
15
10
25
14
35
16
40
383
383388
36) An amount of an ideal gas has a mass of 0.8 x 10-3 kg , a volume of 0.285 x 10-3 m-3,at a temperature of 12o C and under pressure of 105N/m2 . Calculatethe molecular mass of the gas where the universal gas constant equals 8.31 J/˚K.
37) Calculate the mean kinetic energy of an oxygen molecule at a temperature of50oC, where Boltzmann's constant is equal to1.38 x 10-23 J/˚K.
38) If the temperature at the surface of the Sun is 6000oK, find the root mean squarespeed of hydrogen molecules at the surface of the Sun, knowing that the hydrogen is inits atomic state. Its atomic mass =1, Avogadro's number (NA)=6.02 x 1023, andBoltzmann's constant=1.38 x 10-23 J/˚K .
39) Give reason : Efficiency of the battery increases by the decrease of its internalresistance.
40) In electric circuits connected in parallel, thick wires are used at the ends of thebattery,but at the ends of each resistor less thick wires are used. Why?
41) What is meant by: a) the effective value of AC. b) eddy current. c) the sensitivity of a galvanometer. d) the efficiency a transformer.
42) What is the physical concept for the operation of the following devices:galvanometer – transformer – current divider (or shunt)– potential multiplier.
43) Give reason :The step- down transformer increases the current, and the
382
384391
b) the efficiency of a transformer=90%.c) eddy currents. d) the effective value of an AC current =2A.
55) A step -down transformer of efficiency 100% is to be used to light a lamp ofpower 24 W at a potential difference 12 V. If the power source applied to thetransformer is 240 V, the number of turns of the secondary coil is 480 turn. 1) calculate the current passing through each of the primary and secondary coils 2) the number of the turns of the primary coil.
56) When an electric current is flowing through a perpendicular wire in a uniformmagnetic field, the wire is affected by a force. Which of the following instruments isbased on this principle:(1) electromagnet.(2) motor.(3) generator.(4) transformer.
57) Calculate the emf of a source if the work done to transfer 5C is 100 J.
58)Three resistors 10 , 20 , 30 are connected to a power supply . If the currentsare 0.15 A , 0.2 A , 0.05 A, respectively. Calculate the equivalent resistor for thiscircuit, and illustrate your answer with a labeled diagram.
59) Two resistors 400 and 300 are connected in series to a 130V power supply.Compare between the readings of a voltmeter of resistance 200 when connectedacross each resistor seperately ( neglecting the internal resistance of the power supply).
60) A wire has length 2 m and cross sectional area 0.1 m2 is connected to a sourcewith emf 10 V. Calculate the resistivity and conductivity of its material if it carries a
385
385390
0.8 V, when a current of 2A passes through it. Calculate the conductivity of the wirematerial.
50) A rectangular coil of N turns and surface area A is placed parallel to the lines of aregular magnetic field of flux density B Tesla. If the coil starts rotation from thisposition with a regular angular velocity ω, until it compelets half a revolution. Clarifywith a labeled diagram how the value of the emf changes with the rotational angleduring this time, and what is the maximum value of the induced emf generated in thiscoil.
51) A galvanometer has a resistance 40 Ω and reads up to 20 mA. Calculate theresistance of the shunt required to convert it into an ammeter, reading up to 100 mA. Ifthe coil of the galvanometer is connected to a potential multiplier with resistance 210Ω.Calculate the maximum potential difference to be read.
52) Compare between each ofa) a step -down transformer and a step- up transformer in terms of function, use, andnumber of turns of the secondary coil.b) dynamo and motor in terms of function and use.
53) Why does the transmission of the electric power from a generating station requirewires under high voltage?Choose the correct answer and give account
a) to be able to use the transformers .b) to insure that the current will flow for a long distance.c) to minimize the loss in the electric energy .d) to minimize the resistance of the wires.
54)What is meant by :a) the coefficient of mutual induction between two coils =2H.
384
386387
387386
388389
389388
390391
391390
392393
393392
394395
395394
396403
R = VI
= RAL
4 10 .m5
R
e
e
= =
= × ××
= ×
−
−
−
0 82
0 4
0 4 0 3 1030 10
4
2
. .
. .ρ
ρ Ω
Ω
RI RI I
= 200 10
V V R I
sg g
g
-3
g m g
=−
× ×× − ×
== +
= × × + × ×=
− −
− −
400100 10 20 10
10
20 10 40 210 20 105
3 3
3 3
Ω
( ) ( )V V
l
49)
51)
55) I PV
VV
I
VV
N
sw
s
s
p
p
s
p
p
= = =
=
=
= =
=
=
= × =
2412
2
12240 2
24240
0 1
12240
480
480 24012
9600
A
II
I
A
NN
N
p
s
p
s
p
p
.
397
397402
32) A superconductive material is used in making satellite’s antenna, because it has no electrical
resistance. This makes it easily affected by the weakest of electromagnetic waves.
33) Meissner effect appears only in superconductive materials, because they offer no
electrical resistance to the electrons. They are easily affected by external magnetic fields
and retain the kinetic energy acquired due to this field without any loss in the form of
thermal energy.
Hence, current flows continuously ,leading in turn to a counter magnetic field , so that the
net field inside is zero (diamagnetic) .This is why a superconducting magnet is always re-
pulsive to the external magnetic field .
34) The temperature of argon gas is higher than that of helium gas, because the mass of the
argon atom is more than that of a helium atom , so the kinetic energy of an argon atom is
more than that of a helium atom.
35)
39) As the internal resistance of the battery decreases, the lost work done (wasted energy) is
reduced during operation.
40) Because the current intensity in parallel circuits is greater at the input and output compared
to the current in each branch.
12
32
12
32
6.21 x 10-21 x 229.1 x 10-31
12
mv2 = kT
mv2 = x 1.38 x 10-23 x 300
= 6.21 x 10-21 J
v = ( ) = 1.17 x 105 m/s
396
398381
399404
W Q
100
5
VBR+r
122+0.5
2.412
57) VB = = =20V
64)The wasted potential difference is the potential (voltage) drop on the internal resistance of the cell .
I = = =4.8
Ir = 4.8 X 0.5 =2.4V
Percent drop = X 100 = 20%
398
407
serial quantity symbol unit
J
W , Js-1 (watt)
NsCelsius, Fahrenheit, Kelvin
mole
pascal , Nm-2
pascal , Nm-2
J
J kg-1 ˚K-1
JK-1
J kg-1
J kg-1
per degree rise
per degree rise
kg/s
m3/s
Ns m-2
______
C (Coulomb)
C
V (Volt)
V
PEPw
Iimp
t˚C , t˚F , T˚KnPPa
Qth
Cth
qth
Bth
Lth
αV
βPQm
QV
ηvs
ηQ,q
eVVB
potential energy
power
impulse
temperature
quantity of matter
pressure
atmospheric pressure
quantity of heat
specific heat
heat capacity
latent heat for evaporation
latent heat for fusion
volume expansion coefficient
pressure expansion coefficient
mass rate of flow
volume rate of flow
viscosity coefficient
efficiency
electric charge
electron charge
potential difference
battery voltage
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
401
406
serial
Appendix 1
Symbols and Units of Some Physical Quantities
quantity symbol unit
x,y,z,d
A
Vol
t
T
v
α,θ,φω
m,Mme
ρagPL
FFg
τ
WE
KE
m (meter)
m2
m3
s (second)
s
m s-1
deg , rad
rad s-1
kg
kg
kg m-3
m s-2
m s-2
kg m s-1
N , kg ms-2
N(Newton)
Nm
J(Joule)
J
J
displacement
area
volume
time
periodic time
velocity / speed
angle
angular velocity
mass
electron mass
density
acceleration
acceleration due to gravity
linear momentum
force
weight
torque
work
energy
kinetic energy
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
400
409
Appendix 2Fundamental Physical Constants
value
Gk
NA
RKµce
me
mp
hu
RH
mn
g
reMeMmrm
6.677x10-11 N m2 kg-2
1.38x10-23 JK-1
6.02x1026 Molecule.kmol-1
8.31x103 J.kmol-1 K-1
9x109 Nm2C-2
4 x10-7 Weber m-1A-1
3x108 m.s-1
1.6x10-19 C
9.1x10-31 kg
1.79x1011 C.kg-1
1.673x10-27 kg
6.63x10-34 Js
1.66x10-27 kg
1.096x107 m-1
1.675x10-27 kg
22.4x10-3 m3
9.8066 ms-2
6.374x106 m
5.976x1024 kg
7.35x1022 kg
3.844x108 m
eme
symbol Physical Constant1-Universal gravitation constant2-Boltzmann constant 3-Avogadro’s number4- Universal gas constant5-Coulomb’s law constant6-Permeability of free space7-Speed of light in vacuum8-Elementary charge9- Electron rest mass 10-Specific charge of electron11-Proton rest mass12-Planck’s constant13-Atomic mass unit14-Rydberg constant15-Neutron rest mass16-Molar volume of ideal gas at S.T.P17-Standard gravity at the Earth’s surface18-Equatorial radius of the Earth19- Mass of the Earth20-Mass of the Moon 21- Mean radius of the Moon’s orbitaround the Earth
403
408
serial quantity symbol unit
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
V
Vm-1
Gauss
A (Ampere)
Ω (Ohm)
Ω m
Ω-1 m-1
______
Am-1
Tesla , Wb m-2
Wb (Weber)
H (Henry)
H
Weber A-1 m-1
Nm Tesla-1
ms-1
Hertz (Hz)
Hz
m______
______
emfεφe
IRρe
σαe , βe
HBφm
LMµ
md
cνfλn
ωα
electromotive force (emf)
field intensity
electric flux
electric current
electrical resistor
resistivity
conductivity
transistor gain
magnetic field intensity
magnetic flux density
magnetic flux self inductance
mutual inductance
permeability
magnetic dipole
speed of light
frequency of wave
frequency of electric current
wave length
refractive index
dispersive power
402
411
Standard Prefixes
name
Yocto
Zepto
Atto
Femto
Pico
Nano
Micro
Milli
Centi
Deci
___
Deka
Hecto
Kilo
Mega
Giga
Tera
Peta
Exa
Zetta
Yotta
10-24
10-21
10-18
10-15
10-12
10-9
10-6
10-3
10-2
10-1
100
101
102
103
106
109
1012
1015
1018
1021
1024
Appendix 3
power of 10
405
410
value
22-Mass of the Sun 23- Mean radius of the Earth’s orbit around the Sun
24-Period of the Earth’s orbit around the Sun
25- Diameter of our galaxy26- Mass of our galaxy
27- Radius of the Sun
28- Sun’s radiation intensity at the Earth’s surface
symbol Physical ConstantMsresyr
__
__
__
__
1.989x1030 kg1.496x1011 m
3.156x107 s
7.5x1020 m
2.7x1041 kg
7x108 m
0.134 J cm-2 s-1
404
413
Appendix 5Gallery of Scientists
A pioneer in medicine and the discoverer of the laws of motion
Ibn Malka(1072 -1152 )
A pioneer in astronomy and the inventorof the simple pendulum.
Ibn Unis(952 -1009 )
A pioneer in geography and astronomy.Al Baironi(973 - 1048 )
A pioneer in mathematics, astronomy,medicine and the founder of optics.
Ibn Al-Haytham(965 - 1040)
A pioneer in philosophy, physics ,particularly optics.
Al Kindy(800 - 873)
The inventor of the phonograph and theelectric lamp, and other inventions “1000".
Edison (Thomas)(1847-1931)
The discoverer of the ratio of the radius ofa circle to its circumference, buoyancy and
the reflecting mirror.
Arkhimêdês
(287 -212 BC)
The discoverer of the molcular theory Avogadro (Amedeo)(1776 - 1856)
607
412
Greek AlphabetAppendix 4
606
415
The inventor of the barometerTorricelli (Evangelista)
(1608 - 1647)
The inventor of the telescope and thediscoverer of accelration due to gravity.
Galileo (Galilei)(1564 - 1642)
The discoverer of the electric charge inmuscles.
Galvani (Luigi)(1737 - 1798)
The discoverer of the law of mixinggases.
Dalton (John)(1766 - 1844)
The discoverer of radioactivty.Rutherford (Ernest)
(1871 - 1937)
The discoverer of the induction coil.Ruhmkorff (Heinrich)
(1803 - 1877)
The discoverer of X-rays.Rontgen (Wilhelm)
(1845 - 1923)
The discoverer of Quantum Mechanics.Schrodinger (Erwin)
(1887 - 1961)
A pioneer in hydrostatics.Al-Khazin
609
414
He was awarded Nobel prize in 1921 forhis explanation of the photoelectric effect,
the founder of the theory of relativity
Einstein (Albert)(1879 - 1955)
He performed studies on electricity,telegraph and magnetism.
Ampére (André - Marie)(1775 - 1836)
The founder of the theory ofelectromagnetism in 1820
Oersted (Christian)(1777 - 1851)
The discoverer of Ohm’s lawOhm (George)(1789 - 1854)
The discoverer of Pascal’s rule.Pascal (Blaise)(1623 - 1662)
A pioneer in fine mechanics and waterclocks.
Al Joazri
The founder of X-ray diffraction.Bragg (William)
(1862 - 1942)
He produced a model for the atom.Bohr (Neils)(1885 - 1962)
The discoverer of Boyle’s law.Boyle (Robert)(1627 - 1691)
608
417
The discoverer of Lenz’s rule.Lenz (Heinrich)(1804 - 1865)
The discoverer of the photon and theblackbody radiation.
Planck (Max)(1858 - 1947)
The discoverer of Maxwell’s equations.Maxwell (James)
The discoverer of the laws of motion,gravity and colors.
Newton (Isaac)(1642 - 1727)
The discoverer of the electromagneticwaves
Hertz (Heinrich)(1857 - 1894)
He proposed the secondary sources inthe from of a wave.
The discoverer of interference. Young (Thomas)
(1773 - 1829)
Huygens (Christian) (1629 - 1695)
411
416
The discoverer of the laws ofelectromagnetics.
Faraday (Michael)(1791 - 1867)
The discoverer of Van Der Waals’effect.
Van Der Waals (Johannes)(1837 - 1923)
He interpreted the atomic spectra anddiffraction
Fraunhofer (Joseph Von)(1787 - 1826)
The inventor of the battery.Volta (Alessandro)(1745 - 1827)
He contributed to the atomic bomb.Fermi (Enrico) (1901 - 1954)
The discoverer of liquid helium.Kamelingh (Onnes)(1853 - 1926)
The discoverer of the laws of planetarymotion.
Kepler (Johannes)(1571 - 1630)
He proved that the Earth rotates aroundthe Sun.
Copernicus (Nicolas)(1473 - 1543)
The discoverer of Kirchhoff’s law.Kirchhoff (Gustav)(1824 - 1887)
410
381
418
Selected Physics Sites on the Internet
Appendix 6
http://www.dke-encyc.com
http://imagine.gsfc.nasa.gov
http://csep10.phys.utk.edu
http://www.howstuffworks.com
http://www.colorado.edu/physics/2000/index.pl
http://scienceworld.wolfram.com/physics
http://www.physlink.com
http://www.intuitor.com/moviephysics
http://www.newport.com/spectralanding
http://www.mathpages.com/home/iphysics.htm
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
http://www.smsec.com
412