particulate nature of light

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3.1 PARTICULATE NATURE OF ELECTROMAGNETIC RADIATION

The photo-electric effect had initially been described by Hertz in 1887, and developed further by another German, Hallwachs. A negatively charged zinc plate would discharge when exposed

to ultra-violet light; a positively charged plate would not. Also the plate would not discharge in bright red light, but it would in dim UV light.

The photo-electric effect is the emission of electrons from metal surfaces when electromagneticradiation of high enough frequency falls on them. The effect is shown by zinc when exposed to

X-rays or UV radiation. Sodium gives emissions with X-rays, UV and all colours of light except

orange and red.

Detailed investigation of the photoelectric effect gave the following, main observations:

a) No electrons are emitted if the frequency of the light is below a minimum frequency

{called the threshold frequency }, regardless of the intensity of light

b) Rate of electron emission {ie photoelectric current} is proportional to the light intensity.

c) {Emitted electrons have a range of kinetic energy, ranging from zero to a certainmaximum value. Increasing the freq increases the kinetic energies of the emitted

electrons and in particular, increases the maximum kinetic energy.} This maximum

kinetic energy depends only on the frequency and the metal used {ϕ}; the intensity

has no effect on the kinetic energy of the electrons.

d) Emission of electrons begins instantaneously {i.e. no time lag between emission &

illumination} even if the intensity is very low.

NB: (a), (c) & (d) cannot be explained by Wave Theory of Light; instead they provide

evidence for the particulate/particle nature of electromagnetic radiation.

The Inability of the wave theory (classical physics) to explain the

photoelectric effect According to the wave theory of light the energy of the incident radiation is distributed uniformly over thewavefront. Thus when a metal is irradiated each electron in the surface would absorb an equal share of the radiant energy. Wave theory relates light intensity to the amplitude of the electric/magnetic fields that

make up the light wave. Wave theory’s predictions for the above observations are:

a) Electrons should be ejected from the metal at any incident light frequency, as long as the lightintensity is high enough, because energy is transferred to the metal regardless of the incidentlight frequency

b) The higher the light intensity the more electrons can absorb enough energy to escape the metalsurface so the higher the rate of emission will be.

c) There should be no relationship between the frequency of the light and the electron kineticenergy. The kinetic energy should be related to the intensity of the light. Increasing the light



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intensity means greater electric field amplitude, and the greater electric field should ejectelectrons with higher speeds (higher kinetic energies).

d) At low light intensities, a measurable time interval should pass between the instant the light isturned on and the time an electron is ejected from the metal. This time interval is required for theelectron to absorb the incident radiation before it acquires enough energy to escape from themetal.

In Summary:

Wave theory suggests that Supported or Contr adicted?

Any frequency can emit electrons

Contradicted!

Current (rate of electron emission)

depends on intensity

Supported

Maximum energy is independent of

frequency

Contradicted!

Maximum energy would depend onintensity

Contradicted!

Einstein’s explanation of the photoelectric effect

Study of black body radiation provided results that could not be predicted by classical physics.

The formulae worked for long wave radiation, but not short. The German physicist Max Planck (1858 – 1947) came to the conclusion that classical physics does not always apply at the

atomic level, and used a radical new idea to solve the problem.

He then proposed that energy was radiated in discrete energy packets called quanta! The

energy of a quantum of energy is given by

hf E Where h = Planck’s constant (= 6.63 10-34

Js)

f = the frequency of the radiation.

In 1905 Albert Einstein (1879 – 1955) extended the idea by suggesting that when a quantum of

energy is emitted by an atom it continues to exist as a concentrated packet of energy. The energyof the packets (called photons) each has energy hf . A beam of light was considered as a stream

of particles (photons). At long distances the intensity was low because the photons were spread

apart. However the energy of each photon was undiminished.

Einstein then went on to state that

When a photon collides with an electron, it must either be reflected with no loss of energy, or lose all its energy to the electron.

Each photon interacted with one electron only. So it provided energy for ONE electronto escape (The electron that the photon hits absorbs all this energy)

The intensity of a beam of light is proportional to the number of photons per unit cross-section of the beam per unit time. Therefore the number of electrons being emitted was



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proportional to the number of photons that landed on the surface (i.e. the intensity of the

radiation).

Electron cannot escape if the energy in the photon is not sufficient (Photon energy, E =hf ). Some of the energy provided by the photon is used by the electron to overcome theattractive forces of the nuclei’ so freeing it from the metal surface. The remainder of the

energy becomes the kinetic energy of the ejected electrons.o The energy of that photon must be greater than or equal to the energy used by the

electron to overcome the attractive forces of the nuclei (hf o) else no

photoelectrons will be produced. This frequency f o is the cutoff (or thresh hold)

frequency.

o KE max increases linearly with increasing frequency.

Electrons are emitted almost instantaneously, regardless of intensity, because the lightenergy is concentrated in packets rather than spread out in waves. If the frequency is

high enough, no time is needed for the electron to gradually acquire sufficient energy toescape the metal.

This is summarized by Einstein’s Photoelectric Equation

PHOTON ENERGY = MAXIMUM KINETIC ENERGY + WORK FUNCTION

Φmv 2

1 hf

2

Where hf = the energy of each incident photon of frequency, f ½ mv

2 = the maximum kinetic energy of the emitted electron

Φ = the work function of the metal surface = the minimum energy to take an electron

out of a metal against the attractive forces of the surrounding positive ions. It is directly linked to

the threshold frequency

o hf Φ

Threshold frequency, f o , is the minimum frequency at which electrons will be emitted.The maximum wavelength which corresponds to this threshold frequency is the cut-off

wavelength, λ o. Therefore we can write

o

c h Φ

where c is the speed of light.

The electron-volt

The electron-volt (eV) is a unit of energy. One electron-volt is the work done by an electron

when it moves through a potential difference of one volt. The charge on an electron is 1.6 × 10 –

19C, so by using W = Q × V, we have

1 eV = J.VC. 1919 106111061



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Example

The work function for a particular metal is stated as 5.10 × 10-19

J.

a) Calculate the minimum frequency of radiation that will emit photoelectrons from the metal.

Minimum frequency = f o = Hz 10 x697 10 x636

10 x105 14

34

19

..

.

h

o

b) If radiation of frequency 8.45 × 1014

Hz is incident on the same metal, calculate the

maximum kinetic energy gained by each photoelectron.

Photon energy E = hf = 6.63 × 10-34

× 8.45 × 1014

= 5.60 × 10

-19 J

Kinetic energy E k = E – Φo = 5.60 × 10-19

– 5.10 × 10-19

= 0.5 × 10

-19 J

= 5.0 × 10-20

J

c) Express the answers in b) in eV.1 eV = 1.6 × 10

– 19 J

Photon energy =5.60× 10-19

J × 1eV/1.6 × 10 – 19

J = 3.5 eV

Kinetic energy = 5.0 × 10-20

J × 1eV/ 1.6 × 10 – 19

J =0.31 eV

Experiment to demonstrate the Photoelectric effect

The effect can be shown with this experimental arrangement:



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The photocathode is made from a reactive metal such as caesium. A reactive metal loses its

outermost electrons most easily.

The experiment showed that if light had a lower frequency (i.e. longer wavelength) than the

threshold frequency, no current at all was observed, however bright the light was. Dim green or

blue light would work for caesium, bright red light will not.

Variation of intensity of the incident

radiation

For a frequency greater than the threshold

frequency, an increase in intensity (more photons)

will produce an increase in photoelectric emission.

The photoelectric current is directly proportional to

the intensity of the incident radiation.

Measuring Stopping Potential & Deducing EKmax and Work function

Using this apparatus we can measure the stopping potential, V s – the minimum potential which

reduces the photocurrent to zero :

The emitter gives out electrons. So we call it a cathode.

If the emf of the power supply is initially zero, the circuit works just like the one above this. As

the supply is turned up, the cathode becomes more positive (because it is connected to the

positive terminal of the supply).



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So electrons leaving it are attracted back towards it. If they leave with enough energy they can

overcome this attraction and cross to the anode. If they do not have enough energy, they cannot

cross the gap.

By increasing the emf of the supply you can find the pd at which no electrons are able to cross

the gap, even those with the maximum energy, Ekmax.

At that point, the energy needed to cross the gap = maximum Ek of the electrons.

Recall that, work done moving a charge through a pd is W = QV and in this case charge, Q = e,

the charge on an electron,

then 2

max S K mv

2

1 V e E

Using this information you can calculate the maximum energy and the maximum speed of the

photoelectrons emitted from the metal.

We can plot the data on a graph:

From EK = hf - we see that EK = h(f – f o).

We can plot the kinetic energy for several different metals against the frequency. We find thefollowing:

The gradient is the same, whatever the metals;

The gradient is always h, 6.63 10-34

Js; Each metal has a different value for its threshold frequency; The most reactive metals have the lowest threshold frequency; At the threshold frequency, E K = 0 (i.e. X-intercept = f o );

The intercept on the vertical axis gives the work function (Y-intercept = – ).

Different metals have different threshold

frequencies, as shown in this graph. Which

metal has the highest threshold frequency?

Which metal has the highest work function?

Which metal has the longest cut-off wavelength?

Rewrite the photoelectric equation



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S eV hf e

f e

hV S

The experimental results agreed with Einstein’s explanation. This experiment was good

evidence that electromagnetic radiation was particulate in nature.

Other resultsThis graph shows that the stopping potential, VS, is independent of

intensity.

To generate this graph, pd was

varied at two different light

intensities and the photocurrent

measured.

The frequency was kept constantand above the threshold value.

W AVE-P ARTICLE DUALITY

All experimental observations pertaining to the photoelectric effect are well explained by

Einstein’s photon model. This model provides good evidence that light (and electromagneticradiation in general) has particle nature.

However Young’s slits showed that light was a wave. And because light is a wave motion it can

be reflected, refracted, diffracted and produce interferences effects.

Light travels as a wave and shows wave behaviour. It is emitted and absorbed as photons. This

explanation is called Wave-Particle Duality of light.

Louis de Broglie (1892 – 1987), a Belgian Prince, wrote his PhD thesis in 1924 stating that if

light waves showed particle properties, it was very reasonable to state that particles should showwave properties. Any particle of mass m would have an associated wavelength that could beworked out by:



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The wave behaviour of electrons, which are, of course, particles, can be shown with this

experiment:

Electrons are diffracted at certain angles by a very thin layer of graphite to produce rings. The

ring spacing fits very well the model predicted by the Bragg Equation: = 2d sin

This equation was worked out by the father and son team of William and Lawrence Bragg. It

applied to diffraction of X-rays, which are, of course, electromagnetic waves. Therefore we cansay that electrons are showing wave-like properties.

We can find the momentum of the electrons easily enough:

Kinetic energy = electrical energy supplied = charge voltage

1/2 mv2

= eV

mv2 = 2eV (mv)

2= 2meV

p = mv = √ (2meV)

We can combine this with the de Broglie relationship:

A Young's slit type of experiment can be carried out with electron beams. And it is found that

there is an interference effect. If we regard an electron as merely a particle, it is difficult to see it passing through two places at once. But if we take it as having a wave property, then there is no

difficulty.

In 1927, Davisson and Germer did experiments which demonstrated electron diffraction patterns.

They used their results to calculate the electron wavelength and found a good agreement with the

de Broglie wavelength of electrons.



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Diffraction of electrons provides evidence for the wave nature of particles. Thus particles have a

dual nature also.

Emission spectra

An emission spectrum is the range of colours given out (emitted) by a light source. There aretwo kinds of emission spectra: continuous spectra and line spectra. To view spectra produced by various sources, a spectroscope or spectrometer can be used.

Continuous spectra

If a beam of white light from a tungsten filament lamp, say, is passed through a prism or grating,then it splits up to form a continuous spectrum of light from red through to violet. All

frequencies of radiation (colours) are present in the spectrum.

The continuous spectrum colours are red, orange, yellow, green, blue, indigo, violet.

A hot solid, liquid or gas at very high pressure has a continuous spectrum as shown:

There is energy at all wavelengths.

Line spectra

Some sources of light such as mercury and sodium vapour lamps do not produce continuous

spectra when viewed through a spectroscope. They produce line spectra – coloured lines spacedout by different amounts. Only specific, well-defined frequencies of radiation (colours) are

emitted.

For example, the spectrum of sodium has two, very bright, yellow lines close together as well assome other fainter lines. The yellow lines are known as the sodium doublet, or sodium D lines.

A gas at low pressure and high temperature will produce emission lines.



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There is energy only at specific wavelengths.

Explanation of emission spectra–

the Bohr model

Using the ideas of Planck and Einstein, Neils Bohr was able to extend the Rutherford model of

the atom. By suggesting that electrons are confined to certain orbits (or shells) around thenucleus, the Bohr model is able to explain emission spectra.

The electrons have different energies in different orbits.

There is a minimum number of electrons for each orbit.

Electrons tend to occupy the lowest available energy levels (closest to the nucleus). Theycan move between levels, but cannot stop between them.

It is easier to represent the orbits in the form of an energy level diagram.



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The lowest energy level E1 is called the ground state.

An electron which moves from its usual energy level to a higher energy level is said to be in an

excited state.

An electron needs to absorb the correct amount of energy to move up one or more levels. It will

later return to a lower level by emitting energy in the form of a photon.

The energy of the emitted photon is equal to the difference in energy between the two levels.

If the difference in energy levels is denoted as E, then the frequency of the emitted photon will

be expressed by the equation below.

E = hf where f = c/λ so that

The frequency and wavelength of the emitted photon is therefore determined by the magnitude of the energy change.



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The coloured arrows for the Balmer series

indicate that this series results in the emissionof visible light.

The emission spectrum of hydrogen shown

below includes four prominent lines that occur at wavelengths of 656.3 nm, 486.1 nm, 434.1

nm, and 410.2 nm, respectively. These are theBalmer lines.

The shorter wavelength black lines are part of the

Lyman series. They occur in the ultra-violet region of

the electromagnetic spectrum and cannot be seen by the

human eye.

Absorption spectraWhen light is passed through a medium containing a gas, then any photons of light which have

the same frequency as the photons emitted to produce the emission spectrum of the gas, are

absorbed by the gas. This is because the energy of the photons of light (hf ) is the same as theenergy difference required to cause an electron to be moved from the lower to the higher energy

level. The energy is then absorbed by the electron and that photon is ‘removed’ from the

incident light.

A gas at low pressure in front of a hot continuum causes absorption lines.

Dark lines appear on the continuous spectrum



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PRINCIPLES OF THE PRODUCTION OF X-RAYSX-rays are produced by bombarding metal targets with high-speed electrons. A typical spectrum

of the X-rays produced is shown in Figure1.

Figure 1

The spectrum consists of two components.There is a continuous distribution of wavelengths with

a sharp cut-off at short wavelength and also a series of high-intensity spikes that are

characteristic of the target material.

The Continuous Background: Whenever a charged particle is accelerated, electromagnetic

radiation is emitted. The greater the acceleration, the shorter is the wavelength of the emitted

radiation. When high-speed electrons strike a metal target, large decelerations occur and the

radiation produced is in the X-ray region of the electromagnetic spectrum. This radiation isknown as Bremmstrahlung radiation (which means braking radiation). Since the electrons have

a continuous distribution of decelerations, a continuous distribution of wavelengths of X-rays is

produced. There is a minimum wavelength (a cut-off wavelength) where the whole of the energy

of the electron is converted into the energy of one photon. That is,kinetic energy of electron = eV = hc / λ ,

where e is the charge on the electron that has moved through a potential difference V , h is the

Planck constant, c is the speed of light and λ is the wavelength of the emitted X-ray photon.

The relative intensity of the X-ray depends on the initial kinetic energy of the colliding electrons.

Electrons with higher initial kinetic energy produce spectra with shorter minimum wavelengths.Figure 2 shows examples of the spectra produced when different potential differences are used to

accelerate the electrons towards the target in an X-ray tube.



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Figure 2

The Line Spectrum (characteristic spectrum) consists of the sharp peaks that are observed.These peaks correspond to the emission line spectrum of the atoms of the target. The bombarding

electrons cause the innermost electrons of the target material to be ejected from their parent

atoms. This allows an electron in a high energy level to make a transition to this now emptylower energy level. The energy difference between the levels is equal to the energy of the

emitted X-ray photon. X-rays produced in this way have definite energies producing spectra

similar to the visible line spectra resulting from energy-level transitions by an atom’s outer electrons. These are called characteristic X-rays since they have energies determined by the

atomic energy levels in the target atom.

An Alternative explanation of how line spectrum is formed

The electrons in the beam that bombard the target collide with electrons in atoms of the target

metal. Electrons in low energy levels are excited to higher energy levels and X-ray photons areemitted when electrons fall to fill the "space" in the lower energy levels. The wavelengths at

which the peaks in the spectrum occur thus depend on the material of the target. These "lines" inthe spectrum are named after the energy level to which an electron falls, see Figures 3 and 4.

Figure 3



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Figure 4: Line spectrum superimposed on continuous spectrum.

X-ray Tubes

A simplified diagram of a modern form of X-ray tube is shown in Figure 5.

Figure 5: Diagram of an X-ray tube

A tungsten filament cathode is heated to around 2000 °C; this causes it to emit electrons throughthe process called thermionic emission. The electrons are accelerated through a large potential

difference (20 kV → 100 kV for diagnosis) before bombarding a metal anode.

The X-rays produced leave the tube via a ‘window’. Since the majority of the energy of theelectrons in transferred to thermal energy in the metal anode, the anode is either water-cooled or

is made to spin rapidly so that the target area is increased. The anode is held at earth potential.

The intensity of the X-ray beam is determined by the rate of arrival of electrons at the metal

target, that is, the tube current . This tube current is controlled by the heater current of thecathode. The greater the heater current, the hotter the filament and hence the greater will be the

rate of emission of thermo-electrons.

The hardness of the X-ray beam (the penetration of the X-rays) is controlled by the

accelerating voltage between the cathode and the anode. More penetrating X-rays have higher

photon energies and thus a larger accelerating potential is required. Referring to Figure 1, it can



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be seen that longer wavelength X-rays (‘softer’ X-rays) are always also produced. Indeed some

X-ray photons are of such low energy that they would not be able to pass through the patient.

These ‘soft’ X-rays would contribute to the total radiation dose without any useful purpose.Consequently, an aluminium filter is frequently fitted across the window of the X-ray tube to

absorb the ‘soft’ X-ray photons.

Use of X-rays in imaging internal body structuresX-ray radiation affects photographic plates in much the same way as visible light. A photographic plate, once exposed, will appear blackened after development. The degree of

blackening is dependent on the total X-ray exposure.

X-ray photons also cause fluorescence in certain materials. The mechanism is similar to that by

which visible light is produced on the screen of a cathode-ray oscilloscope.

X-ray beams are used to obtain ‘shadow’ pictures of the inside of the body to assist in thediagnosis or treatment of illness. If a picture is required of bones, this is relatively simple since

the absorption by bone of X-ray photons is considerably greater than the absorption bysurrounding muscles and tissues. X-ray pictures of other parts of the body may be obtained if

there is sufficient difference between the absorption properties of the organ under review and thesurrounding tissues.

The quality of the shadow picture (the image) produced on the photographic plate depends on itssharpness and contrast. Sharpness is concerned with the ease with which the edges of structures

can be determined. A sharp image implies that the edges of organs are clearly defined. An image

may be sharp but, unless there is a marked difference in the degree of blackening of the image

between one organ and another (or between different parts of the same organ), the informationthat can be gained is limited. An X-ray plate with a wide range of exposures, having areas

showing little or no blackening as well as areas of heavy blackening, is said to have goodcontrast.

In order to achieve as sharp an image as possible, the X-ray tube is designed to generate a beam

of X-rays with minimum width. Factors in the design of the X-ray apparatus that may affect

sharpness include the area of the target anode, as illustrated in Figure 6,

Figure 6



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the size of the aperture, produced by overlapping metal plates, through which the X-ray

beam passes after leaving the tube (see Figure 7),

Figure 7

the use of a lead grid in front of the photographic film to absorb scattered X-ray photons,

as illustrated in Figure 8.

Figure 8

In order to improve contrast, a ‘contrast medium’ may be used. For example, the stomach may

be examined by giving the patient a drink containing barium sulphate. Similarly, to outline bloodvessels, a contrast medium that absorbs strongly the X-radiation would be injected into the

bloodstream.

The contrast of the image produced on the photographic film is affected by exposure time, X-ray

penetration and scattering of the X-ray beam within the patient’s body. Contrast may beimproved by backing the photographic film with a fluorescent material.

Attenuation of X-rays in matterThe energy of an X-ray beam radiates from the source in all directions in a vacuum. Its intensity

decreases in proportion to the inverse of the square of the distance from the source. This is a

consequence of the energy being ‘spread’ over the surface of a sphere of radius r having surface

area 4πr 2. Thus, in a vacuum, I I o/r

2. The law also applies approximately to X-rays in air since

there is little absorption of X-rays by air.

In a medium where absorption processes are occurring, the intensity I of a parallel beam

decreases by a constant fraction in passing through equal small thicknesses of the medium. This

gives rise to an exponential decrease in the intensity of the transmitted beam. For a parallel beam

of radiation of initial intensity I o passing through a thickness x of a medium, then the transmittedintensity I is given by

I = I 0exp ( –μx ),



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where μ is a constant for the medium that is dependent on photon energy. The unit of μ is mm – 1

or cm – 1

or m – 1

. μ is referred to as the linear absorption coefficient or linear attenuation

coefficient .

The variation with thickness x of an absorber of the percentage transmission of a parallel beam of

X-ray radiation is illustrated in Figure 9.

Figure 9

The thickness of the medium required to reduce the transmitted intensity to one half of its initial value is a constant and is known as the hal f-value thickness x ½ or HVT . The half-value

thickness x½ is related to the linear absorption coefficient μ by the expression x ½ × μ = ln2. In practice, x½ does not have a precise value as it is constant only when the beam has photons of

one energy only.

Computed (Axial) Tomography or CAT scanningThe image produced on an X-ray plate is a ‘flat image’ and does not give any impression of

depth. That is, whether an organ is near to the skin or deep within the body is not apparent.

Tomography is a technique by which an image of a slice, or plane, of the object may be obtained.In this technique, a series of X-ray images are obtained from different angles through a slice but

in the plane of the slice, as illustrated in Figure 10.



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+

++

+

-

+

+

+

++ -

-

-

-

--

--

-

-

-

Figure 10

Computer techniques make it possible to combine these images to give an image of the slice. The

technique is called computed (axial) tomography or CAT scanning .

The CAT scanner is shaped like a doughnut. The X-rays beam moves around the patient,

scanning from hundreds of different angles. The computer takes the information to build an

image up. The patient lies down on a platform, which slowly moves through the hole in themachine. The X-ray tube is mounted on a moveable ring around the edges of the hole. The ring

also supports an array of X-ray detectors directly opposite the X-ray tube. A motor turns the ring

so that the X-ray detectors revolve around the body. Each full revolution scans a narrow,horizontal “slice” of the body. The control system moves the platform farther into the hole so the

tube and detectors can scan the next slice.

Images of successive slices can be combined to give a three-dimensional image. The three-

dimensional image can be rotated and viewed from any angle on the computer. ________________________________________________________________

3.2 STRUCTURE OF THE ATOM

The plum pudding model of the atom was proposed by J.J Thomson. It states that the atom is

composed of electrons surrounded by a positively charged soup or cloud which balances the

negative charge on the electrons, like negatively charged plums surrounded by a positively

charged pudding.

The plum pudding model

(J. J. Thomson)

Figure 1.1



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According to Thomson the electrons are arranged non-randomly, in rotating rings throughout the

cloud of positive charge.

The Geiger- Marsden Experiment

This was carried out by Hans Geiger and Ernest Marsden working under the supervisionof Ernest Rutherford.

Apparatus:

Alpha particles were directed onto a thin gold foil which was placed in the centre of anevacuated vessel as shown in the apparatus above

They used a radon tube in a lead case as a source of particles and limited the particlesto a narrow beam

The scattering of the particles after passing through the foil was observed on a fluorescent

screen. Scintillations were seen whenever the screen was struck by particlesResults

Majority of the particles passed through undeflected

A few were deflected through large angles

Even fewer bounced back

Conclusion

The gold atoms are mostly empty space occupied by light electrons

Most of the mass and positive charge is concentrated in a small core called the nucleus Electrons are found in the space outside the nucleus.



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Rutherford pictured the atom as a tiny solar system in which the nucleus plays the role of the

Sun, and the electrons the role of orbiting planets.

________________________________________________________________________

To describe the atom, we use the standard nuclide notation, for example:

The top number (A) is called the nucleon number (as it is the number of things in thenucleus of the atom – protons and neutrons) or the mass number (as it is the mass of the

atom).

The bottom number (Z) is called the proton number (as it is the number of protons) or the

atomic number (as it is the number that tells you which element the atoms belongs to).

The letter (X) is the symbol of the element.The relationship between the mass number and the nucleons may be represented by:

NZA (i.e. Mass number = Atomic number + Neutron number)

IsotopesThe number of protons in an atom is crucial. It gives you the charge of the nucleus and

therefore it gives you the number of electrons needed for a neutral atom. And the number of

electrons governs how an atom behaves and reacts chemically with other atoms. In other words,it gives you its properties. So the number of protons makes the atom belong to a particular

element. Change the number of protons and you change the element.

The number of neutrons in the nucleus is less crucial. You can change the number of neutrons

without changing the chemical properties of the atom. So it behaves in the same way. Atomswith the same proton number but different numbers of neutrons are called isotopes .

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Energy Levels

All charged particles emit radiation when they are accelerated.

Electrons orbiting nuclei do not.Bohr suggested that electrons could travel in certain orbits without losing energy, those orbits are

called energy levels.

Evidence for his theory came from emission spectra and can be observed using vapours that haveelectric currents passed through, using a diffraction grating . A series of bright emission lines areobserved at certain frequencies for different elements.

Spectral lines of atomic hydrogen:



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These bright emission lines are caused by electrons moving from higher energy levels to lower energy levels emitting photons of light in the process.

Energy levels in a hydrogen atom:



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The energy is expressed as a negative number because the electron is bound to the atom. Energy

needs to be transferred to the electron to free it from the atom. At infinity the energy possessed

by the (free) electron is zero.

Energy of emitted photon = E1 – E2 = ΔE = hf or hf = E1 – E2

Electrons can only move between these allowed energy levels, they can only gain or lose specific

amounts of energy (i.e., the energy is quantized).

Different elements have different energy levels and, therefore, have different spectral lines.

Millikan’s Oil Drop Experiment

Robert Millikan used a simple but famous experiment that served to confirm the unit electronic

charge as 1.6 × 10-19 C. He sprayed oil drops into a space between two charged plates. Eachtiny oil droplet was charged up by friction as it left the sprayer. The theory was simple; theattractive electrostatic force between the droplet and the positively charged plate would balance

out the weight of the droplet. His apparatus was like this:

He would select a particular oil drop and hold it stationary by altering the voltage between the

two plates.



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Question 1 What are the forces acting on the plates? What is the resultant force?

Gravity acts downwards ()

Force of electrostatic attraction acts upwards ()

The resultant force is zero ()

Because the two forces are equal and opposite. ()The forces on the stationary drop are like this:

We know that:the electric force = electric field × charge ( F = Eq)

the electric field strength in a uniform field, E = V/d

It doesn't take a genius to see that:

Question 2 How might you find the weight of the drop? We could measure the diameter of the sphere ()

And from that work out the volume. ()

Then use mass = density × volume to get the mass ()

Then multiply the mass by g to get the weight ().

This is not a very satisfactory way because the uncertainty is too great. So another method was

used. Millikan turned off the plates and watched the oil drop. Very quickly the oil drop reached

terminal speed.

Question 3 What forces are acting on an object at terminal speed? What is the resultant force?



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Weight acting downwards ()

is balanced ()

By viscous drag acting upwards. (

)

Resultant force is zero. ()

So we can write:

mg = drag force

The drag force can be worked out indirectly using Stoke's Law, which describes the force

acting on a sphere falling at terminal speed through a viscous fluid. Normally we think of

viscous fluids as thick gooey oils; for a tiny oil droplet, air is pretty viscous. The equation thatdescribes Stoke's Law is:

[r - radius of the sphere, v - terminal speed (m/s).]

The strange looking symbol, , is "eta", a Greek lower case letter long 'ē', the Physics Code for

the coefficient of the viscosity of a fluid. The units for are N s m-2

. For air, = 1.8 × 10-5

N sm

-2.

In question 2 we looked at the intuitive way of finding the weight by:

Weight = density volume g

We can write this as:

So we can bring in the Stoke’s Law equation in by writing:

Rearranging and cancelling out gives us:



26

So by observing the terminal speed, we can work out the radius. From that, we can work out

the volume, hence the mass and weight. Although it seems long-winded, this method produces

much less uncertainty than attempting a direct measurement of the radius.

The important finding from this experiment was that the charge was always a whole number

multiple of 1.6 10

-19

C. Therefore the electron charge came in definite amounts (quanta);therefore it is said to be quantised. Electrons are indivisible, i.e. fundamental particles of matter.

SummaryWeight of the oil drop is balanced by force from electric field.

The radius is too small to be measured directly.

Oil drop is allowed to fall (and quickly reaches its terminal speed).

Speed is measured.

Stoke’s Law is used to measure radius by viscous drag.

Radius is used to work out the volume

Mass is found by volume density

Weight is determined.

Weight = force due to electric field.

Eqmg or qd

Vmg

V

dmgq

3.3 NUCLEAR ENERGY Mass Defect of the Nucleus and Nuclear Binding Energy The nucleons are held tightly together in a stable nucleus. The energy required to separate it into

its constituent protons and neutrons is called the binding energy of the nucleus.



27

In Einstein’s theory of special relativity, mass and energy are equivalent. A change in

mass m is equivalent to a change in energy E0 of the rest mass of the system.2

mcE

Where: c is the speed of light in a vacuum,

E is the binding energy

Therefore separated nucleons have greater masses than when bound to form a nucleus.

The difference in mass m is known as the mass defect of the nucleus.

Mass defect of the nucleus = mass of the separated nucleons mass of the nucleus

Obviously, the bigger the nucleus you make the more energy you get out (every time you addanother proton or neutron (smack!) you get more mass going missing). So the bigger nuclei have

bigger mass defects. But if you look at the binding energy per nucleon involved in the nucleus,

you notice that certain atoms have higher values than others.

Top of the tree is iron. Each nucleon has to give out a larger proportion of itself (and therefore

convert more of itself into energy) to make iron than to make any other nucleus. That means that

you have to put in more energy per nucleon to break up iron than any other nucleus. It is

therefore the most stable nucleus.

Binding Energy Per Nucleon

Binding Energy per Nucleon = Binding Energy

Mass Number

It is a measure of the stability of a nucleus

Graph of binding energy per nucleon against mass number

+ Binding Energy

Nucleus

(small mass)

Protons and Neutrons

(greater mass)



28

The nuclides of intermediate mass number have the largest values for the binding energy per

nucleon and the most stable is Fe56

26 has a value 8.8 MeV. It is the most stable since it needs the

most energy to disintegrate.

The smaller values of binding energy per nucleon for higher and lower mass numbersimply that they are potential sources of nuclear energy.

The nuclides with low mass numbers can produce energy by fusion, when two light nuclei have

fused to produce a heavier nucleus.

In contrast heavier nuclides (i.e. higher mass numbers) can produce energy by fission(disintegration) of their nuclei into two lighter nuclei.

In both cases, nuclei are produced having greater binding energy per nucleon and therefore

1. more stable nuclei are formed 2 there is consequently a mass transfer (mass loss) during their formation

Unified Atomic Mass UnitIf another unit of energy is needed, then one may use the unit of mass since mass and energy are

interchangeable. The unified atomic mass unit (u) is defined as 1/12th of the mass of the carbon-

12 atom C12

6 . That is 1u = 1.66 10-27

kg.

Using 2mcE and JMeV

13106.11 , the unified atomic mass may be approximated as

follows:

MeVu 9311 Consider as an example, the helium nucleus He

4

2 . This has 4 nucleons, 2 protons and 2 neutrons.

The mass of a proton is 1.0073u and the mass of a neutron is 1.0087u.

total mass of 2 proton plus 2 neutron = 2 1.0073 + 2 1.0087 = 4.0320u.But the helium nucleus has a mass of 4.0015u.

Binding energy = mass difference of nucleons and nucleus = 4.0320 – 4.0015

= 0.0305u = 0.0305u 931MeV = 28.4MeV

The binding energy per nucleon is the binding energy divided by the total number of nucleons. In

this case it is 28.4/4 or 7.1MeV.

NUCLEAR FISSION. This is disintegration of heavy nuclei into two lighter nuclei. This decreases nucleon potential energy and

releases it in other forms.

energyn3Kr BanU 1

0

92

36

141

56

1

0

235

92

This type of reaction takes place in nuclear reactors.

Use the Data given and calculate the energy released in from this reaction.



29

U235

92 =235.0439299u n1

0= 1.0086649u Ba

141

56 = 140.88589u Kr 92

36 = 91.926156u

NUCLEAR FUSION: is combining of two light nuclei to form heavy nucleus.

This decreases the average nucleon potential energy (andincreases the binding energy per nucleon), and the PE lost

is emitted in other forms such as gamma-ray photons or kinetic energy of particles.

energynHeHH 1

0

3

2

2

1

2

1

This type of reaction takes place on the Sun. Very high temperatures are required tomake the reaction

Use the Data given and calculate the energy released in from this reaction.

2H = 2.01412u,

3He = 3.01605u &

1n = 1.00867u

You can use the Binding Energy curve to calculate the amount of energy released by a reaction by considering the total binding energy of all the nucleons involved before the reaction

(remember – the curve give energy values per nucleon, not per nucleus) and compare it to the

total binding energy of all the nucleons after the reaction. The difference is the energy released.

Note –

The amount of energy released in a single nuclear reaction is equal to the increase in

binding energy (reduction in mass) that occurs. Hence the amount of energy

released per unit mass of ‘fuel’ is equal to the increase in binding energy per unit. The nuclear reactions which will be most efficient at generating energy will be those

that result in the biggest increase in binding energy per nucleon. 3.4 RADIOACTIVITY

Radioactivity refers to the spontaneous and random decomposition of unstable atoms. Three

types of radiation are emitted from the nuclei of unstable atoms. They are alpha (α) particles, beta (β) particles and gamma (γ) rays.

The stability of the nucleus is dependent upon the neutron-to-proton ratio. As Z increases the

number of neutrons needed also increases but not in a linear relationship. When a graph of

neutron number versus proton number is plotted for all elements, the stable nuclides fit into a

region called the band of stability.



30

Notes to Consider:

1. All nuclides with Z>83 are unstable

2. Z values lower than 20 haveneutron/proton = 1

3. As Z increases above 20, the

neutron/proton ratio increases for example90

Zr = 1.25,120

Sn = 1.4,200

Hg = 1.5

4. Region above the band represents excess

neutrons

5. Region below the band represents excess

protons

Elements above the band have high neutron/proton ratios and will undergo beta Emission to

gain stability. Beta emission increases Z while ‘A’ remains the same.

Elements with Z > 83 are all unstable. Alpha particles are emitted to reduce both ‘A’ and Z andform more stable atoms.

Elements below the band of stability have low neutron/proton ratios and will undergo Positron

emission or electron capture to become stable. Both positron emission and electron capture

decrease Z while ‘A’ remains the same.

The main features of these radiations are outlined in the following table:

particles -particles - Radiation

Contain He²+ nuclei Contain stream of high-energy

electrons similar to cathode ray.

E.M radiation of very

short wavelength

Have a Range of 3 – 5 cm in air.Most are stopped by ordinary

thickness of paper or by a very

thin sheet of aluminium.

Have a range of about 15 cm inair. They are able to penetrate

several mm thickness of

aluminium. Stopped by lead or thick aluminium.

Have Infinite range in air.Their intensity is greatly

reduced (i.e., they are

absorbed) by thick lead.

Speed vary up to 7% of light. Speed vary up to 99% that of light.

Speed of light

Least penetrating. More penetrating than -

particles.

Most penetrating

Most ionising. More ionising than rays Least ionising

Deflected by electric and

magnetic field

Deflected by electric and

magnetic field

Not deflected by electric

or magnetic field.



31

Effect of Electric and Magnetic fields

Electric FieldsThe effect of the field depends on the charge of the radiation.

Alpha particles are positively charged and are therefore attracted to the negative plate in an

electric field.Beta particles are negatively charged and are therefore attracted to the positive plate in anelectric field.

Gamma rays are unaffected.

Figure: Effect of electric field on nuclear radiation

Magnetic Fields

Use Fleming’s Left Hand Rule (see Electromagnetism) to predict behaviour in magnetic fields.

The “current” (second finger flow of charged particles) is the beam of radiation. Remember, thesecond finger shows conventional current flow so for beta particles point it in the reverse

direction to the beam. For alpha particles it points in same direction as beam. Gamma rays have

no charge so experience no force.

Figure: Effect of magnetic field on nuclear radiation



32

Radioactive DecayRadioactive Decay Equations These equations are actually based on a few very simple ideas.

1) Radioactive decay is a spontaneous event and cannot be predicted. However, if there are

many of these decays, then the rate of decay can be predicted using statistics as described in the

decay equations.2) The number of decays you will measure each second (the activity, A) from a sample depends

on the number of atoms in the sample, N. Note that the activity of a sample is measured in Becquerel, Bq.

- Decay

The Parent nucleus has lost two (2) protons and two (2) neutrons or moves two (2) up the

periodic table.

HeYX 4

2

4A

2Z

A

Z

Parent Daughter particle

nucleus nucleus nucleus

HeRnRa 4

2

222

86

226

88

- Decay

a neutron get converted into a proton and -particle is emitted

eYX 0

1

A

1Z

A

Z

Parent Daughter particle

nucleus nucleus nucleus

0

1

24

12

24

11 Mg Na



33

- Decay

The nucleus, like the orbital electrons, exists only in discrete energy states or levels. When anucleus changes from an excited energy state (denoted by an asterisk *) to a lower energy state, a

photon is emitted.

XX AZ

*AZ

CoCo 60

27

*60

27

The Activity (A) (or rate of disintegration) of a given unstable element is proportional to thenumber of atoms N present at the time

NAor Nt

NA

Nt

NA

is called the decay constant.

The S.I. unit of Activity (A) is the Becquerel (Bq) 1 Bq = 1 disintegration per sec

Activity is also measured in Currie1 C = 3.7 1010

s-1

If the N0 is the number of radioactive atom present at time t=0 and N the number at the time t,then it can be shown by using calculus that

N = No et

Since the activity A N a similar expression can be written

A = A0 et

The activity is proportional to the count rate per second, which can be denoted as C or R

R = R 0 et

A general expression then would be

The decay X = X0 et

State that a radioactive substance service decay exponentially - particle and -ray are emitted with a single energy or with a small number of discrete valuesof energy.

Half Life (T1/2)

This is time taken for the Number atoms present decrease to half of it initial value.

This is time taken for the Activity to decrease to half of its initial value.



34

This is time taken for the Mass to decrease to half of its initial value.

This is time taken for the Count Rate to decrease to half of its initial value.

This is time taken for the Activity to decrease to half of its initial value.

Starting with N = N0 e-t

At time t = 0, N = N0

At time t = T½ (half-life), N = ½N0

2lnT

T2ln

eln2

1ln

e2

1

e N N2

1

21

21

T

T

T

00

21

21

21



35

Radiation Detectors

1. The Spark CounterAn early form of detector, the Spark Counter is another instrument that uses the ionizing

effect of radioactivity, and for this reason it works best with particles.

A high voltage is applied between the gauze and the wire, and adjusted until it is just belowthe voltage required to produce sparks. When a radioactive source is brought near, the air

between the gauze and the wire is ionised, and sparks jump where particles pass

2. The Geiger Counter

The ionising effect of radiation is used in the Geiger-Muller (GM) tube as a means of

detecting the radiation. The GM tube is a hollow cylinder filled with a gas at low pressure.The tube has a thin window made of mica at one end. There is a central electrode inside the

GM tube. A voltage supply is connected across the casing of the tube and the central

electrode as shown in the following diagram. Two diagrams are presented, choose the oneyou like:

When an alpha or beta or gamma radiation enters the tube it produces ions in the gas. The

ions created in the gas enable the tube to conduct. A current is produced in the tube for ashort time. The current produces a voltage pulse. Each voltage pulse corresponds to one

ionising radiation entering the GM tube. The voltage pulse is amplified and counted.



36

3. The Scintillation counter

Scintillation Detectors" work by the radiation striking a suitable material (such as Sodium

Iodide), and producing a tiny flash of light. This is amplified by a "photomultiplier tube"which results in a burst of electrons large enough to be detected. Scintillation detectors form

the basis of the hand-held instruments used to monitor contamination in nuclear power

stations. They can recognise the difference between a, b and g radiation, and make differentnoises (such as bleeps or clicks) accordingly.

4. Cloud Chamber:Consists of a gas at its liquefaction point. The gas condenses to leave a

trace of the particles path. There are two types of cloud chamber: the "expansion" type and

the "diffusion" type. In both types, or particles leave trails in the vapour in the chamber,similar to how high-altitude aircraft leave trails in the sky. The chamber contains a

supersaturated vapour (e.g. methylated spirits), which condenses into droplets when

disturbed and ionised by the passage of a particle (alpha particles are best for this).

You can clearly see the direction and energy of the particles (low energy particles only leave

short trails). Occasionally, a particle collides with an air molecule and changes direction. Acloud chamber also shows the randomness of radioactive emissions clearly.

5. Bubble Chamber: Consists of a liquid just at its boiling point. In a superheated liquid,

bubbles start to form on nucleation centres - surface irregularities, dirt or positive ions - inthe liquid. Bubbles can therefore form along the track of a charged particle, and make it

visible.

A similar idea to an expansion cloud chamber, particles leave trails of tiny bubbles in aliquid. This used to be the main instrument for tracking the results of collisions in particle

accelerators.

The chamber would be surrounded by powerful magnets, so any charged particles passingthough the chamber would move in curved paths. The shapes of the curves tell us about the

charge, mass and speed of each particle, so we can work out what they are - otherwise one

line of bubbles looks pretty much like another.

Particle tracks in a bubble chamber

6. Photographic FilmIn 1896, Henri Becquerel, working in Paris, discovered that Uranium compounds would

darken a photographic plate, even if the plate were wrapped up so that no light could get in.

Radioactivity will darken ("fog") photographic film, and we can use this effect tomeasure how much radiation has struck the film. Workers in the nuclear industry wear

"film badges" which are sent to a laboratory to be developed, just like your photographs.

This allows us to measure the dose that each worker has received (usually each month).

http://www.darvill.clara.net/nucrad/people.htm#becquerel






37

The badges have "windows" made of different materials, so that we can see how much of theradiation was alpha particles, or beta particles, or gamma rays

Practical Activity: Radioactive Decay of Thoron Gas

Aim :To measure the half-life of Thoron gas.

Method : Thoron gas is an isotope of radon Rn86

220produced in the radioactive series that starts

with a long half-life isotope of thorium Th90

232. All the other nuclides in the series have

half-lives either much longer or much shorter than thoron gas so they do not contribute to the

activity of the sample of the gas. The thorium is in powdered form in a sealed plastic bottle

and the thoron gas is produced in the air space above the powder.

Set the scaler timer to “count”. Find the background count -rate by switching on the counter for 100s.

This value is used to correct the count-rates in the thoron decay.

Using two-tubes with one-way valves, the radon gas can be transferred into a bottle containing the end

of a Geiger-Muller tube by squeezing the thorium bottle a few times. The whole system is sealed and

should be quite safe but to make sure, keep all the windows open and if any leak occurs, evacuate the

room, and report to your teacher or the laboratory technician immediately.

When the gas is transferred, switch on the counter and start timing. Record the count every 20

seconds for about 5 minutes. From these readings the number of decays in each 20s interval can be

found and hence the count-rate at 10s, 30s, 50s.

Plot a graph to show how the activity varies with time and use the graph to obtain a value for the half-

life of the radon.

The activity of the thoron (radon-220) will decay exponentially so you should be able to derive an equation

suitable for plotting a linear graph from which the half-life may also be found. By selecting the part of the graph

before the decay gets too random a more precise value than the first one may be obtained.

particulate nature of light

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