chapter 4 atomic structure - universitetet i...
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Radio city broadcasts on a frequency of 5,090 KHz.What is the wavelength of electromagnetic radiation emitted by the transmitter?
Illustrative problem 1
cλ =
ν8
3
3 105090 10
×λ =
×20.589 10= ×
58.9 m=
s/m103isc 8×
The ratio of the energy of a photon of 2000 Å wavelength radiation to
that of 6000 Å radiation is
(a) ¼ (b) 4
(b) ½ (d) 3
Illustrative Problem 2
hcE h= ν =λ 1 2
1 2
hc hcE E= =λ λ
1 2
2 1
EE
λ∴ =
λ
0
0
6000 A 32000 A
= =
Hence, answer is (d).
Solution:
Illustrative Problem 3
The energy of the electron in the second and third Bohr orbits of the hydrogen atom is -5.42 X 10-12 and –2.41 X 10-12 respectively. Calculate the wavelength of the emitted radiation, when the electron drops from third to second orbit.
Solution 2 1E E E∆ = −
According to Planck’s quantum theory
12 125.42 10 ( 2.41 10 )ergs− −= − × − − ×123.01 10 ergs.−= − ×
E h= υ
cE h=λ
27h 6.6 10 ergs−= ×
Solution
56.6 10 cm−λ = ×
3 06.6 10 A= ×
0 81A 10 cm−=
27 8
126.62 10 3 10
3.01 10
−
−× × ×
=×
chE
∴ =λ
Class Exercise - 1 Which of the following fundamental particles are present in the nucleus of an atom? (a) Alpha particles and protons (b) Protons and neutrons (c) Protons and electrons (d) Electrons, protons and neutrons
Solution
The nucleus of an atom is positively charged and almost the entire mass of the atom is concentrated in it. Hence, it contains protons and neutrons.
Hence, answer is (b).
Class Exercise - 2 The mass of the proton is (a) 1.672 × 10–24 g (b) 1.672 × 10–25 g (c) 1.672 × 1025 g (d) 1.672 × 1026 g
Solution
Hence, answer is (a).
The mass of the proton is 1.672 × 10–24 g
Class Exercise - 3 Which of the following is not true in case of an electron? (a) It is a fundamental particle (b) It has wave nature (c) Its motion is affected by magnetic field (d) It emits energy while moving in orbits
Solution
Hence, answer is (d).
An electron does not emit energy while moving in orbit. This is so because if it would have done that it would have eventually fallen into the nucleus and the atom would have collapsed.
Class Exercise - 4 Positive charge of an atom is (a) concentrated in the nucleus (b) revolves around the nucleus (c) scattered all over the atom (d) None of these Solution
Hence, answer is (a).
Positive charge of an atom is present entirely in the nucleus.
Class Exercise - 5 Why only very few a-particles are deflected back on hitting a thin gold foil?
Solution
Due to the presence of a very small centre in which the entire mass is concentrated.
Class Exercise - 6 Explain why cathode rays are produced only when the pressure in the discharge tube is very low.
Solution
This is happened because at higher pressure no electric current flows through the tube as gases are poor conductor of electricity.
Class Exercise - 7 If a neutron is introduced into the nucleus of an atom, it would result in the change of (a) number of electrons (b) atomic number (c) atomic weight (d) chemical nature of the atom Solution
Hence, answer is (c).
Neutrons contribute in a major way to the weight of the nucleus, thus addition of neutron would result in increase in the atomic weight.
Class Exercise - 8 The concept of stationary orbits lies in the fact that (a) Electrons are stationary (b) No change in energy takes place in stationary orbit (c) Electrons gain kinetic energy (d) Energy goes on increasing
Solution
Hence, answer is (b).
When an electron revolves in a stationary orbit, no energy change takes place. Energy is emitted or absorbed only when the electron jumps from one stationary orbit to another.
Class Exercise - 9 What is the energy possessed by 1 mole of photons of radiations of frequency 10 × 1014 Hz?
Solution
E = hv E = 6.6 × 10–34 × 10 × 1014
E = 66 × 10–20 = 6.6 × 10–19 joules
∴ energy of 1 mole of photons = 6.6 × 10–19 × 6.023 × 1023
= 39.7518 × 104
= 397.518 kJ/mol
17
The Atomic Weight Scale and Atomic Weights
isotopeCu isotopeCu 6563
amu) .9(0.309)(64 amu) .9(0.691)(62 weight atomic +=
18
The Atomic Weight Scale and Atomic Weights
copperfor amu 63.5 weight atomic
amu) .9(0.309)(64 amu) .9(0.691)(62 weight atomicisotopeCu isotopeCu 6563
=
+=
Atomic Number Atomic number (Z) of an element is the number of protons in the nucleus of each atom of that element.
Element # of protons Atomic # (Z)
Carbon 6 6
Phosphorus 15 15
Gold 79 79
Question 2 The table on the next slide shows the five isotopes of germanium found in nature, the abundance of each isotope, and the atomic mass of each isotope.
The wavelength of the peak in the intensity graph is given by Wien’s law (T must be in kelvin):
Wien’s Displacement Law
The Electron Volt • Consider an electron
accelerating (in a vacuum) from rest across a parallel plate capacitor with a 1.0 V potential difference.
• The electron’s kinetic energy when it reaches the positive plate is 1.60 x 10−19 J.
• Let us define a new unit of energy, called the electron volt, as 1 eV = 1.60 x 10−19 J.
Example 1 Write out the electron configuration of the elements (i) carbon (Z = 6) (ii) magnesium (Z = 12) (iii) argon (Z = 18) Solution (i) 1s2 2s2 2p2 (ii) 1s2 2s2 2p6 3s2 (iii) 1s2 2s2 2p6 3s2 3p6 Example 2 Determine which of the following are in the excited state (i) 1s2 2s2 2p5 3s1 (ii) 1s2 2s2 2p6 3s2 3p6 4s1 (iii) 1s2 2s2 2p3 (iv) 1s2 2s2 2p6 3s1 3p5 Solution (i) and (iv) are in excited states
Mass Number Mass number is the number of protons and neutrons in the nucleus of an isotope: Mass # = p+ + n0
Nuclide p+ n0 e- Mass #
Oxygen -
10
- 33 42
- 31 15
8 8 18 18
Arsenic 75 33 75
Phosphorus 15 31 16
Information in the Periodic Table
The number at the bottom of each box is the average atomic mass of that element.
This number is the weighted average mass of all the naturally occurring isotopes of that element.
Question 1 How does the atomic number of an element differ from the element’s mass number?
Answer The atomic number of an element is the number of protons in the nucleus. The mass number is the sum of the number of protons and neutrons.
The Atomic Weight Scale and Atomic Weights
The atomic weight of an element is the weighted average of the masses of its stable isotopes
Example 5-2: Naturally occurring Cu consists of 2 isotopes. It is 69.1% 63Cu with a mass of 62.9 amu, and 30.9% 65Cu, which has a mass of 64.9 amu. Calculate the atomic weight of Cu to one decimal place.
Calculating Atomic Mass
Copper exists as a mixture of two isotopes.
The lighter isotope (Cu-63), with 29 protons and 34 neutrons, makes up 69.17% of copper atoms.
The heavier isotope (Cu-65), with 29 protons and 36 neutrons, constitutes the remaining 30.83% of copper atoms.
Calculating Atomic Mass
The atomic mass of Cu-63 is 62.930 amu, and the atomic mass of Cu-65 is 64.928 amu.
Use the data above to compute the atomic mass of copper.
Calculating Atomic Mass First, calculate the contribution of each isotope to the average atomic mass, being sure
to convert each percent to a fractional abundance.
Calculating Atomic Mass
The average atomic mass of the element is the sum of the mass contributions of each isotope.
Can you tell me: how many atoms are there in 43.2 grams of iron? Using unit analysis: The unit of the answer is atoms. The unit of the information given is grams. But there is no conversion factor for grams ---> atoms. Reason: the mass of an atom is different for every element. Solution: use the mole concept.
43.2g Fe x 1 mol Fe55.85 g Fe
x 6.02x1023 atoms Fe1 mol Fe
= 4.66x 1023 atoms Fe
Electron Orbits Consider the planetary model for electrons which move in a circle around the positive nucleus. The figure below is for the hydrogen atom.
Coulomb’s law:
2
204C
eFrπε
=
Centripetal FC:
2
2CmvFr
=
2 2
204
mv er rπε
= Radius of Hydrogen atom 2
204er
mvπε=
FC +
-
Nucleus
e- r
Example 1: Use the Balmer equation to find the wavelength of the first line (n = 3) in the Balmer
series. How can you find the energy?
2 2
1 1 1 ; 32
R nnλ
= + =
R = 1.097 x 107 m-1
2 2
1 1 1 1(0.361); 2 3 0.361
R RR
λλ
= + = =
7 -1
10.361(1.097 x 10 m )
λ = λ = 656 nm
The frequency and the energy are found from:
c = fλ and E = hf
Example 2: Find the radius of the Hydrogen atom in its most stable state (n = 1).
2 20
2nn hr
meε
π=
m = 9.1 x 10-31 kg
e = 1.6 x 10-19 C
2
22 -12 34 2Nm
C-31 -19 2
(1) (8.85 x 10 )(6.63 x 10 J s)(9.1 x 10 kg)(1.6 x 10 C)
rπ
− ⋅=
r = 5.31 x 10-11 m r = 53.1 pm
Total Energy of an Atom The total energy at level n is the sum of the kinetic and potential energies at that level.
221
20
; ; 4
eE K U K mv Urπε
= + = =
Substitution for v and r gives expression for total energy.
2
02nev
nhε=
2 20
2nn hr
meε
π=
But we recall that:
4
2 2 208nmeE
n hε= −
Total energy of Hydrogen atom for level n.
Energy for a Particular State It will be useful to simplify the energy formula for a particular state by substitution of constants.
m = 9.1 x 10-31 kg
e = 1.6 x 10-19 C
εo = 8.85 x 10--12 C2/Nm2
h = 6.63 x 10-34 J s
2
2
4 -31 -19 4
2 2 2 -12 2 2 -34 2C0 Nm
(9.1 x 10 kg)(1.6 x 10 C)8 8(8.85 x 10 ) (6.63 x 10 Js)n
meEn h nε
= − = −
-18
2
2.17 x 10 JnE
n= − 2
13.6 eVnE
n−
=Or
Balmer Revisited 4
2 2 208nmeE
n hε= −
Total energy of Hydrogen atom for level n.
Negative because outside energy to raise n level.
When an electron moves from an initial state ni to a final state nf, energy involved is:
4 4
0 2 2 2 2 2 20 0 0
1 1; 8 8f
f
hc me meE E Ehc h n h nλ λ ε ε
−= = − = +
4 4
2 3 2 2 2 2 3 20 0
1 1 1 ; If 8 8f f i f
me meRh cn n n h cnλ ε ε
= − =
Balmer’s Equation: 7 -1
2 20
1 1 1 ; 1.097 x 10 mf
R Rn nλ
= − =
Energy Levels We can now visualize the hydrogen atom with an electron at many possible energy levels.
Emission
Absorption
The energy of the atom increases on absorption (nf > ni) and de-creases on emission (nf < ni).
Energy of nth level: 2
13.6 eVEn
−=
The change in energy of the atom can be given in terms of initial ni and final nf levels:
2 20
1 113.6 eVf
En n
= − −
Spectral Series for an Atom The Lyman series is for transitions to n = 1 level.
The Balmer series is for transitions to n = 2 level.
The Pashen series is for transitions to n = 3 level.
The Brackett series is for transitions to n = 4 level.
n =2
n =6
n =1
n =3 n =4
n =5 2 20
1 113.6 eVf
En n
= − −
Example 3: What is the energy of an emitted photon if an electron drops from the n = 3 level
to the n = 1 level for the hydrogen atom?
2 20
1 113.6 eVf
En n
= − −
Change in energy of the atom.
2 2
1 113.6 eV 12.1 eV1 3
E = − − = −
∆E = -12.1 eV
The energy of the atom decreases by 12.1 eV as a photon of that energy is emitted.
You should show that 13.6 eV is required to move an electron from n = 1 to n = ∞.
Summary (Cont.)
Balmer’s Equation: 7 -1
2 20
1 1 1 ; 1.097 x 10 mf
R Rn nλ
= − =
653 nm 486 nm 410 nm
434 nm Spectrum for nf = 2 (Balmer)
n = 3 n = 4 n = 5 n6
The general equation for a change from one level to another:
Emission spectrum
Questions
a. Sketch a cathode-ray tube and label it with the following parts: partially evacuated glass vessel, electrically charged plates, cathode, anode, fluorescent screen, high voltage source.
b. Explain how it works.
1.
Questions
2. Do you think the hypothesis inferring from the observations are reasonable? Why or why not? Explain in each hypothesis
Questions
3. How can the observations from the cathode ray tube experiment lead to the new atomic model “the plum pudding model”
Plum Pudding Model • in 1904 J.J. Thomson proposed the plum pudding model of
atom.
• According to this model the electrons were randomly stuck into a ball of positively charge matter.
positively charge matter
electrons © 2004 by Pearson Education Inc.
• A more modern name for this model might be chocolate-chip muffin model.
57 Discovering the Electron
The Discovery of the Electron
Thomson found q/m for cathode ray particles to be 1.76 x 1011 C/kg This is close to 2000 x q/m for a hydrogen ion Thomson’s conclusion: cathode-ray particles (today called electrons) are approximately 1/2000 the mass of a hydrogen ion (today called protons)
The Discovery of the Electron
Since Thomson didn’t know the charge or mass of these particles couldn’t use to determine their speed He balanced so that particle path was undeflected
= 212qV mv
versus e mF F
The Discovery of the Electron
Then, Using only a magnetic field,
=
=
=
e mF F
q E qv B
Ev
B=
=
=
2
m cF F
vqv B m
rq vm Br
The Discovery of the Electron
Examples: Practice Problem 1, Practice Problem 1,
446.0 10
2.4 102.50
NC m
s
Ev
TB×
= = = ×
681.0 10
1.0 101.0 0.0100
ms C
kgq vm Br T m
×= = = ×
×
6.3 Volts
120 Volts Plate is heated and electrons boil off. Velocity= 0 Potential Energy= ½ mv^2
Electrons are attracted to positively charged plate. They accelerate towards it and small percentage escape the plate through small hole, creating electron beam.
The potential energy of electrons is converted to kinetic energy
Since change in energy is the voltage times the charge then ½mv²=qV Therefore v= √(2qV/m)
√
We now know that v= √(2qV/m), so we can now easily find the velocity of our beam of electrons.
q(charge) of an electron= -1.6•10^-19
V(volts)=120
m(mass) of an electron=9.11•10^-31 kg
Therefore: v=√(2)(-1.6•10^-19)(120)/(9.11•10^-31)
v=√4.215•10^13
v=649•10^6 m/s
In order to predict the angle at which the electrons are deflected, we must first measure the force that the magnetic field inserts upon the beam
To do this, we use the equation: F=qvB
Magnetic field
Electrons
Like Solar Wind, the electrons in the CRT beam are deflected when entering a magnetic field, therefore the electron beam “bends.”
The force is always Perpendicular to the magnetic field And the velocity of the electrons
In order to find the force of the magnetic field, we must first calculate its strenghth.
mass= 9.11•10^-31 kg
velocity= 6.492•10^6 m/s
And if the measured distance of the electron beam from the magnets Is .075 meters
Then B= (9.11•10^-31)(6.492•10^6)/(1.6•10^-19)(.075)
B=2.772•10^-6 tesla
Since F=qvB and, according to Newton’s second law, F=m•v²/r, we can deduce that qvB=m•v²/r Or B=mv/qr
Charge= 1.6•10^-19 C
Now that we know the strength of the magnetic field at the electron beam, we can Calculate the force which the field exerts upon the electrons.
F=qvB
F=(6.49•10^6) (1.6•10^-19)(2.772•10^-6
F=2.879•10^-18 N
Examples 1. In a Cathode ray tube an electron travels through an electrical
field of 5.80 x 103 N/C and a magnetic field of 6.60 x 10-3 T undeflected.
a. what is the velocity of the electron?
b. If the electric field is turned off the magnetic field causes the electron to travel in a circular path, what is the radius of the circular path of the electron?
2. An electron travels through a CRT and a magnetic field of 4.20 x 10-3 T, the electron travels a circular path with a radius of 4.90 x 10-3 m. a. What is the velocity of the electron?
3. An unknown particle travels in a CRT tube through a magnetic field of 4.60 x 10-2 T at a velocity of 5.20 x 105 m/s, the magnetic field causes the particle to travel in a circular path with a radius of 0.235 m.
What is the charge to mass ratio of this particle?
Deflection of Electrons in a Uniform Electric Field (1)
Consider an electron beam directed between two oppositely charged parallel plates as shown below.
With a constant potential difference between the two deflecting plates, the trace is curved towards the positive plate. +
-
d
Deflection of Electrons in a Uniform Electric Field (2)
The force acting on each electron in the field is given by
deVeEF P==
where E = electric field strength, V = p.d. between plates, d = plate spacing.
p
Deflection of Electrons in a Uniform Electric Field (3)
The vertical displacement y is given by
22 )(21
21 t
mdeV
aty p==
2
2
)(21
vx
mdeVp=
This is the equation for a parabola.
Deflection of Electrons in a Uniform Magnetic Field
The force F acting on an electron in a uniform magnetic field is given by
BevF =
Since the magnetic force F is at right angles to the velocity direction, the electron moves round a circular path.
Deflection of Electrons in a Uniform Magnetic Field
The centripetal acceleration of the electrons is
mBeva =
Hence m
Bevr
va ==2
which gives
eBmvr =
Determination of Specific Charge Using a Fine Beam Tube
and the kinetic energy of the electron provided by the electron gun is
eVmv =2
21
Where V is the anode voltage.
mBerv = (For an electron moving in a
uniform magnetic field) Since
Determination of Specific Charge Using a Fine Beam Tube (3)
Rearrange the equation gives
22
2rBV
me
=
The value of the specific charge of an electron is now known accurately to be
1110)000003.0758803.1( ×± C/kg
eVm
Berm =2)(21So
Thomson’s e/m Experiment (1)
Thomson’s apparatus for measuring the ratio e/m
× × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × ×
+
-
v
Thomson’s e/m Experiment (2)
A beam of electron is produced by an electron gun with an accelerating voltage V.
The electron beam is arranged to travel through an electric field and a magnetic field which are perpendicular to each other.
The apparatus is set-up so that an electron from the gun is undeflected.
Thomson’s e/m experiment (3)
As the electron from the gun is undeflected, this gives
BE FF =
i.e. BeveE =BEv =⇒
On the other hand, eVmv =2
21
Combining the equations, we get 2
2
2VBE
me
=
Bev
eE
v
A B
d
+V
F +Q
position d
To move the charge from A to B a force equal to F would be needed. The work done moving Q from A to B is W.
W = Fd
But E = F/Q or F = EQ
So, W = EQd
Potential difference is defined as work done per unit charge:
V = W/Q
So: V = W/Q = EQd /Q V = Ed
E = V/d
V
(Uniform fields)
Acceleration: gravitational and electrical
A falling ball An accelerating charge
starts at highgravitationalpotential energy
gravitationalfield g
height h
gainskineticenergy
mass m
ends at lowgravitationalpotential energy
gravitationalpotential
ball falls down gravitationalpotential hill
gain of kinetic energy = loss of potential energy = mgh
electricpotential
charge ‘falls down’ electricalpotential hill
gain of kinetic energy = loss of potential energy = qV
starts at highelectricalpotentialenergy
potentialdifference V
gainskineticenergy
electricfield
ends at lowelectricalpotential energy
+
–
+
An electric field accelerates a charge as a gravitational field accelerates a mass
+charge q
ΔEp=mgΔh ΔW = QΔV
Draw the forces acting on the drop
Can you come up with an expression for the charge on the droplet?
E
Iden
tify
the
forc
es o
n a
char
ge in
an
elec
tric
field
Draw the forces acting on the drop
Can you come up with an expression for the charge on the droplet?
F= Eq
F= mg
q = mg E
Iden
tify
the
forc
es o
n a
char
ge in
an
elec
tric
field
1) electric force upward = weight downward EQ = mg Q = mg/E = 1.8 x 10-15 x 9.81 / 1.9 x 104 = 9.4 x 10-19 C
Quantization of Charge
Example: Practice Problem 1, page 763 (Pearson Physics) 2
5 14
19
19
19
5.0 10 2.4 10 9.81
4.7 10
4.7 103
1.60 10
e g
N mC s
Ce
F F
E m g
kg
q
q
q C
Ce
−
−
−
−−
−
=
=
× × = × ×
= ×
×=
×
Example
A beam of electrons travels an undeflected path in a cathode ray tube. E is 7.0 x 103 N/C. B is 3.5 x 10-2 T. What is the speed of the electrons as they travel through the tube?
What we know: E = 7.0 x 103 N/C B=3.5 x 10-2 T
Equation: v = E/B
Substitute: v = (7.0 x 103N/A s) / (3.5 x 10-2 N/A m)
Solve! v = 2.0 x 105 m/s
Example An electron of mass 9.11 x 10-31 kg moves with a speed of 2.0 x
105 m/s across a magnetic field. The magnetic induction is 8.0 x 10-4 T. What is the radius of the circular path followed by the electrons while in the field?
What we know: M = 9.11 x 10-31 kg B=8.0 x 10-4 T v=2.0 x 105 m/s
Equation: Bqv = mv2/r so r = (mv) / (Bq)
Substitute: R = (9.11x10-31kg)(2.0x105 m/s) (8.0x10-4N/Am)(1.6x10-19As)
Solve! r = 1.4 x 10-3 m
Modifications of the Bohr Theory – Elliptical Orbits
Sommerfeld extended the results to include elliptical orbits Retained the principle quantum number, n
Determines the energy of the allowed states Added the orbital quantum number, ℓ
ℓ ranges from 0 to n-1 in integer steps All states with the same principle quantum number are said
to form a shell The states with given values of n and ℓ are said to form a
subshell
Example
A negatively charged polystyrene sphere of mass 3.3 x 10-15 kg is held at rest between 2 parallel plates separated by 5.0 mm when the potentialdifference between them is 170 V. How many excess electrons are on the sphere?
Solution: Using qV/d = mg gives q = 9.5 x 10-19 C Taking e to be 1.6 x 10-19, q corresponds to 5.9 excess electrons Because we cannot have 0.9 of an electron, the answer must be 6 electrons, the discrepancy must be due to experimental inaccuray or rounding errors
Example
Electrons are accelerated to a speed of 8.4 x 106 m/s. they then pass into a region of uniform magnetic flux density 0.50 mT. The path of the electrons in the field is a circle with a radius 9.6 cm. Calculate:
a) the specific charge of the electron b) the mass of the electron, assuming the charge on the electron is 1.6 x
10-19 C Solution a) e/m = v/Br = 1.8 x 1011 C/kg b) using e/m, m = 9.1 x 10-31 kg
Example
It is required to select charged ions which have a speed of 4.2 x 106 m s-1. the electric field strength in the velocity selector is 3.2 x 104 V m-1. Calculate the magnetic field strength required.
Solution v = E/B = 7.6 x 10-3 T
Example
A charged particle has mass 6.7 x 10-27 kg and charge +3.2 x 10-19 C. It is travelling at speed 2.5 x 108 m s-1 when it enters a region of space where there is a uniform magnetic field of flux density 1.6 T at right angles to its direction of motion. Calculate
a) the force on the particle due to the magnetic field b) the radius of its orbit in the field
Solution a) F = Bqv sin θ = 1.3 x 10-10 N b) centripetal force is provided by the electromotive force mv2/r = Bqv r = 3.2 m
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