NAROK UNIVERSITY COLLEGE
(A CONSTITUENT COLLEGE OF MOI UNIVERSITY)
UNIVERSITY EXAMINATIONS 2011/2012
SECOND YEAR SECOND SEMESTER EXAMINATION SCHOOL OF SCIENCE
UNIVERSITY EXAMINATIONS FOR THE DEGREE OF BACHELOR OF
SCIENCE BACHELOR OF EDUCATION (SCIENCE)
COURSE CODE: PHY212
COURSE TITLE: MODERN PHYSICS
DATE:
APRIL, 2012
TIME: 3 HOURS
INSTRUCTIONS


Answer Question ONE and any other TWO.
Use of sketch diagrams where necessary and brief illustrations are encouraged.
 Read the instructions on the answer booklet keenly and adhere to them.
PHYSICAL CONSTANTS
o
o
o
o
o
o
o
o
o
o
o
o
o
o
Planck constant ; h = 6.626x10-34Js
Charge of proton; e = +1.602x10-19C
Mass of electron; me = 9.109x10-31kg
Acceleration due to gravity, g = 9.81 m s-2
Avogadro constant; NA = 6.023x1023mol-1
Stefan constant ; σ = 5.670x10-8Wm-2K-4
Rydberg constant; Rhc = 1.097x107m-1
Speed of light in vacuum; c = 2.998x108ms-1
Charge of electron; e = -1.602x10-19C
Mass of proton; mp = 1.67 x10-27kg
Rest energy of electron; Ee = 0.511MeV
One atomic mass unit; u= 1.66x10-27kg
Atomic mass unit energy equivalent ; a.m.u = 931.5MeV
One ElectronVolt ; 1eV = 1.602 x 10-19 J
This paper consists of 8 printed pages.
Page 1 of 8
QUESTION ONE: [24 marks]
a) State the two postulates of Special relativity theory as postulated by Einstein in 1905. [2]
 1st postulate: The laws of physics have the same form in all inertial frames of reference.
 2nd postulate: Light propagates through empty space with a definite speed c independent of the
speed of the source or observer
b) Assuming that the tungsten filament of a light bulb is a blackbody, determine its peak
wavelength if its temperature is 2 900 K.
[3]
Recall Wein;s law: λmax T = 0.2898 x 10-2 m • K. It can be used to solve for lmax:
0.2898  102 m  K
max 
T
0.2898  102 m  K 

 9.99  107 m  999nm
Thus max 
2900 K
This is infrared, so most of the electric energy goes into heat!
c) Photons of light have zero mass. How is it possible that they have momentum? [2]
 A photon transports energy. The relativistic equivalence of mass and energy means
that is enough to give it momentum.
d) What is the energy released when uranium 238 decays by alpha emission to Thorium 234?
Uranium 238 = 238.0508 u ; Thorium 234 = 234.0436 u; Alpha particle = 4.0026 u [4]
 Mass loss (mass defect) : = 238.0508 - (234.0436 + 4.0026) = 0.0046 u
 Since 1 u = 931 MeV; Energy released (MeV) = 0.0046x931 = 4.3
 Answer: energy released from uranium 238 decay is 4.3 Mev
e) Estimate the surface temperature of the Sun from the following information. The Sun’s
radius is given by Rs = 7.0 x 108 m. The average Earth–Sun distance is R = 1.5 x 1011 m. The
power per unit area (at all frequencies) from the Sun is measured at the Earth to be 1400
W/m2. Assume that the Sun is a blackbody. [4]
 For black body, we take ε = 1 and so total (Rs)  T4 where the notation ρ total(Rs) stands
for the total power per unit area at the surface of the Sun. Because the problem gives the
total power per unit area at the Earth, ρ total(R), we need the connection between ρ total(R)
and ρ total(Rs)

(R)x R 2
 total (R s )x4R s2  total (R)x4R 2 or total (R s )  total
R 2s
1/ 4
2 1/ 4  1400x (1.5x1011 )2


(R)x
R



T   total

 5800K
2

8
8
2





R
5.6x10
x(7.0x10
)
s
s



f) The Balmer series for the hydrogen atom corresponds to electronic transitions that terminate
in the state of quantum number n = 2.
(i) Find the longest-wavelength photon emitted and determine its energy [3]
Page 2 of 8
(ii) Find the shortest-wavelength photon emitted in the Balmer series.
[2]
(i) Solution The longest-wavelength (least-energetic) photon in the Balmer series results from the
transition from n = 3 to n = 2. This gives
1
1
1
1
1
1
5
6
 R(

)
 R(  )  R and  max 
 656.3nm

max
36
5R
n f 2 ni 2
22 32
 This wavelength is in the red region of the visible spectrum. The energy of this photon is

Ephoton  hf 
hc

6.63x1034 x3x108
 3.03x1019  1.89eV

9
656.3x10
max
 We could also obtain the energy of the photon by using the expression hf = E3 - E2, where E2 and
E3 are the energy levels of the hydrogen atom, which can be calculated. Note that this is the lowestenergy photon in this series because it involves the smallest energy change.
(ii) The shortest-wavelength (most-energetic) photon in the Balmer series is emitted when the electron
makes a transition from n =∞ to n = 2. Therefore,
1
1 1
R
4
4
 R(  )    min  
 364.6nm

min
4
R 1.097x107
22 
 This wavelength is in the ultraviolet region and corresponds to the series limit.
g) Explain the physical meaning of the square of the wave function as postulated by Born [2]
 The value of Ψ2 at some location at a given time is proportional to the probability of finding
the particle at that location at that time
h) Explain why electrons are emitted from the surface in photoelectric effect is almost
instantaneously, even at low intensities
[3]
 The effect is instantaneous since there is a one-to-one interaction between the photon
and the electron
i) Show that the photons in a 1240 nm infrared light beam have energies of 1.00 eV .[3]
6.63x1034 x2.998 x108
E  hf 
 1.602 x1019 J  1/ 00eV
9
1240 x10
j) The speed of light in water is 230 Mm/s. Suppose an electron is moving through water at 250
Mm/s. Does this violate the principle of relativity?
[2]
 No. The principle of relativity implies that nothing can travel faster than the speed of light in a
vacuum, which is 300 Mm/s. The electron would emit light in a conical shock wave of radiation.
Page 3 of 8
QUESTION TWO:
[20 marks]
a) Briefly explain the term ‘ black body’
[4]
A black body is a theoretical object that absorbs 100% of the radiation that hits it.
Therefore it reflects no radiation and appears perfectly black.
b) Explain two applications of the study of black body.
[4]
 The black body is importance in thermal radiation theory and practice.
 The ideal black body notion is importance
in studying thermal radiation and
electromagnetic radiation transfer in all wavelength bands.
 The black body is used as a standard with which the absorption of real bodies is
compared.
c) Discuss the Rayleigh-Jeans Law of black body radiation.
[6]
2ckT
 Its relation is given by I( , T ) 
4
 It agrees with experimental measurements for long wavelengths.
 It predicts an energy output that diverges towards infinity as wavelengths grow
smaller.
 The failure has become known as the ultraviolet catastrophe.
d) Explain the Ultraviolet catastrophe





[6]
Classical theory did not match the experimental data
At long wavelengths, the match is good
At short wavelengths, classical theory predicted infinite energy
At short wavelengths, experiment showed no energy
This contradiction is called the ultraviolet catastrophe
Page 4 of 8
QUESTION THREE: [20 marks]
a) Explain the difference between fusion and fission
[4]
 Nuclear fission in the process whereby a nucleus, with a high mass number, splits into 2
nuclei which have roughly equal smaller mass numbers.
 In nuclear fusion, two nuclei with low mass numbers combine to produce a single nucleus
with a higher mass number.
b) Define the term atomic mass unit (u), and show that 1amu = 1.49 x 10 -10 J. [3]
 By definition 1 u is equal to 1/12 the mass of a carbon-12 atom.
 1 atomic mass unit = 1u = 1.66x 10-27 kg = 931MeV = 1.49 x 10-10 J
c) Calculate the binding energy (in MeV) of an alpha particle from the following information:
(ans. to 1 d.p.). Take Mass of a proton = 1.0076 u ; Mass of a neutron = 1.0090 u ; Mass of an
alpha particle = 4.0028 u
[4]
 Mass of a proton = 1.0076 u ; Mass of a neutron = 1.0090 u ; Mass of an alpha particle = 4.0028 u
 1 u = 931 MeV ;Mass of particles in alpha particle = 2 protons + 2 neutrons
 = 2(1.0076) +2(1.0090) = 4.0332 u ; Mass loss (mass defect) = 4.0332 - 4.0028 = 0.0304 u
 Since 1 u = 931 MeV,; Binding energy = 0.0304 x 931 = 28.3 MeV
 Answer: binding energy of an alpha particle is 28.3 MeV
d) Consider the nuclear fission below and determine the energy released. [7]
Element
Atomic Mass (kg)
235
-25
U
3.9014 x 10
Cs
2.2895 x 10
Rb
1.5925 x 10
92
138
-25
55
96
37
-25
1
-27
n
1.6750 x 10
0
Soln
Page 5 of 8













Calculate the total mass before and after fission takes place.
The total mass before fission (LHS of the equation):
3.9014 x 10-25 + 1.6750 x 10-27 = 3.91815 x 10-25 kg
The total mass after fission (RHS of the equation):
2.2895 x 10-25 + 1.5925 x 10-25 + (2 x 1.6750 x 10-27) = 3.9155 x 10-25 kg
The total mass before fission =3.91815 x 10-25 kg
The total mass after fission =3.91550 x 10-25 kg
total mass before fission > total mass after fission
mass difference, m = total mass before fission – total mass after fission
m = 3.91815 x 10-25 – 3.91550 x 10-25= 2.65 x 10-28 kg
This reduction in mass results in the release of energy.
The energy released can be calculated using the equation: E = mc2
E = 2.65 x 10-28 x (3 x 108)2=E = 2.385 x 10-11 J
e) The energy released from this fission reaction does not seem a lot. Explain. [2]
 This is because it is produced from the fission of a single nucleus.
 Large amounts of energy are released when a large number of nuclei undergo fission reactions.

QUESTION FOUR: [20 marks]
a) State the de Broglie postulate
[3]
 In 1924, Louis de Broglie postulated that because photons have wave and particle
characteristics, perhaps all forms of matter have both properties
h
h

 The de Broglie wavelength of a particle is  
p mv
b) X-rays of wavelength λ= 0.200 nm are aimed at a block of carbon. The scattered x-rays are
observed at an angle of 45.0o to the incident beam. Calculate the increased wavelength of the
scattered x-rays at this angle. [4]
 The shift in wavelength of the scattered x-rays is given by taking θ= 45.0o, we find
h
6.63x1034
(1  cos ) 
(1  cos 45o )  7.11x1013 m  0.00071nm
  

31
8
me c
9.11x10
x3x10
 Hence the wavelength of the scattered X-rays at this angle is λ = Δλ + λo = 0.200711nm
c) Why are x-ray photons used in the Compton experiment, rather than visible-light photons?
To answer this question, we shall first calculate the Compton shift for scattering at 90o from
graphite for the following cases:
(i) Very high energy γ-rays from cobalt, λ = 0.0106 Å;
(ii) x-rays from molybdenum, λ = 0.712 Å; and
(iii) green light from a mercury lamp, λ = 5461 Å. [5]
 In all cases, the Compton shift formula gives
  1  o  0.0243(1  cos 90o )  0.0243 A  0.00243nm
Page 6 of 8
 That is, regardless of the incident wavelength, the same small shift is observed. However, the
fractional change in wavelength,  / o is quite different in each case:
γ-rays from cobalt:
 0.0243
 0.0243

 2.29 ; X-rays from molybdenum:

 0.0341

0.0106

0.712
 0.0243

 4.45x106

5461
 Because both incident and scattered wavelengths are simultaneously present in the beam, they can
be easily resolved only if  / o is a few percent or if λo ≤1 Å.
Visible light from mercury:
d) The so-called free electrons in carbon are actually electrons with a binding energy of about 4
eV. Why may this binding energy be ignored for x-rays with λo= 0.712 Å? [3]
hc 1240eV

 17400eV
 The energy of a photon with this wavelength is E  hf 

0.712
 Therefore, the electron binding energy of 4 eV is negligible in comparison with the incident x-ray
energy.
f) An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the
lighter fragment is 2.50 × 10–28 kg, and that of the heavier fragment is 1.67 × 10–27 kg. If the
lighter fragment has a speed of 0.893c after the breakup, what is the speed of the heavier
fragment?
[5]
 Momentum must be conserved, so the momenta of the two fragments must add to zero. Thus, their
magnitudes must be equal, or
p2 = p1 = γ1 m1 v1
 2.50 ×10
=

kg  0.893c 
1-  0.893
 For the heavier fragment,
3.37  v c   1   v c 
-28
2
2
1.67  10
27

kg v
1  v c
2


= 4.96 ×10-28 kg c


 4.96  1028 kg c which reduces to
and yields v  0.285c
Page 7 of 8
QUESTION FIVE: [20 marks]
a) State the correspondence principle according to Bohr. [3]
 This principle states that predictions of quantum theory must correspond to the predictions
of classical physics in the region of sizes where classical theory is known to hold.
b) Explain four failures of the Bohr postulates. [8]





(c)
By the early 1920s scientists recognized that the Bohr theory contained many
inadequacies:
It failed to predict the observed intensities of spectral lines.
It had only limited success in predicting emission and absorption wavelengths for multi-electron
atoms.
It failed to provide an equation of motion governing the time development of atomic systems
starting from some initial state.
It overemphasized the particle nature of matter and could not explain the newly discovered wave–
particle duality of light.
It did not supply a general scheme for “quantizing” other systems, especially those without
periodic motion.
A particle of charge q and mass m is accelerated from rest through a small potential
difference V.
(i) Find its de Broglie wavelength, assuming that the particle is not relativistic. [4]
(ii) Calculate λ if the particle is an electron and V = 50 V.
[5]
(i)
When a charge is accelerated from rest through a potential difference V, its gain in
1
kinetic energy, 1/2mv2, must equal the loss in potential energy qV. That is, mv2  qV .
2
p2
 qV or p  2mqV . Substituting
Because p = mv, we can express this in the form
2m
h
h
this expression for p into the de Broglie relation λ = h/p gives   
p
2mqV
(ii)
The de Broglie wavelength of an electron accelerated through 50 V is

h
h
6.63x1034


 1.7x1010 m  1.7A
p
2mqV
2(9.11x1031 x1.6x1019 x50)
 This wavelength is of the order of atomic dimensions and the spacing between atoms in a solid.
Such low-energy electrons are routinely used in electron diffraction experiments to determine
atomic positions on a surface.
The End
Page 8 of 8