Name _________________
Solutions to Test 2
October 14, 2015
This test consists of three parts. Please note that in parts II and III, you can skip one question of
those offered. The equations below may be helpful with some problems.
Constants
h  6.626  1034 J  s
h  4.136 1015 eV  s
  1.055  1034 J  s
16
 =6.582  10 eV  s
k B  1.3807  1023 J/K
k B  8.6173  105 eV/K
k  8.988  109 N  m 2 / C2
e  1.602 1019 C
1 eV  1.602  1019 J
ke 2
1


c 137
Hydrogen-Like Atoms
c2  2 Z 2

k 2e4  Z 2
E

2 2 n 2
2n 2
 13.60 eV  Z 2
E
n2
Wave
Relationships
2

k

1
 f 
2
T
Hydrogen Spectrum
1 
 1
   91.17 nm   2  2 
n m 
Reduced Mass
mM

mM
Compton Effect
h
   
1  cos  
mc
h
 2.426 1012 m
mc
Black Bodies
U
 2  k BT 
15  c 
4
3
maxT  .002898 m  K
Rutherford Scattering
kqQ
 
cot  
b
2
m v
2
Part I: Multiple Choice [20 points]
For each question, choose the best answer (2 points each)
1. One of the uncertainty relations says that
A) E t  12 
B) E t  12 
C) t  12 E
1
D) t  12 E
R
2Ze 2 k
E
E) none of these
2. Why was there a problem in statistical mechanics when studying a thermal distribution of
electromagnetic waves?
A) The energy in each mode could be anything, so there was no idea how to figure out how
much there was
B) Long wavelengths had most of the energy, leading to the infrared catastrophe
C) Short wavelengths had most of the energy, leading to the ultraviolet catastrophe
D) The predictions could not explain the photoelectric effect
E) There was already ample evidence that light was made of particles (photons)
3. When you have a wave packet, the speed of the entire packet is governed by the ________
velocity, while the speed of individual waves within it are governed by the _______ velocity.
A) Group, wave B) Wave, group C) Group, phase D) Phase, group E) Phase, wave
4. Which hypothesis is attributed to deBroglie?
A) Electrons have wave-like properties
B) Electrons must satisfy Schrödinger’s equation
C) Electrons orbit the nucleus on circular paths
D) Electrons have angular momentum that is a multiple of 
E) Electrons satisfy an uncertainty relation on their position and momentum
5. If   4  3i , what is  ?
2
A) 7
B) 25
C) 7 + 24i
D) 25 + 24i
E) None of these
6. If f  x, y   A  x 2  y 2  , then f x 
A) Ax
B) 2Ax
C) A(x + y)
D) 2A(x + y)
E) 2Ay
7. In which of the following ways is an atom very much like the solar system?
A) The main force involved is electromagnetism in both cases
B) The main force involved is gravity in both cases
C) The force between two of the orbiting objects is negligible compared to the force
between one of the orbiting objects and the central object
D) The angular momentum for a single electron or planet is usually a small multiple of 
E) The central object (Sun or nucleus) is much smaller than the size of the whole object
(solar system or atom)
8. A dispersion relation is a relation between which two quantities?
B)  and f
C)  and T
D)  and k
A)  and k
E) f and T
9. Rutherford scattered alpha particles off of gold atoms, and discovered, to his surprise, that
A) Most of them went straight through, suggesting the atom was mostly empty space
B) The gold atoms were often knocked out, suggesting they were loosely bound
C) The electrons did little to stop them, showing that electrons were light
D) The alpha particles sometimes scattered at large angles, suggesting there was a lot of
mass and charge concentrated in the “nucleus”
E) The gold atoms would often be transmuted to other metals, like lead
10. Wave effects in quantum mechanics are not noticeable in the everyday world because
A) Quantum mechanics doesn’t apply to composite objects, like all everyday objects
B) Quantum effects only describe wave packets, which don’t apply to particles
C) Quantum effects are so slow that you can’t observe them
D) The wavelength is enormous compared to the size of everyday objects
E) The wavelength is tiny compared to the size of everyday objects
Part II: Short answer [20 points]
Choose two of the following questions and give a short answer (2-3 sentences) (10 points
each).
11. Explain, in the photoelectric effect, why there is a minimum frequency which suffices to
free an electron from a metal. When you are above that frequency, give a formula for
the maximum energy with which an electron can be ejected.
The photons have energy E = hf. There is a minimum amount of work  (called the work
function) for any given metal that must be input in order to free the electron. If hf < , then no
electrons can be freed. If hf > , then electrons can be freed, and those electrons can have at
most an energy of Emax = hf –  .
12. Since an electron is attracted to the nucleus, it should have a minimum energy when it
is exactly at the nucleus and has zero velocity. Explain why this can’t happen in
quantum mechanics. You should have at least one mathematical relation in your
answer.
The minimum energy would correspond to position x = 0 and momentum p = 0.
However, specifying both of these simultaneously is impossible for quantum waves, because the
product of the uncertainties are contrained by the inequality  x  p   12  .
13. When electrons are accelerated through a thin gas, and the voltage is gradually
increased, at first the current will increase, then at certain threshold velocities, the
current will suddenly drop and the gas begins to glow. Explain qualitatively what is
happening.
This is a description of the Franck-Hertz experiment. Because atoms have only certain
levels that are energies, there is a minimum amount of energy that can be absorbed by the atoms.
Hence when electrons with less than this energy collide, they must collide elastically,
maintaining their speed. But when we reach this threshold energy, the atoms suddenly can
absorb the energy, slowing down the electrons and ultimately decreasing the current. But the
atom is now in an excited state, and it emits a photon (it glows) when it falls back to its ground
state.
Part III: Calculation: [60 points]
Choose three of the following four questions and perform the indicated calculations (20
points each).
14. One important event in the early universe is called recombination, and occurred when
the temperature of the universe was approximately T = 2970 K. The universe was filled
with almost perfect black body radiation at this time.
(a) Find the energy density of the blackbody radiation at this time in J/m3.
The energy density is given by
U
 2  k BT 
15  c 
 2 1.3807 1023 J/K   2970 K  
4
3

15 1.055 10
34
4
J  s  2.998  10 m/s  
8
3
 0.0588 J/m3 .
(b) Find the wavelength where the blackbody radiation was strongest at this time, in
nm.
We use Wien’s Law, maxT  .002898 m  K , to find
maxT 
.002898 m  K
 9.758 107 m  975.8 nm.
2970 K
(c) Find the energy for a single photon corresponding to the wavelength you found in
part (b).
The energy of a photon is given by E  hf , where f can be found from f   c , so
E  hf 

hc

 6.626 10

34
J  s  2.998  108 m/s 
9.758  10
7
m
 2.036 1019 J
19
2.036 10 J
 1.271 eV
1.602  1019 J/eV
(d) Using the energy density you found in part (a) and the energy you found in part (c),
estimate the number of photons per unit volume (in m-3) at this time.
The energy density is equal to the energy per photon times the number density of
photons, so a simple estimate would yield
n
U
0.0588 J/m3

 2.88 1017 m 3 .
E 2.036 1019 J
There is a flaw in this argument, in that we used the energy of the peak wavelength photon. The
actual number is more like 5.311017 m 3 instead.
15. A N+6 ion has atomic number Z = 7, one electron, and is in the n = 7 state initially.
(a) What is the energy of this state?
The energy is given by
13.60 eV  Z 2

E
n2
13.60 eV  7 2


 13.60 eV .
72
(b) A photon of energy E = 13.056 eV is absorbed by this atom. What is the new energy
of the atom? What is the value of n now?
Since a photon is absorbed, the energy increases to
E f  Ei  E  13.60 eV +13.056 eV  0.544 eV
The final state can then be found from
13.60 eV  Z 2

0.544 eV  E  
13.60 eV  7 2

,

n2
13.60 eV  7 2

2
n 
 1225 ,
0.544 eV
n  35 .
n2
(c) The same atom, still at n = 7, instead emits a photon of energy E = 13.056 eV. What
is the new energy of the atom? What is the value of n now?
This time the energy is lost, so
E f  Ei  E  13.60 eV  13.056 eV  26.656 eV
The final state can then be found from
26.656 eV  E  
13.60 eV  Z 2

13.60 eV  7 2 ,
n2
13.60 eV  7 2  25 ,
n2 
26.656 eV
n  5.
n2
(d) What is the wavelength of light corresponding to an energy of E = 13.056 eV?
The wavelength can be found from f   c , and the frequency from E  hf , so
c hc hc  4.136 10
 


f hf
E
15
eV  s  2.998  108 m/s 
13.056 eV
 9.49 108 m  94.9 nm .
16. The 2015 Nobel Prize in physics went to experimenters who demonstrated that one of
the neutrinos has mass. Based on the measured mass, a long wavelength neutrino
would have a dispersion relation of the form   Ak 2 where A  6650 m 2 /s .
(a) Find a formula for the phase velocity and the group velocity in terms of k.
The phase velocity and group velocity are given by

Ak 2
 Ak ,
k
k
w 
vg 
  Ak 2   2 Ak  2v p .
k k
vp 

(b) A typical neutrino left over from the Big Bang might have k  851 m 1 . For this
wave, find the wavelength  , the angular frequency  and the frequency f.
Each of these quantities is easily computed from the relevant formulas:

2
2

 7.4 103 m  7.38 mm ,
k
851 m 1
  Ak 2   6650 m 2 /s  851 m 1   4.816 109 s 1 ,
2
f 
 4.816 109 s 1

 7.665 108 s 1  766.5 MHz .
2
2
(c) Find the group and phase velocity for the neutrino from part (b) in m/s.
We simply substitute into the formulas from part (a) to yield
v p  Ak   6650 m 2 /s  851 m 1   5.66 106 m/s ,
vg  2 Ak  2  6650 m 2 /s  851 m 1   1.132 107 m/s .
(d) Find the energy in eV of the neutrino you found in part (b).
We use the simple formula
E  hf   4.136 1015 eV  s  7.665 108 s 1   3.17 106 eV  3.17  eV .
17. The wave function for a particle is given by
 Nx 2a  x  a ,
elsewhere.
0
– 2a
  x  
Na
–a
(x)
a
– Na
– 2Na
x
where N and a are positive constants. The wave is
sketched at right.
(a) Where in the allowed region 2a  x  a is the particle most likely to be? Where is
it impossible for it to be? You can rely on the sketch if you wish.
It is most likely to be where the wave function has the largest magnitude (positive or
negative), so this is clearly at x  2a . It is impossible for it to be where the wave function
vanishes, which is at x = 0.
(b) What is the normalization constant N?
We find the normalization constant by demanding that the probability that it is
somewhere be given by 1. We therefore have

1     x  dx  

2
a
2 a
3
 13 N 2  a 3   2a    13 N 2 a 3 1  8   3 N 2 a 3 ,


1
N2  3 ,
3a
1
N
3a 3
N 2 x 2 dx  13 N 2 x3
a
2 a
We could have chosen N negative, but the problem says N is positive.
(c) If the particle’s position is measured, what is the probability that it will have x > 0?
This is simply given by

a
a
2
1 1
1     x  dx   N 2 x 2 dx  13 N 2 x3  13 N 2 a 3   3 a 3  19  11.1% .
0
0
0
3 3a