Name _________________
Solutions to Test 2
October 15, 2014
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.
Black Bodies
Constants
h  6.626  10 J  s  4.136 1015 eV  s
34
  1.055 1034 J  s  6.582 1016 eV  s
k B  1.3807 1023 J/K  8.6173 105 eV/K
U
 2  k BT 
15  c 
4
3
maxT  2.898 103 m  K
k  8.988 10 N  m / C
Rutherford Scattering
e  1.602 1019 C
kqQ
 
1 eV  1.602  1019 J
cot  
b
2
m v
2
ke 2
1

 0.00729735 
2 Ze 2 k
137
c
R
E
Compton Effect
Hydrogen-Like Atoms
h
c2  2 Z 2
Wave

k 2e4  Z 2
   
1  cos  
E

2 2
2
mc
Relationships
2 n
2n
h
2
12
 13.6 eV  Z 2
 2.426 10 m

E

mc
k
n2

1
Hydrogen Spectrum
 f 
Reduced
Mass
2
T
1
1 
 1
mM
   91.17 nm   2  2 

n m 
mM
9
2
2
Part I: Multiple Choice [20 points]
For each question, choose the best answer (2 points each)
1. If   ei , what is  ?
2
A) e 2i
B) e  i
C) e
2
D) e 
2
E) 1
2. For light in vacuum, the phase velocity is __ and the group velocity is __?
A) c, c/2
B) c/2, c
C) c, c
D) c, 2c
E) 2c, c
3. The Millikan oil drop experiment allowed Millikan, for the first time, to measure
A) The size of the atom
B) The size of the nucleus
C) The wavelength of electrons
D) The charge of electrons
E) The size of the electrons
4. When an X-ray scatters from an electron at rest in an atom, the final photon will have
a _____ wavelength and a ________ frequency
A) longer, lower
B) longer, higher
C) shorter, lower
D) shorter, higher
E) unchanged, unchanged
5. Which assumption did Planck have to make to account for blackbody radiation?
A) The electron orbits the nucleus with angular momentum a multiple of 
B) The energy of electromagnetic waves comes in multiples of hf
C) The wavelength of an electron of momentum p is h/p
D) An electron is described in terms of a wave function, not position and velocity
E) When electrons move from one level to another in an atom, the energy comes out
as a single photon
6. An expression for  is
A) 2 h
B) h  2 
C) 2 h
D)  1  2 h  E) none of these
7. Which of the following is approximately the radius of a typical atomic nucleus?
A) 3 106 m B) 3 109 m C) 3 1011 m D) 3 1015 m E) 3 1018 m
8. What evidence convinced Rutherford that the positive charge and most of the mass of
the atom was concentrated in a small region, rather than spread out
A) Alpha particles, when scattered off of atoms, sometimes rebounded at large
angles
B) He could measure the change in charge when a single nucleus left or was added to
a small piece of gold foil
C) He was able to focus the alpha particles as they scattered from the nucleus to
produce an image of the nucleus
D) He could see the nuclei because they were knocked out of the gold foil by the
alpha particles
E) The electrons knocked out of atoms had angles indicating they were coming from
a point charge
9. When electrons in the innermost level of an atom change levels, you can get the
approximate energy by treating it as if the nucleus is Z  1 instead of Z. Why is this
appropriate?
A) The electron is moving at high speed, which due to relativistic effects, reduce the
effective nuclear charge
B) The electron is partly inside the nucleus, which means some of its charge is
cancelled
C) The other electrons partly shield and cancel out the nuclear charge
D) The Coulomb force gets weaker at short distance, effectively reducing the charge
E) There are magnetic effects as well as electric forces, which partly overcome the
attraction of the nucleus
10. To make a wave packet, what must you do?
A) Include both cosines and sines in the wave
B) Use a complex wave, eikx
C) Include one wave moving to the right and one to the left
D) Include a potential that traps the particle in a small region
E) Add up many waves with different wavelengths
Part II: Short answer [20 points]
Choose two of the following questions and give a short answer (2-3 sentences)
(10 points each).
11. There is background light left over from the big bang event that formed the
universe. Explain, in principle, how one could measure the temperature of this
light. Include at least one equation
One way would be to measure the spectrum as a function of wavelength, and find
the wavelength for the peak of the spectrum, and then use maxT  2.898 103 m  K to
find the temperature. You could also measure the total energy density and then use
4
3
U   2  k BT  15  c  .
12. Explain qualitatively why an electron confined to a small region automatically
has a lot of energy. At least on equation or inequality is appropriate.
If you are confining an electron to a small region, you are effectively demanding
that it have a low uncertainty x in its position. Because of the uncertainty principle,
x  p  12  , this implies a large uncertainty in the momentum, and hence it has to have
a lot of momentum. Since the kinetic energy is E  p 2  2m  , this implies a lot of
energy.
13. Explain impact parameter b for a collision between two objects is. It might help
to sketch a diagram.
The impact parameter is the amount by which a particle would miss making a
head-on collision with its target if there were no forces on it, as sketched below:
b
Part III: Calculation: [60 points]
Choose three of the following four questions and perform the indicated
calculations (20 points each).
14. When silicon is irradiated with light, it is observed that electrons are released,
but only if the light has a frequency f > 1.1121015 s-1
(a) Explain why there is a minimum frequency. Find the work function  for
silicon.
In order to liberate an electron from silicon, a minimum amount of work  must
be done. Low frequency photons have less energy than high frequency, according to
E  hf . Since the minimum frequency is 1.1121015 s-1, the work function must be
  hf min   4.136 1015 eV  s 1.112 1015 s 1   4.60 eV=7.37 1019 J .
(b) Silicon is irradiated with light with frequency f = 2.431015 s-1 and electrons
are found to escape from the surface. Predict the maximum resulting voltage
Vmax that these electrons can overcome after escaping the surface
We use the formula
eVmax  hf     4.136 1015 eV  s  2.43 1015 s 1   4.60 eV  5.45 eV ,
Vmax  5.45 V .
(c) Light of unknown frequency and wavelength is found to produce electrons
that have 1.60 eV of kinetic energy after having been ejected from silicon.
What is the frequency and wavelength of this light?
We turn the equation around, keeping in mind that eVmax is the kinetic energy, so
we have
hf  eVmax    1.60 eV    4.60 eV   6.20 eV ,
f 
6.20 eV
 1.50  1015 s 1
15
4.136 10 eV  s
The corresponding wavelength can then be found from  f  c , so we have

c 2.998 108 m/s

 2.00  107 m  200. nm .
f 1.499  1015 s 1
15. An atom with unknown atomic number Z has a single electron in the orbit n = 5.
The electron then falls down the smallest possible amount to the next level.
(a) Find a formula for the initial energy and for the final energy, in eV, in terms
of Z, as well as their difference
The level n can only take on integer values. The smallest possible amount to drop
would be to n = 4 (lower values of n have more negative energy, or lower energy). The
energy for level n is given by E   13.6 eV  Z 2 n 2 , so we have
E5  
13.6 eV  Z 2
 0.544 Z 2 eV ,
52
13.6 eV  Z 2

E4  
 0.850 Z 2 eV ,
2
4
E  E5  E4   0.544 Z 2 eV    0.850 Z 2 eV   0.306 Z 2 eV .
(b) The photon released is found to have an energy of 2.75 eV. What is Z?
We simply equate the two energies and solve for Z:
2.75 eV  0.306 Z 2 eV ,
2.75 eV
Z2 
 8.99 .
0.306 eV
Since Z must be an integer, we conclude Z = 3, and any discrepancy can be attributed to
rounding error. We are working with a lithium atom.
(c) The atom is returned to the n = 5 state, and we wish to now increase its
energy by the smallest possible amount, so it goes up one level. What is the
energy difference this time?
We now wish to go up one level, to n = 6, so we have
E  E6  E5
13.6 eV  32  13.6 eV  32 


 



52
62


 3.40 eV    4.90 eV   1.50 eV .
(d) What wavelength of light should you bathe the atom in if you want it to
increase its level as described in part (c)?
We use the formulas E = hf and f   c , so
c hc  4.136  10
  
f
E
15
eV  s  2.998 108 m/s 
1.50 eV
 8.27  107 m  827 nm .
16. A one-dimensional crystal has sound waves with dispersion relation
  0 sin  kd  .
(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
vp =


0 sin  kd 
,
k
k
d
vg 
 0 d cos  kd 
dk
(b) Evaluate both velocities if k  1.80  109 m 1 , d  2.00  1010 m ,
0  6.00 1012 s 1 . Don’t forget to work in radians!
We simply substitute into the previous expressions.
vp =
0 sin  kd 
 6.00 10

k
vg  0 d cos  kd 
12
s 1  sin 1.80  109 m 1  2.00  1010 m 
1.80  109 m 1
 1174 m/s ,
  6.00 1012 s 1  2.00 1010 m  cos 1.80  109 m 1  2.00 1010 m   1123 m/s .
(c) Just as light waves are quantized, so also are sound waves. Find the energy
of one phonon (quantum of sound) in eV using the same numbers given in
part (b).
We use the quantum equation E   , using our formula for the frequency from
above
E    0 sin  kd 
  6.582 1016 eV  s  6.00 1012 s 1  sin 1.80  109 m 1  2.00 1010 m 
  0.00395 eV  sin  0.36   0.00139 eV .
2N
N
17. The wave function for a particle is given by
0  x  3a
 2N

  x    3 N
 0

 (x )
x
3a  x  4a
-N
otherwise
a
2a 3a 4a
-2N
where N and a are positive constants. The
wave is sketched at right.
(a) What is the normalization constant N?
-3N
We use the normalization condition:

1     x  dx  
2

3a
0
2 N dx  
2
4a
3a
3 N dx  4 N 2 x 0  9 N 2 x 3a  12aN 2  9aN 2
3a
2
4a
 21aN ,
2
N
1
.
21a
(b) Which of the regions has the highest probability density of finding the
particle?
The probability density is just given by   x  which is larger in the region
2
3a  x  4a . So the region on the right has the highest probability density.
(c) If the particle’s position is measured, what is the probability that it will be
found to have a value in the range 0  x  a ?
We calculate this in the usual way:
P  0  x  a      x  dx  4 N 2 x 0  4 N 2 a 
a
0
2
a
4a
4

 19.05%
21a 21