Name _________________
Solutions to Test 2
October 18, 2013
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 J  s  4.136 1015 eV  s
Black Bodies
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
k  8.988 10 N  m / C
e  1.602 1019 C
1 eV  1.602 1019 J
ke 2
1

 0.00729735 
137
c
Compton Effect
Wave
h
Relationships

  
1  cos  
mc
2

h
12
k
 2.426 10 m
mc

1
 f 
2
T
Hydrogen Spectrum
9
1 
 1
 2
2
n m 
   91.17 nm  
2
1
2
U
 2  k BT 
15  c 
4
3
maxT  2.898 103 m  K
Rutherford Scattering
kqQ
 
cot  
b
2
m v
2
R
2Ze 2 k
E
Hydrogen-Like Atoms
c2  2 Z 2

k 2e4  Z 2

E
2 2 n 2
2n 2
 13.6 eV  Z 2
E
n2
Part I: Multiple Choice [20 points]
For each question, choose the best answer (2 points each)
1. What is the smallest positive angle  such that ei  1 ?
A) 14 
B) 12 
C) 
D) 2
Reduced Mass
mM

mM
E) 4
2. Which of the following is not true about the wavelengths of light that are emitted
from atoms when, for example, they are excited by putting a spark through them?
A) There is a simple formula for the spectral lines for hydrogen
B) There is a simple formula for the spectral lines for heavy elements, like iron
C) The wavelengths coming out can be used to identify which element the atoms are
D) The combination of wavelengths depends primarily on which element it is
E) The wavelengths correspond to the energy given off when the electron(s) move
from one level to another within the atom
3. If we have a perfect black body distribution, if we double the temperature, the energy
density is multiplied by a factor of ___ and the peak wavelength by a factor of ____
D) 16, ½
E) ½, ½
A) 2, 2
B) 2, ½
C) 16, 2
4. Which types of objects, under appropriate circumstances, can sometimes act like
waves and sometimes like particles?
A) Photons (only)
B) Electrons (only)
C) Atoms (only)
D) Photons and electrons, but not atoms
E) Photons, electrons, and atoms
5. The reason you can get very strong X-ray scattering at certain angles with certain
wavelengths from a crystal is because
A) There is constructive interference adding together waves scattering from
many layers of the atoms
B) These angles are the angles that the electrons are aligned along inside the atoms
C) These X-rays have the exact right energy to be absorbed and then re-emitted from
the crystals
D) The wavelength of the X-rays match exactly the wavelength of the electrons in
the crystal
E) The angles are those so that the X-rays can perfectly slip between atoms without
ever bumping into a nucleus
6. If an atom were the size of the circle at right, how large would the nucleus be?
A)
B) 
C) 
D) 
E) Smaller than those
7. The momentum of a particle with wavelength  is given by
B) h 
C)  h
D) 
E)  
A) h
8. In the Franck-Hertz experiment, increasing the voltage accelerating electrons through
a thin gas causes the current to increase, until a certain threshold is reached, when
there is a sudden decrease in the current. What is causing this decrease?
A) When the electrons reach this energy, they start producing electron-positron pairs,
which slows them down
B) The electrons have grown so energetic they actually quantum tunnel through the
collector plate
C) The electrons have enough energy to bump up atoms they collide with to a
higher energy, so the electrons can now lose energy
D) The high voltage makes their momentum high, and therefore very definite,
making the actual position of the electrons highly uncertain
E) The voltages are so strong, the thin gas starts moving, which is much slower than
the electrons
9. Which of the following is a consequence of one of the classical uncertainty
principles?
A) A sufficiently short musical note does not have a definite pitch
B) An object that is constrained in a box of size L has a position uncertainty of about
1
4 L
C) It is never possible to know very precisely the momentum of a particle
D) The process of measuring a particle’s wavelength changes that wavelength
E) I am entirely uncertain of the answer to this question, so mark it wrong
10. Which of the following best explains why the harmonic oscillator has a minimum
energy that is greater than zero?
A) Because a particle is a wave, its momentum cannot be exactly zero
B) Because a particle is a wave, its position cannot be exactly zero
C) Because of the uncertainty principle, the position and momentum cannot
simultaneously forced to be zero, their minimum value
D) Because of the uncertainty principle, the energy cannot be precisely known in
finite time
E) The process of measuring the harmonic oscillator forces its energy to become
positive
Part II: Short answer [20 points]
Choose two of the following questions and give a short answer (2-3 sentences)
(10 points each).
11. Light with frequency f shines on a metal. Explain under what conditions it will
manage to dislodge an electron from the metal, and if it does, how much voltage
V the resulting electron will be able to go against before being stopped. At least
one formula is highly advised.
There is a certain amount of work required to remove an electron from the metal.
This is called the work function . If the frequency of the photon f is high enough such
that the energy of the photon, hf, exceeds , then the electron can be ejected. The
leftover energy allows the electron to overcome a potential up to Vmax, where
hf    eVmax .
12. Explain qualitatively (or with a picture) the difference between group velocity
and phase velocity. For a quantum wave packet (like an electron), which is the
speed of the electron?
The group velocity is how quickly a wave packet moves. The phase velocity is
how quickly the individual waves move within the wave packet. These are generally
different. A physical particle like an electron moves at the group velocity.
13. What did Rutherford learn by scattering -particles off of atoms? Also, what in
particular was he able to measure by using the highest energy -particles
available and using atoms with small total positive charge Z?
First, he learned that all of the positive charge and most (essentially all) of the
mass was located in a small object in the center, called the nucleus. By using the highest
energy -particles available and using Aluminum, he was actually get the -particles
close enough to touch the nucleus, and measured its size.
Part III: Calculation: [60 points]
Choose three of the following four questions and
perform the indicated calculations (20 points each).



14. An experimenter is scattering X-rays of unknown wavelength  off of electrons,
and measuring the wavelength   when they scatter to an arbitrary angle  , as
sketched above. At an angle of   90 , the observed wavelength is
   1.083 1011 m .
(a) What is the incoming wavelength  ?
We use the formula     
h
1  cos   . We therefore have
mc
h
1  cos    10.83 pm    2.426 pm 1  cos 90 
mc
 10.83 pm    2.426 pm 1  0   8.40 pm .
   
(b) At what angle would the observed scattered wavelength be
   1.204 1011 m ?
Again, we use the formula for Compton scattering. We have
h
   
1  cos   ,
mc
12.04 pm    8.40 pm    2.426 pm 1  cos   ,
3.64 pm
 1  cos  ,
2.426 pm
3.64
 1  1.5004  0.5004 ,
cos   1 
2.426
  120.
(c) What is the frequency for the incoming X-rays? What is the energy in eV?
We calculate the frequency using c  f  , so that
f 
c


2.998 108 m/s
 3.569  109 s 1
8.40 1012 m
We then go on to find the energy using E = hf, so that
E  hf   3.569 1019 s 1  4.136 1015 eV  s   1.476 105 eV  147.6 keV .
15. Pionic hydrogen consists of a proton (mp = 938.3 MeV/c2) electromagnetically
bound to a   meson (m = 139.6 MeV/c2). Except for the replacement of the
electron with a pion, it is otherwise just like ordinary hydrogen.
(a) What is the reduced mass for this system, preferably in MeV/c2?
We use the reduced mass formula,
938.3 MeV / c 2 139.6 MeV / c 2 

mM


 121.5 MeV / c 2 .
2
2
m  M  938.3 MeV / c   139.6 MeV / c 
(b) What is the energy of the n’th state of a pion bound in pionic hydrogen?
Hydrogen has Z = 1. We can’t use the formula with 13.6 eV in it, since this
formula assumes we are using an electron. We therefore work it out in terms of the
reduced mass, so we have
 c 
E
2
2n
2
2
Z2
121.5 MeV  0.0072974 

2n
2
2 2
1

0.003236 MeV
3.236 keV

2
n
n2
(c) If the pion fell from the n = 20 to n = 17 state, what would be the energy, in
eV, of the resulting photon that would be emitted?
The energy of the photon is the energy lost, or the initial energy minus the final
energy, and therefore is
E  
3.236 keV 3.236 keV

  3.236 keV  0.0009602   0.003107 keV  3.107 eV .
202
17 2
(d) Find the wavelength in nm of the photon you found in part (c).
Knowing the energy, we can find the frequency from E = hf, and then the
wavelength from f   c . Putting this all together, we have
c hc  4.136  10
  
f
E
15
eV  s  2.998 108 m/s 
3.107 eV
This is visible, but near the violet end of the spectrum.
 3.990  107 m  399.0 nm .
16. A proton is believed to have three objects called quarks inside of them. A proton
may be thought of as a bag approximately 1.76×10–15 m in diameter, inside
which must be found the quarks.
(a) According to Carlson’s rule, what is the approximate uncertainty in the
position of the quark?
Carlson’s rule says that the uncertainty in the position is one-fourth of the size of
the region to which it is restricted, so
x  14 L 
1
4
1.76 10
15
m   4.40 1016 m .
(b) What is the corresponding minimum uncertainty in the momentum of the
quark?
The uncertainty principle states that xp  12  , so we have
p 

1.055  1034 J  s

 1.199 1019 kg  m/s .
16
2x 2  4.40  10 m 
(c) Using non-relativistic formulas, find the corresponding uncertainty in the
velocity of the quark in m/s. Assume a quark has a mass
mq  13 m p  5.58  1028 kg .
The momentum and velocity are related by p = mv, so we have p  mv , so
v 
p 1.199  1019 kg  m/s

 2.15  108 m/s .
m
5.58 1028 kg
(d) Find v c . If you followed directions in part (c), you used non-relativistic
formulas. Do you think this was a good idea? Why or why not?
As directed, we find
v 2.15  108 m/s

 0.717 .
c 2.998  108 m/s
Since it is moving at about 72% of the speed of light, according to this crude
approximation, it is certainly very foolish to ignore relativity.
17. The wave function for a particle in the n = 3
state of the infinite square well has wave
function
 N sin  3 x a 
  x  

0  x  a,
otherwise .
0
where N and a are positive constants. The wave is sketched at right. Possibly
helpful integrals are listed below.
(a) Based on the graph estimate the most likely values of x/a to find the particle.
You are not expected to do any calculations.
The most likely places are when the wave function is most positive or most
negative. From the graph, it is clear this is around x/a = 0.16, 0.50, and 0.84. The actual
values, it is not hard to show, are x  16 a, 12 a, 56 a .
(b) What is the normalization constant N?
The normalization condition is

1     x  dx  
2

a
a
0
a
 3 x 
 6 x  
2 x
N sin 
sin 
 dx  N  

 a 
 a 0
 2 12
2
2
2
a
a
 N a
sin  6   
,
 N2  
2
 2 12

N
2
a
(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  14 a ?
We simply redo the calculation from before, but substitute the found value of N
and the appropriately modified integrals, so we have
1a
4
x
a
2 a
a
 6 x  
 3
P  0  x  a      x  dx  N  
sin 
  
sin 


0
a  8 12
 a  0
 2
 2 12
1 1
1 1
 
 1    30.31% .
4 6
4 6
1
4
1a
4
2
2
 sin  x  dx    cos  x  ,
Indefinite Integrals:
 cos  x  dx   sin  x  ,
1
1
 sin  x  dx 
 cos  x  dx 



2
1
2
x  41 sin  2 x  ,
2
1
2
x  41 sin  2 x  .