Name _________________
Solutions to Test 2
October 17, 2012
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
k  8.988 109 N  m 2 / C2
e  1.602 1019 C
1 eV  1.602 1019 J
ke 2
1

 7.29735 103 
137
c
Compton Effect
Wave
h
Relationships
   
1  cos  
mc
2


h
k
 2.426 1012 m
mc

1
 f 
2
T
Hydrogen Spectrum
1 
 1
   91.17 nm   2  2 
n m 
1
U
 2  k BT 
15  c 
4
3
maxT  2.898 103 m  K
Rutherford Scattering
kqQ
 
cot  
b
2
m v
2
2Ze 2 k
R
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
Reduced Mass
mM

mM
Part I: Multiple Choice [20 points]
For each question, choose the best answer (2 points each)
1. If the wave function is given by   x   2ie   x , what is the probability density?
A) 4e 2 x
B) 4ie2  x
C) 4e 2 x
2. The formula for group velocity is
B)  k
C)  k
A) f 
D) 4e   x
2
D) d  dk
E) 4e   x
2
E) f 
3. In the Frank-Hertz experiment, what is the source of the light that is being emitted?
A) The voltage is getting large enough to cause arcing, which is causing the glow
B) The accelerated electrons are getting converted directly into photons
C) Accelerated electron pairs collide to produce light
D) The electrons collide with the anode, which becomes warm and glows
E) The atoms that have been impacted by electrons fall back to a lower energy
state
4. Why do different isotopes of hydrogen have actually very slightly different spectral
lines?
A) The charges of the nuclei are very slightly different
B) You have to work with the reduced mass , rather than the electron mass,
which is a tiny bit different for different isotopes
C) The photons emitted are partially absorbed by the nucleus, losing some energy
D) The atoms are in motion, causing Doppler shift, but the heavier isotopes are
moving more slowly
E) Magnetic interactions with the nucleus cause a shift in the energies
5. Suppose you know the speed of an electron. What can you learn by studying the
curve it makes in a magnetic field?
A) Its mass m
B) Its charge e
C) The ratio of its mass and its charge, e/m
D) The product of its mass and its charge, me
E) None of these can be learned from this experiment
6. The fundamental assumption Planck had to make to explain the black body spectrum
was
A) Light acts as if it comes in chunks of energy proportional to the frequency
B) Light acts as if it comes in chunks of energy proportional to the wavelength
C) The black body spectrum must have a peak wavelength max
D) The angular momentum of photons was always a multiple of 
E) Light is actually waves, not particles as previously thought
7. Rutherford scattering with -particles allowed Rutherford to
A) Measure the angular momentum of the electrons in an atom
B) Measure the energy levels of the electrons in an atom
C) Measure the charge of the electrons in an atom
D) Measure the size of atoms
E) Measure the size of the nucleus
8. Which of the following was a success of the Bohr model of the atom?
A) It correctly predicted the approximate radius of hydrogen (only)
B) It correctly predicted the wavelengths of the light emitted from hydrogen (only)
C) It correctly predicted the wavelength of light emitted from other atoms, like
neutral helium (only)
D) A and B are true, but not C
E) A, B, and C are all true
9. What is the correct relationship between h and  ?
h
1
2
C)  
D) h 
A)   2 h B)  
2

h
E)  
1
2 h
10. How did deBroglie’s relation p  h help
explain the Bohr model?
A) It explained diffraction for electrons,
which was assumed by Bohr
B) It explained why only circular orbits
worked in the Bohr model
C) It explained why energy always was
emitted as a single photon
D) It explained why the angular
momentum was always multiples of 
E) It explained the isotope effects of
different isotopes of hydrogen
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 qualitatively how one can use the
spectrum of light from a distant star to
determine the star’s temperature,
assuming that spectrum is a black body distribution.
The color of a star gives an indication of the surface temperature of a star. For a
perfect black body distribution, you could find the wavelength where the most power is
produced, and then use Wien’s Law, maxT  2.898  103 m  K , to find the surface
temperature.
12. When a photon scatters from an electron at rest, how does the scattered
photon’s wavelength, momentum, and energy differ (increase or decrease) from
the initial photon?
According to the Compton scattering formula, the wavelength gets longer.
According to the deBroglie relation, this means that its momentum decreases, which
implies that the energy decreases as well.
13. Classically, it was not understood why the electron did not simply fall to the
position of the nucleus in an atom. Explain, in terms of the uncertainty
principle, why this doesn’t happen.
If we demand that the electron be at the position of the nucleus, we are specifying
its position very accurately, and according to the uncertainty principle, px  12  , this
means its momentum is very uncertain. With a large momentum uncertainty, the
momentum cannot be made small, and therefore it will have a lot of kinetic energy. To
minimize the energy you instead want to only make the electron near the nucleus.
Part III: Calculation: [60 points]
Choose three of the following four questions and perform the indicated
calculations (20 points each).
14. Ytterbium (Yb) has a work function   2.60 eV .
(a) What is the lowest frequency f that can liberate an electron from ytterbium?
A photon can only liberate an electron if it has more energy than the work
function. Since the energy of a photon is hf, this implies hf   , so we must have
f 

h

2.60 eV
 6.29 1014 Hz .
15
4.136 10 eV  s
(b) Suppose light with wavelength   257 nm impacts some ytterbium. What is
the voltage Vmax that the liberated electrons from ytterbium can overcome?
We first need to calculate the frequency, which is
f 
c


2.998 108 m/s
 1.167 1015 s 1 .
9
257 10 m
We then calculate
eVmax  hf     4.136 1015 eV  s 1.167 1015 s 1   2.60 eV  2.23 eV ,
Vmax  2.23 V .
(c) When an unknown light is shone on ytterbium, it is found that the electrons
can only overcome a voltage of Vmax = 1.75 V. What is the frequency of the
light?
Rearranging the same equation, we have
hf  eVmax    e 1.75 V   2.60 eV  1.75 eV  2.60 eV  4.35 eV ,
f 
4.35 eV
 1.052  1015 s 1 .
15
4.136 10 eV  s
15. A certain element, when all but one electron has been removed, emits light with
a wavelength of 7.60 nm when an electron falls from n = 2 to n = 1.
(a) What is the energy of this photon?
The energy can be found from the frequency, so we first need the frequency,
which is given by
f 
c


2.998  108 m/s
 3.945 1016 s 1
9
7.60  10 m
The energy is then
E  hf   4.136 1015 eV  s  3.945 1016 s 1   163.2 eV .
(b) What is the Z-value of this element?
The energy of an electron in an arbitrary level is
 13.6 eV  Z 2
E
n2
The difference photon’s energy is the difference between the n = 1 and n = 2 levels, so
13.6 eV  Z 2 13.6 eV  Z 2

E  

2
2
2
1
 13.6 eV  3.40 eV  Z 2  10.2 Z 2 eV .
Equating this to the photon energy, we have
10.2 Z 2 eV  163.2 eV ,
163.2 eV
Z2 
 16.00 ,
10.2 eV
Z  4.
Since Z has to be an integer, we conclude that it is exactly 4.
(c) If this atom had an electron fall from n = 9 to n = 8, what would be the
energy of the resulting photon?
We simply use the same formula again, which gives us
E  
13.6 eV  42  13.6 eV  42
92
82
 0.714 eV .
16. Neutrons with a mass of
m  1.675 1027 kg are being fired at
neutrons
a narrow slit of width
w  14.0 nm with a kinetic energy of
E  4.86 1023 J .
(a) What is the momentum of these neutrons?

The kinetic energy can be written in terms of the momentum as E  p 2 2m .
Solving for the momentum, we have
p 2  2mE ,
p  2mE  2 1.675 1027 kg  4.86 1023 kg  m 2s 2   4.035 1025 kg  m/s .
(b) By Carlson’s rule, what is the uncertainty in their vertical position based on
the fact that they went through the slit? Based on the uncertainty principle,
what is the corresponding uncertainty in their vertical momentum p ?
Carlson’s rules says that the position is uncertain by one-fourth of the width, so
y  w  3.50 nm . By the uncertainty principle, this means there will be a
corresponding uncertainty in the momentum in this direction, with a minimum value of
1
4

1.055 1034 J  s
p y 

 1.508 1026 kg  m/s .
9
2y 2  3.50  10 m
(c) Assume the total momentum p is the same after they pass through the slit,
but they have now acquired a vertical momentum p . How much has their
direction changed by passing through the slit?
p
If the total momentum remains p and they have a vertical

momentum p , it is easy to see from the sketch at right that they
are going at an angle given by sin   p y p . Hence the angle is given by
 p y
 p
  sin 1 
26

kg  m/s 
1  1.508  10
1
sin



  sin  0.0374   2.14 .
25


4.035
10
kg
m/s



p
17. The wave function of a particle is given by
 N  a 3 x  x 4  0  x  a,
  x  
0
otherwise .


Na 4
where N and a are positive constants. The
wave is sketched at right.
(a) What is the most likely place to find the
particle?
x a
The most likely place to find the particle is
where the wave function is maximized; that is, when its derivative vanishes. We
therefore have
d
  x   N  a3  4 x3  ,
dx
3
4 x  a3 ,
a
x  3  0.630a .
4
0
(b) What is the normalization constant N?
We demand that the probability that the particle is somewhere be one, so we have
1     x  dx   N 2  a 3 x  x 4  dx  N 2   a 6 x 2  2a 3 x 5  x8  dx

a
2

a
2
0
0
 N 2  13 a 6 x 3  62 a 3 x 6  19 x 9   N 2  13 a 9  13 a 9  19 a 9   19 N 2 a 9 ,
a
0
9
,
a9
N  3a 9 2 .
N2 
(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 3 2 ?
We simply repeat the integration of part (b), but change the limits and use our
value of N from part (b).

P 0 x
1
3
2

a 
a
3
2
0
 9a
9
  x  dx  N
2

1
3
a
6 1
2
a  13 a
3
2
3 1
4

1
3
a x  ax  x
6 3
a  91  18 a
6
3 6
2
6
9

1
9
3
2
9

a
3
2
0
 34  18  12861  78 .
Thus there is an 87.5% chance of finding the particle in this range.