Name _________________
Solutions to Final Exam
December 12, 2014
This test consists of five parts. Please note that in parts II through V, you can skip one question
of those offered.
Part I: Multiple Choice (mixed new and review questions) [50 points]
For each question, choose the best answer (2 points each)
1. In what way is gravity different from other forces, like electromagnetism, and this clue
helped Einstein come up with his general relativity?
A) Electromagnetism is much stronger, in general, than gravity
B) Electromagnetism has vector fields like the electric and magnetic fields, unlike gravity
C) Electromagnetism binds atoms together; gravity does not
D) Gravity falls off with the square of distance, electromagnetism has a different power
E) All objects accelerate exactly the same way in a gravitational field, unlike from
other forces
2. What is the strongest piece of evidence demonstrating the presence of dark matter in spiral
galaxies?
A) The obstructed view of large portions of the galaxy, the light blocked by the dark matter
B) Measuring the rotational velocities of gas using the 21 cm line of hydrogen
C) Measuring the orbital periods of stars around the center of the galaxy
D) The focusing of light by MACHO’s in the galactic halo
E) The clumping of stars caused by concentrations of dark matter
3. What is the significance of the event horizon, or Schwarzschild radius, for a black hole?
A) It is the point of no return; beyond this, nothing can leave the black hole
B) It is the last radius at which an object can orbit in a circle around the black hole
C) It is the point at which tidal forces would rip any physical object apart
D) It is the point at which gravitational forces become infinite
E) It is the point at which spacetime around a black hole begins to be curved
4. Under what circumstance can you tell that event A occurred before event B, as agreed by all
observers?
A) If the time of A is before the time of B
B) If the time of A is before the time of B, and the events are timelike separated
C) If the time of A is before the time of B, and the events are spacelike separated
D) If the time of A is before the time of B, and they occur at the different locations
E) It is impossible to ever determine which of two events occurs first
5. If an object of mass m is accelerated to near the speed of light, what is the upper limit of its
total energy?
A) 12 mc 2
B) mc 2
C) 32 mc 2
D) 2mc 2
E) 
6. The largest amount of our galaxy’s mass is in the
A) Disk
B) Bulge
C) Halo
D) Nucleus
E) Fast food restaurants
7. In one half-life, fifty percent of the original atoms are gone. What percentage of the original
atoms are gone after two-half lives?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%
8. What assumption did Planck have to make to avoid the ultraviolet catastrophe?
A) The number of polarizations of light at high energy had to be modified
B) An upper limit had to be put on the frequencies of light that could occur
C) The energy in each mode had to come in multiples of a constant times the frequency
D) The energy in each mode had to come in multiples of a constant times the wavelength
E) Light had to be in circular orbits around atoms
9. The stress-energy tensor measures the presence of matter and other sources of gravity.
According to general relativity, what is the stress energy tensor proportional to?
A) The metric, or local distance formula
B) The geodesic, or formula for the path of a particle
C) The Einstein tensor, a measure of space-time curvature
D) The local velocity of light
E) The torsion field, or measure of the twisting of space-time
10. What property of the hydrogen atom made it possible to factor the wave function into an
angular part and a radial part, reducing a 3D problem to a 1D problem?
A) The potential is spherically symmetric
B) The force is an inverse-square law
C) The electron mass is much less than the nuclear mass
D) The nucleus is much smaller than the size of the atom
E) The atom is net neutral, so the nuclear charge cancels the electron’s charge
11. An  particle contains ___ protons and ___ neutrons
A) 0, 0
B) 0, 2
C) 2, 0
D) 2, 2
E) 2, 4
12. Why is it X-rays that produce diffraction patterns when they scatter off of crystals, rather
than some other wavelength?
A) X-rays have a wavelength comparable to the interatomic spacing in crystals
B) Only X-rays have enough energy to knock an inner electron out of the crystal
C) Only X-rays can penetrate deep enough in the crystal to scatter from it
D) Only X-rays feel the effects of the atoms, other wavelengths are immune
E) Only X-rays carry enough momentum to bounce off the atomic layers
13. What is the superiority of the formula c 2 d 2  c 2  dt    dx    dy    dz  used in general
2
2
2
2
relativity compared to the usual formula, c 2  2  c 2  t    x    y    z 
2
A)
B)
C)
D)
E)
2
2
2
It contains curvature, allowing one to move to general relativity
It allows one to calculate proper time along curved paths, not just straight ones
It allows for both spacelike and timelike paths, not just one or the other
It makes it easier to quantize the theory
It has the Latin letter d, which is easier to write than the Greek letter 
14. Which of the following is a correct uncertainty principle?
A) xp  12 
B) xp  12  C) xt  12  D) xt  12 
E) pt  12 
15. How bright are the brightest active galaxies?
A) A bit brighter than our Sun, but not as bright as a hundred Suns
B) Brighter than a hundred Suns, but not as bright as a million Suns
C) Brighter than a million Suns, but much dimmer than our galaxy
D) About as bright as our entire galaxy
E) Considerably brighter than our entire galaxy
16. Which of the following is a surprising but true fact that comes from special relativity?
A) One can determine which of two observers is actually moving by measuring the relative
rate of their clocks
B) One can determine which of two observers is actually moving by measuring the relative
length of their measuring sticks
C) There are no objects, even fundamental particles, that travel at the speed of light
D) There is no such thing as a truly rigid object
E) Matter and energy are the same thing
17. How do we hope to detect gravity waves in the near future, or have they been detected in the
past?
A) By carefully measuring oscillations in the distance of two arms of a large
interferometer
B) By studying the rotation of sensitive gyroscopes in orbit around the Earth
C) By studying the change in the orbit of planets and/or satellites
D) By measuring the bending of starlight around the Sun
E) It is not conceivable that gravity waves will be directly detected in the near future
18. 12B contains twice as many neutrons as protons. The most likely decay for this type of
nucleus would be
A) –
B) +
C) 
D) 
E) electron capture
19. In one version of the barn-pole paradox, a runner with a pole the same length as a barn runs
through a barn, and the farmer then closes the doors both simultaneously, proving the pole is
shorter than the barn. How does this resolve or not resolve the paradox?
A) It resolves it by demonstrating that the runner is truly moving, not the barn
B) It fails to resolve it, because the farmer will fail to close both doors, since the pole is
actually longer
C) It fails to resolve it, because the farmer will see both doors close, but the runner will say
he didn’t close them
D) It fails to resolve it, because the runner will say that the doors were not closed at the
same time
E) It fails to resolve it, because the runner will claim the pole shrank briefly, but only during
the time he was inside the barn
20. Which of the following is close to the mass of a 120Sn atom (atomic number 50)?
A) 50 u
B) 70 u
C) 120 u
D) 190 u
E) None of the preceding
21. Which of the following is true when we do quantum mechanics in 3D?
A) There are separate wave functions for each of the three dimensions
B) The wave function is usually a sum of three wave functions, one for each dimension
C) The energies are usually the product of energies in each of the three dimensions
D) There wave function is never a function of angles, just the distance from the origin
E) The wave function becomes a function of x, y, and z, rather than just x.
22. Why did J.J.Thomson put electrons he liberated in a magnetic field?
A) The field kept the particles trapped long enough that he could study them
B) Magnetic fields allowed him to separate electrons from extraneous particles he wasn’t
interested in
C) Electrons were attracted to the magnet, allowing him to collect them
D) The curvature told him the mass to charge ratio of the electrons
E) He was trying to explain the Earth’s magnetic field, and thought studying electrons might
clarify this
23. At the right is a crude sketch of our galaxy. Where do we
live?
A
B CD E
24. Which types of galaxies typically contain gas, dust, and young stars?
A) Elliptical (only)
B) Spiral (only)
C) Barred spiral (only)
D) Elliptical and spiral, but not barred spiral
E) Spiral and barred spiral, but not elliptical
25. Classically, the minimum energy for a particle in a finite box is zero. When we include
quantum mechanics, the minimum energy is
A) Negative (always)
B) Zero (always)
C) Positive (always)
D) Positive or negative, but not zero
E) Positive, negative, or zero
Part II: Short answer (review material) [20 points]
Choose two of the following three questions and give a short answer (1-3 sentences) (10
points each).
26. If I fire a laser at rest, the light moves at speed c. If I am running forwards at 0.5c, the
laser will fire at speed 1.5c, as viewed by an observer at rest. Explain what, if anything,
is wrong with this conclusion.
The question falsely assumes the addition of velocities formula, and hence draws an
erroneous conclusion. In fact, light in vacuum always travels at speed c, as viewed by all
observers.
27. Explain what it means for a wave function to be normalized, giving the relevant
equation. Suppose we have a solution to Schrödinger’s equation that is not normalized,
but it does have a finite integral. Explain how to fix the problem.
A wave function is normalized if
comes out incorrect, say
  x    x 






  x  dx  1 . If the result of this integration
2
  x  dx  N , this can be fixed by simply replacing
2
N.
28. When a wave with energy E comes across a step potential V0 > E, explain qualitatively
what happens. In particular, what is the probability that it is reflected, and is there any
wave function in the classically forbidden region.
When the wave has insufficient energy to penetrate the barrier, it will be 100% reflected.
However, even though it is completely reflected, there will be a small, exponentially falling
wave function inside the forbidden region.
Part III: Short answer (new material) [30 points]
Choose three of the following four questions and give a short answer (1-3 sentences)
(10 points each).
29. Explain qualitatively why the most stable nuclei have approximately equal numbers of
protons and neutrons. Now explain why in heavy nuclei, there is a preference for
neutrons.
Protons and neutrons each fill similar quantum states in the nucleus, and hence you get
the lowest energies by filling them up to the same level. However, electric repulsion causes
protons to resent the presence of other protons, so for heavy nuclei, this slightly favors neutrons
which have no such repulsion.
30. In – decay, explain at the particle level which particles are being destroyed and
produced. It is observed that in – decay, the light negatively charged particle that
comes out has a range of energy, rather than coming out as a single energy. How do
you account for this range of energy?
At the particle level, a neutron is decaying to a proton, an electron, and an anti-neutrino.
The energy comes out as kinetic energy of the particles, principally the light particles (electron
and anti-neutrino). However, there is no way to control how much comes out as the electron and
how much the neutrino, so the energy in the electron will end up varying in each decay.
31. Explain, giving any relevant equations, why the evidence that certain active galactic
nuclei vary very quickly tells us that they are not large in size.
Even if a large object changed luminosity instantaneously, it would look like it varies in a
time t  d c , just because it takes longer for us to see the distant parts changing than the nearer
parts. Hence something that changes fast must be small.
32. What is a geodesic? When do particles follow geodesics?
In spacetime, a geodesic is the longest proper-time path between two events. Objects
follow geodesics when they have no forces other than gravitational forces on them.
Part IV: Calculation (review material) [40 points]
Choose two of the following three questions and perform the indicated calculations (20
points each)
1
1 v c
2
2
 
A

The card will be Lorentz contracted in the direction of motion by a
factor of  . To make it come out square, it would have to be thrown such
that its 8.9 cm dimension is reduced to 5.7 cm. We therefore would have:

A

33. A standard playing card is about 8.9 cm  5.7 cm (3½ in  2¼ in) in size. A card
thrower attempts to throw the card in such a way that to a stationary observer, it
appears square.
(a) How fast would the card have to be moving to look square?
8.9 cm
 1.561 ,
5.7 cm
1  v 2 c 2  1.561  0.4102 ,
2
v 2 c 2  1  0.4102  0.5898 ,
v  0.5898c  0.768c  2.302 108 m/s .
(b) Which direction would it have to be thrown?
It would have to be thrown in its long direction, so left or right as sketched.
(c) According to an observer moving along with the card, what would the dimensions of
the card be?
There is no Lorentz contraction as viewed by a co-moving observer, so the observer
would see it as 8.9 cm  5.7 cm.
(d) The card contains a timed explosive, designed to go off one second after it is thrown
(1.00 s in the card’s frame of reference). How long, according to a stationary
observer, after it is thrown does the card explode?
Because the card is moving relativistically, the time as we observe it will be longer than
the time as moved by a commoving observer, so t   . We have  from part (a), so our
answer is
t    1.5611.00 s   1.56 s .
34. Sound waves, just like light waves, are quantized, and come in discrete amounts. It also
satisfies the frequency-wavelength relation as light if we replace c by the speed of
sound, typically c  343 m/s . A speaker is putting out 1.00 W = 1.00 J/s of sound power
at middle C, a frequency of 262 Hz.
(a) What is the wavelength for this sound?
Since the problem says it satisfies the same frequency-wavelength relationship, so we
have c   f , and therefore

c 343 m/s

 1.309 m .
f
262 s 1
(b) What is the energy and momentum for one phonon (sound quantum) of this sound,
in J?
We use the usual relations E  hf and p  h  , and we see that
E  hf   6.626 1034 J  s  262 s 1   1.736 1031 J,
p
h


6.626 1034 J  s
 5.062 1034 kg  m/s .
1.309 m
(c) How many phonons per second are coming out of this speaker? What is the total
momentum per second coming from this speaker?
The rate of photons coming out is just

P
1.00 J/s

 5.76 1030 s 1.
31
E 1.736 10 J
If we multiply this by the momentum per phonon, we will get the force:
F  P   5.76 1030 s 1  5.062 1034 kg  m/s   0.00292 N .
35. An electron is in a hydrogen atom in the n = 3 state.
(a) What are the possible values of the angular momentum quantum number l? What
are the corresponding values of the angular momentum squared L2?
The general rule is that l runs from 0 to n – 1, so since n = 3, we can have l = 0, 1, 2. The
corresponding values of L2 are given by L2   2  l 2  l  , so the possible values are
L2  0, 2 2 , 6 2
(b) The actual value of L2 is measured, and discovered to take on its maximum value.
What are the possible values of m? What are the corresponding values of the z-component of angular momentum Lz?
Since it took on its maximum values, the value of l must be 2. In general, m runs from –l
to +l, so the possible values of m are m  2, 1, 0,1, 2 . The value of Lz is given by Lz  m , so
Lz  2, , 0, , 2 .
(c) The atom then loses energy when the electron falls the smallest possible amount to a
different value of n. How much energy, in eV, will the resulting light have?
Recall that for hydrogen, Z = 1.The smallest amount it can fall is from n = 3 to n = 2.
The energy difference comes out as a single photon of light, with energy
E  E2  E3 
13.6 eV 13.6 eV

  1.51 eV    3.40 eV   1.89 eV .
32
22
Part V: Calculation (new material): [60 points]
Choose three of the following four questions and perform the indicated calculations (20
points each)
36. Plutonium is normally thought of as being manmade, but it is likely there is some
natural plutonium present even today. 244Pu is the most stable isotope, with a half-life
of 8.1107 y. Suppose that when the Earth was formed, we had a lump of 1.00 kg of
pure 244Pu.
(a) What is the approximate weight, in u, of 244Pu? How many atoms would be in
1.00kg?
The approximate weight of 244Pu is 244 u. The number of atoms is then
N0 
M 1000 g 1000
1000


NA 
6.022 1023   2.47  1024 atoms

m 244 u
244
244
(b) What is the decay constant  for 244Pu in y-1?
The decay constant can be determined from

ln 2
0.693

 8.56 109 y 1 .
7
t1/2 8.110 y
(c) The Earth is now 4.54109 y old. How many atoms of 244Pu would be left now?
The number of atoms is just
N  N 0e  t   2.47 1024  exp    8.56 109 y 1  4.54 109 y   
  2.47 1024  e 38.86  3.27 107 atoms .
So, only 33 million atoms would remain.
(d) How many radioactive decays per year would be coming from this lump today?
How many decays would have occurred from this lump since you were born, if you
are 21 years old?
The rate is just R   N , so we have:
R   N   8.56 109 y 1  3.27 107   0.28 y 1 .
So, that’s a bit less than one decay every three years! In 21 years, the number of decays would
be
N  Rt   0.280 y 1   21 y   5.9 decays.
37. Photocopied with the equations on the next page is a
portion of Appendix A from the text. 236Np is an
unstable nucleus which decays first by electron
capture, then -decay, then -decay again, and then
– decay, one after the other. Keep in mind that for
each step, the parent isotope is the daughter from
the previous step
(a) What are the isotopes after each of these steps?
mode
e.c.


–
Daughter Q (MeV)
236
U
Th
228
Ra
228
Ac
232
0.929
4.573
4.085
0.046
Electron capture cancels out one unit of charge, while leaving A unchanged, so we first
go to U. Then we have -decay, which removes two units of charge and four units of A, so we
go first to 232Th and then 228Ra. Finally, – decay removes a negative charge (i.e., it increases
the charge by one) while leaving A unchanged, so it ends at 228Ac. These isotopes are all
included in the table above. Actually, it continues its decay, but we won’t be keeping track of
subsequent steps.
236
(b) What is the Q-value for each of these steps?
For the first and last step, the Q-value is just  M P  M D  c 2 , while for -decay it is
 M P  M D  M 4 He  c 2 .
So the energy is
Q   M 236 Np  M 236U  c 2   236.046559  236.045562  uc 2
 0.000977  931.5 MeV   0.929 MeV ,
Q   M 236U  M 232Th  M 4 He  c 2   236.045562  232.038051  4.002602  uc 2
 0.004909  931.5 MeV   4.573 MeV
Q   M 232Th  M 228 Ra  M 4 He  c 2   232.038051  228.031064  4.002602  uc 2
 0.004385  931.5 MeV   4.085 MeV
Q   M 228 Ra  M 228 Ac  c 2   228.031064  228.031015  uc 2
 0.000049  931.5 MeV   0.046 MeV
(c) On the first step, why didn’t it do + decay instead?
The + decay produces the same isotope (236U), but because it produces an anti-electron,
rather than absorbing an electron, it must come up with an additional energy of twice the
electron mass. Hence the energy available, or Q –value, will be
Q   M P  M D  c 2  2me c 2   0.929 MeV   1.022 MeV   0.093 MeV .
Since the result is negative, the process would need an external source for energy, and hence is
in fact impossible. That’s why it did electron capture instead.
38. A ring of atomic hydrogen gas is observed orbiting the
center of a distant galaxy. Normally, this gas
produces a spectral line with a wavelength of 21.10
cm, but in this case it is observed that the spectral line
is shifted to a wavelength that depends on how far out
the gas is from the center, as plotted at right.
(a) Find the red shift for gas at a distance of 200 Tm
from the center, and repeat for gas at a distance of
800 Tm. For each of these locations, find the
velocity of the gas in km/s.
Reading off the graph, we see that the observed wavelength is about 21.50 cm at 200 Tm
and 21.30 cm at 800 Tm. The corresponding redshifts are
200
21.50 cm
1 
 1  0.01896 ,
21.10 cm
0

21.30 cm
 800  1 
 1  0.00948 .
21.10 cm
0
z200 
z200
R (Tm)
200
400
z
0.01896
0.00948
v (km/s)
5680
2840
M (kg)
9.67  1037
9.67  1037
Since these are small numbers, we use the non-relativistic approximation z  v c to get v:
v200  z200 c   0.01896   2.998  105 km/s   5680 km/s ,
v800  z200 c   0.00948   2.998 105 km/s   2840 km/s
(b) Estimate the mass of the source of gravity for each of the distances in part (a) (1 Tm
= 1012 m). You may leave your answer in kg.
Squaring the relevant equation given at the end of the test, we have v 2  GM r , or
M  rv 2 G . We simply use this equation twice to get the mass:
14
6
rv 2  2.00 10 m  5.68 10 m/s 


 9.67 1037 kg ,
G
6.673 1011 m3 / kg / s 2
2
M 200
14
6
rv 2  8.00 10 m  2.84 10 m/s 


 9.67 1037 kg .
11
3
2
G
6.673 10 m / kg / s
2
M 800
All the results are accumulated in a table above.
(c) Based on your computations from part (b), is the source of gravity a single object at
the center, or is it a diffuse source of gravity?
We note that the mass worked out to approximately the same value in each case. This
implies that the gravity source is, indeed, concentrated at the center. If the mass were spread out
more, the second mass would have worked out substantially bigger than the first.
39. A scientist is studying a black hole. She stays safely far away from it, while a probe
travels to a distance of R = 100.00 km. It then sends a radio signal with a frequency f0 =
137.0 MHz up to the scientist, where she observes it at a frequency of f = 112.0 MHz.
(a) What is the mass of the black hole, in solar masses (MSun = 1.9891030 kg)?
We use the frequency formula and solve for the mass:
1/2
 2GM 
f  f0 1  2 
cr 

,
2
1
2GM f 2  112.0 MHz 
 2 
  0.668 ,
c2r
f 0  137.0 MHz 
2GM
 r 1  0.6683  0.3317 100.0 km   33.17 km ,
c2
2.998 108 m/s   33.17 103 m 

c2
M
 2.234 1031 kg ,
 33.17 km  
11
3
2
2G
2  6.673 10 m / kg / s 
2
M   2.234 1031 kg 
M
 11.2 M  .
1.989 1030 kg
(b) There is a timer on the probe so it is supposed to return after one hour. However,
an hour later, the probe is still near the black hole. What went wrong, and when
should the scientist expect the probe to return?
Because the probe is near the black hole, time effectively slows down. Hence the time as
viewed by a distant observer will be increased compared to the probe. The scientist should
expect the probe to return at
 2GM 
t   1  2 
cr 

1/ 2

f0
 137.0 MHz 
 1 h  
  1.22 h  73.4 min.
f
 112.0 MHz 
So, no reason to panic, it will be back 13 and a half minutes late.
(c) The scientist now sends the probe to the Schwarzschild radius. What is this
distance?
This was calculate in part (a), so we do no new work:
Rs 
2GM
 33.17 km .
c2
h  6.626 10
Constants:
34
Equations
J  s  4.136 1015 eV  s
u  931.494 MeV / c 2
  1.055 1034 J  s  6.582 1016 eV  s
u  1.661 1027 kg
G  6.673 1011 m3 / kg / s 2
2me c 2  1.022 MeV
M He  4.002602 u
N A  6.022 1023
Orbits:
 13.6 eV  Z 2
Hydrogen-like atoms energy: En 
n2
GM
v
r
Gravitational time dilation:
Red Shift:


  0 1 
  t 1
2GM 

c2r 
1/2
2GM
c2r
Schwarzschild radius: RS 
1/2
,
 2GM 
f  f 0 1  2 
cr 

Isotope Masses
2GM
c2