Name _________________
Solutions to Final Exam
December 7, 2015
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. Suppose you have a barrier potential with a finite but large width L but height V0 > E, the
energy. On the other side of the barrier, what does the wave look like?
A) There is no wave at all, because the barrier is too high to get across
B) The wave always penetrates the barrier, so it has the same magnitude as it had before
C) The wave continues as a wave, but the magnitude is slightly reduced
D) The wave damps exponentially, falling off to zero eventually
E) The wave has a very small chance of penetrating, coming out as a much smaller
amplitude wave
2. Most of the mass of a typical galaxy is apparently made of
A) Stars
B) Gas
C) Dust
D) Dark matter E) Black Holes
3. According to Einstein’s equations, the curvature of spacetime is related to the
A) Presence of energy, momentum, etc., that makes up the stress-energy-momentum
tensor
B) Gravitational pull from nearby objects
C) Motion of test masses present at that location
D) Integral of mass over the nearby space divided by the distance
E) Number of geodesics passing through a given point
4. If observer A goes to a distant star and returns at high velocity, while observer B remains on
Earth, when they are reunited, how will their relative ages compare?
A) Observer A will be younger
B) Observer B will be younger
C) They will be the same age
D) It depends on which one you ask: each one will claim to be the younger
E) It depends on which one you ask: each one will claim to be the older
5. What was the nature of the ultraviolet catastrophe that Planck resolved? What was wrong
with the classical prediction for black body radiation?
A) It had too much energy at low frequencies/long wavelengths
B) It had too little energy at low frequencies/long wavelengths
C) It had too much energy at high frequencies/short wavelengths
D) It had too little energy at high frequencies/short wavelengths
E) None of the above
6. If the wave function at a point x is given by  , the probability density of finding the particle
at x is given by
A) 
B)  *
C)  2
D)  *
E)  *2
7. The best way to spot atomic hydrogen, and to map out the spiral arms of a galaxy, is by
A) The X-rays caused by collisions of high temperature atoms
B) The radio waves from vibrations of molecules
C) The 21 cm line caused by the spin flip of the electron
D) The glow of stars that have formed from atomic hydrogen
E) The infrared light that can make it through the dust clouds
8. Suppose you took 1 kg of water and heated it from 50 C to 100 C. According to special
relativity, would its mass change?
A) No
B) Yes; it would increase a tiny bit
C) Yes; it would increase a lot
D) Yes; it would decrease a tiny bit
E) Yes; it would decrease a lot
9. Which of the following is true when you are near a mass, such as the Earth?
A) Masses become greater
B) Masses become smaller
C) Time speeds up
D) Time slows down
E) None of the above
10. Which of the following particles experiences the strong force?
A) Protons (only)
B) Neutrons (only)
C) Electrons (only)
D) Protons and neutrons, but not electrons
E) Protons, neutrons, and electrons
11. When you place an object in an infinite square well of finite width, the minimum possible
energy it can have, according to quantum mechanics, is
A) Positive B) Negative C) Zero
D) Infinity
E) Negative infinity
12. In a galaxy like our own, the most likely place to find new stars being born is in the
A) Disk
B) Nucleus C) Halo
D) Bulge
E) None of these
13. Which of the following is not a standard categorization of galaxies?
A) Spindle B) Spiral
C) Elliptical
D) Barred Spiral
E) Elliptical
14. Which of the following is the correct four-dimensional distance formula in special relativity?
2
2
2
A) s 2   x    y    z 
B) s 2   x    y    z 
2
2
2
C) s 2   x    y    z   c 2  t 
2
D) s 2   x    y    z   c 2  t 
2
2
2
2
2
2
2
E) s 2    x    y    z   c 2  t 
2
2
2
2
15. According to quantum mechanics, an object with momentum p will have wavelength
A) p
B)  p
C) hp
D) p h
E) h p
16. At the center of many galaxies, including our own, lies a giant
A) Dust cloud
B) Star C) Molecular cloud D) Neutron star
E) Black hole
17. There are two types of -decay. The visible particle that comes out in the two types is one of
which two particles?
A) Proton or electron
B) Proton or neutron
C) Electron or neutron
D) Proton or anti-electron (positron)
E) Electron or anti-electron (positron)
18. Curvature is a measure of
A) How much your coordinates are curved compared to straight coordinates
B) How much spacetime is actually curved
C) Whether you are using Cartesian or spherical coordinates
D) How much matter there is locally
E) The amount of gravity present
19. When studying the wave function for the ground state of the Harmonic oscillator, it was
2
2
found that   e Ax satisfies Schrödinger’s equation, but ultimately we rejected   e Ax
because
A) It led to infinite energy
B) It was not continuous
C) It was not normalizable
2
D) It is equivalent to   e Ax , and it is easier to work with the minus sign
E) It was not computable
20. A geodesic in four dimensions is a path through spacetime that
A)
B)
C)
D)
E)
Stays at constant space coordinates while time changes
The coordinates change at a constant speed with respect to time
Has the longest proper time between two events
Is not curved in any coordinate system
Changes direction only when encountering a mass
21. Which of the following is not necessarily true in special relativity?
A) Nothing can go faster than light
B) Light in vacuum always goes at the speed of light
C) Energy is conserved
D) Momentum is conserved
E) Mass is conserved
22. Which type of nuclear decay can only happen when a nucleus starts off in an excited state?
A) +
B) –
C) 
D) 
E) Electron capture
23. Which of the following objects is not believed to be subject to quantum mechanics?
A) Electrons
B) Nuclei
C) Atoms
D) Molecules
E) All objects are affected by quantum mechanics
24. For a heavy but fairly stable nucleus, what fraction of the nucleons will typically be protons?
A) 30%
B) 40%
C) 50%
D) 60%
E) 70%
25. According to the equivalence principle, masses in a gravitational field experience much the
same thing as
A) Charges in an electric field
B) Masses as viewed in an accelerating reference frame
C) Objects being pulled by an external force
D) Waves interacting with a background potential
E) Matter absorbing momentum from impacting photons
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. In one version of the barn and pole paradox, a runner carrying a pole insists that the
pole is too long to fit inside a barn, while a farmer standing nearby insists that he can
close the front and back of the pole in the barn at the same moment, proving the pole fit
inside. Explain the resolution of this apparent paradox.
The resolution in this paradox concerns the ambiguity of the term “simultaneous” as
embodied in the words “at the same moment” in the wording. Although the farmer claims that
the front and back doors were closed simultaneously, the runner would claim that one door was
opened and closed before the other, and hence the pole was never actually completely inside the
barn.
27. A hydrogen atom has a nucleus with positive charge and an electron with negative
charge. Why doesn’t the electron simply go to the center of the atom (x = 0) and stay
there (v = 0), according to quantum mechanics? Your answer should have at least one
equation or inequality in it.
According to quantum mechanics, the particle is described not as a particle with a
definite location and velocity, but as a wave, which therefore has indefinite position and
momentum. Indeed, the uncertainty principle states that  x  p   12  , so that it is impossible
to specify both the position and the momentum (which is proportional to the velocity).
28. For a hydrogen-like atom with a single electron, the electron is described by four
numbers, n, l, m, and ms. Explain what physical quantity each of these numbers
describes about the electron. You do not have to give any equations.
The quantum number n tells us the energy of the electron, l tells us the total angular
momentum, m tells us the orbital angular momentum around the z-axis, and ms tells us the spin
around that same z-axis.
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. According to the modern theory of electromagnetism, electric attractions are caused by
the interchange of photons. What particles are interchanged in strong interactions that
hold the nucleus together? Explain qualitatively how this results in a force that is short
range rather than long range.
The strong force involves the transfer of pions, which come in three types, the +, the –
and the 0. Because these particles have a mass (unlike the photon, which is massless), they
have a minimum energy, and by the uncertainty relation between energy and time, they can exist
only for a short time. Therefore, they can only go a short distance, resulting in a short-range
force.
30. List the following four types of radioactive decay from most to least dangerous: alpha,
beta, gamma, and electron capture.
Electron capture involves the emission of only neutrinos, and is therefore harmless.
Therefore, in order from most to least dangerous, we have gamma, beta, alpha, and electron
capture. Alpha emission can be stopped by a piece of paper (or even dead skin), beta can
penetrate some materials but can be blocked by relatively thin layers of metal, and gammas can
penetrate even significant thicknesses of lead.
31. Explain qualitatively or in pictures what causes a blazar, which can vary in intensity
quickly and is very bright, to be different from other types of active galactic nuclei.
A blazar is an active galactic nucleus with a jet, where we happen to be looking directly
along the line of the jet coming from the nucleus. Because the jet beams a lot of its power
straight forwards, a blazar can be very bright. Also, because the jet can temporarily turn on or
off as matter is fed or not fed into it, it can vary on a short time scale.
32. Explain approximately what a gravity wave is. Although they have not been detected
directly, explain how studying pulsars has given us indirect evidence that they do exist.
A gravity wave is a distortion in spacetime that travels at the speed of light from a source.
It causes an oscillating fluctuation in the distance between objects, and can be detected directly
with interferometers. Although they have not been detected this way, we have seen pairs of
orbiting pulsars that gradually are moving together, signaling that they are moving towards each
other from the loss of energy due to these gravity waves.
Part IV: Calculation (review material) [40 points]
Choose two of the following three questions and perform the indicated calculations (20
points each)
33. An atom of the isotope 64Cu, when at rest, can be thought of as a sphere of diameter of
approximately 290 pm, and has a half-life of 12.70 hours.
(a) A group of 64Cu atoms are moving at such high speed that their actual half-life is
instead 18.64 hours. How fast are they moving? You may give the answer in m/s or
as a fraction of c.
The half-life is the life span as measured when you are moving along with the atom. If it
is moving, this time will get dilated to t   . We therefore have
1
1 v c
2
2
 
t


18.64 h
 1.468 ,
12.70 h
1  v 2 c 2  1.4682  0.4642 ,
v 2 c 2  1  0.4642  0.5358 ,
v  0.5358c  0.732c  0.732  2.998  108 m/s   2.194  108 m/s .
(b) What would be the shape of the atoms? Clarify which directions are which.
The atom will Lorentz contract, but only in the direction it is
moving. The “diameter” in this direction will be reduced to
L
Lp


290 pm
 197.6 pm .
1.468
The shape will be an oblate (s quashed) spheroid, with this distance in
the direction of motion and the other two dimensions unchanged, as
sketched at right.
(c) A 64Cu atom has approximately a mass of 59.62 GeV/c2. What is the total energy of
each of these atoms in GeV?
We simply calculate the energy using
E   mc 2  1.468  59.62 GeV   87.51 GeV .
(d) What is the momentum of these atoms in GeV/c?
This is easily found from
p   mv  1.468  59.62 GeV / c 2   0.732 c   64.05 GeV/c .
34. It is found that when light of wavelength  = 237 nm impacts Mg, electrons of energy up
to 1.55 eV are emitted.
(a) What is the frequency and energy for  = 237 nm?
The frequency can be found from f   c , or solving for the wavelength,
2.998 108 m/s
f  
 1.265  1015 s 1  1.265 1015 Hz .
9

237  10 m
c
The energy is then found from E = hf to be
E  hf   4.136 1015 eV  s 1.265  1015 s 1   5.232 eV.
(b) What is the work function for Mg?
If an electron of energy hf impacts Mg, and the work function for Mg is , then the
amount of energy we can get out is at most Emax  hf   . It follows that the work function must
be
  hf  Emax   5.232 eV   1.55 eV   3.68 eV.
(c) What is the longest wavelength of light that can extract an electron from Mg?
The longest wavelength that can extract electrons will be ones with just enough energy to
overcome the work function, so they will have hf   . This allows us to find the frequency,
which is
f 

h

3.68 eV
 8.90  1014 s 1 .
15
4.136 10 eV  s
We then find the wavelength again from f   c , so that

c 2.998 108 m/s

 3.37  107 m = 337 nm.
15 1
f
8.90  10 s
35. A particle is in a 1D harmonic oscillator with angular frequency  = 1.26  1015 s–1 .
(a) If the particle is in the n = 6 quantum state, what is the energy of the harmonic
oscillator, in eV?
The energy in the n’th state is given by
En    n  12    6.582 1016 eV  s 1.26 1015 s 1   6  12   6.5  0.829 eV   5.39 eV .
(b) Suppose it then falls to the n = 3 state. Would it emit or absorb a photon? What
would be the corresponding energy of this photon?
As n increases, the energy increases, so since n decreased, the energy decreased, which
implies energy is emitted. The amount of energy would be given by
E  E6  E3   6  12     3  12    3  3  0.829 eV   2.49 eV .
(c) Suppose, instead, that it absorbed a photon of energy approximately 3.32 eV. What
would be the value of n now?
Since this time it is absorbing a photon, the energy must be increasing. The final energy
must be the initial energy plus the energy added, so
En  E6  E   5.39 eV    3.32 eV   8.71 eV.
We can then solve for the final value n by equating this to the formula for En:
  n  12   En ,
En
8.71 eV

 10.51 ,
 0.829 eV
n  10.01.
n  12 
Since n must be an integer, we assume the discrepancy is due to rounding error, and hence it
ends at level n = 10.
Part V: Calculation (new material): [60 points]
Choose three of the following four questions and perform the calculations (20 points each)
36. Potassium is commonly incorporated into rocks. Argon, by comparison, almost never
is present when rocks form. Consider a rock that is formed with 1.60  108 atoms of
40
K, an isotope with half-life of t1/2 = 1.248  109 y.
(a) What is the decay constant  for 40K in y-1?
We use the formula  t1/2  ln 2 to find

ln 2
ln 2

 5.54  1010 y.
9
t1/2 1.248 10 y
(b) Suppose a rock is 3.74  109 y old. How many atoms of 40K would remain today?
We use the formula N  N 0 e  t to find
N  N 0 e  t  1.60 108  exp    5.55 1010 y 1  3.74  109 y    1.60 108  e 2.077  2.00 107 .
(c)
40
K has multiple decay modes, but approximately 10.72% of the time, it decays to
Ar. Assuming these atoms are trapped in the rock, how many atoms of 40Ar will
there be?
40
Since initially, there were 1.60 108 atoms, but all but 2.00  107 atoms have decayed, it
follows that the number of decays must be N d  1.60 108  2.00 107  1.40 108 atoms. Since
10.72% of them decayed to argon, the number of argon atoms must be
N Ar  N d  0.1072   1.40 108   0.1072   1.50 107.
(d) Suppose another sample of rock had equal numbers of 40Ar and 40K. Would you
predict the rock is younger than 3.74  109 y, or older, or what?
After 3.74  109 y, there is still more 40K than 40Ar, though the numbers are now
comparable. But the 40K is continuously dropping and the 40Ar is rising, so if we wait a bit
longer they will become equal. Although the amount of each of these elements will depend on
the initial amount of 40K, the ratio of these two isotopes will not, which suggests that this
argument is independent of the initial concentrations. In summary, a rock with equal amounts of
these two isotopes must be somewhat older than 3.74 billion years. Indeed, it would be about
4.21 billion years old.
37. Photocopied with the equations on the next page is a portion of Appendix A from the
text. 215Po is a nucleus that might decay
mode Daughter Q (MeV) Possible?
by one of the modes listed at right. You
211

Pb 7.528 yes
may use the table at right to summarize
electro
your answers, if you wish.
no
215
n
Bi -2.252
(a) For each of these, what is the
capture
daughter isotope?
215
+
Bi -3.274 no
For -decay, the value of Z decreases by
215
–
At 0 727 yes
two, and the value of A by four, so we end up
with 211Pb. For electron capture and + decay, Z decreases by one while A is unchanged, so we
end with 215Bi. Finally, for – decay, Z increases by one while A is unchanged, so we have 215At.
(b) What is the Q-value for each decay?
For -decay, we use the formula
Q   M P  M D  M He  c 2   214.999418  210.988734  4.002602  uc 2
 0.008082  931.5 MeV   7.528 MeV .
For electron capture and – decay, the formulas look very similar:
Qec   M P  M D  c 2   214.999418  215.001836  uc 2  0.002418  931.5 MeV   2.252 MeV,
Q    M P  M D  c 2   214.999418  214.998638  uc 2  0.000780  931.5 MeV   0.7266 MeV.
The only difference was the daughter isotope. The formula for + decay looks a little different,
but the daughter isotope is the same as for electron capture, so it is quickly worked out.
Q    M P  M D  c 2  2me c 2   2.252 MeV   1.022 MeV   3.274 MeV .
All the values have been incorporated into the table above.
(c) Which of the modes can actually occur?
The only modes that are possible are those with positive Q-values, so  and – are both
possible. Because of the smallness of the energy available for – decay, we would guess
(correctly) that it is rare. In fact, this decay occurs only about 0.023% of the time.
38. A large central galaxy A has a smaller satellite galaxy B orbiting
it at a distance of 1.25 Mly (1 Mly = 9.46  1021 m). The
hydrogen- line, normally at a wavelength of 656.28 nm, is
observed to be at 678.35 nm coming from A and 678.02 nm
coming from galaxy B.
(a) What is the approximate radial velocity of each of
these galaxies towards or away from us, in km/s?
1.25 Mly
B
A
We first need the z-value for each of these, which are given
by
To Earth
A  0 678.35 nm  656.28 nm

 0.03363
656.28 nm
0
  0 678.02 nm  656.28 nm
zB  B

 0.03313
656.28 nm
0
zA 
Since these are both pretty small numbers, we can use the simple formula z  v c to estimate
each of these velocities.
v A  z Ac  0.03363  299,800 km/s   10082 km/s
vB  z B c  0.03313  299,800 km/s   9931 km/s
Since they came out positive, they are both moving away from us.
(b) Assume galaxy B is in circular orbit around galaxy A, such that we are viewing the
orbit edge on (see sketch at right). Which direction, and at what velocity, is galaxy
B orbiting?
Since galaxy B is moving more slowly away from us than galaxy A, it is in fact orbiting
clockwise, or towards us at the moment. The orbital velocity is just
v  vA  vB  10082 km/s  9931 km/s  151 km/s
(c) Estimate the mass of galaxy A, in solar masses (1 MSun = 1.989  1030 kg).
We take the formula for orbital velocity and rearrange it:
GM
,
r
v 2 r  GM
v
5
21
v 2 r 1.51 10 m/s  1.25  9.46  10 m 
4.04  1042 kg


 2.03  1012 M 
M
3
2
30
11
6.673 10 m / kg / s
1.989 10 kg/M 
G
2
39. A probe is studying a neutron star at a distance of r = 10.72 km. The probe is designed
to communicate using a laser of wavelength 437 nm, but distant observers see a
wavelength of 532 nm.
(a) What is the mass of the neutron star, in solar masses (MSun = 1.9891030 kg)?
The wavelength shift due to gravitational interactions is given on the formula sheet. We
rearrange this formula to get the mass of the neutron star:


  0 1 
2GM 

c2r 
1/2
,
2GM 0
 ,
c2r

2

2GM
1  02  2 ,
cr

1
8
c 2 r  02   2.998 10 m/s  10720 m   437 2 
2.348 1030 kg
M
1
1




 1.181 M  .



2 
30
2G   2 
2  6.673 1011 m3 / kg / s 2   532  1.989 10 kg/M 
2
(b) After 7.00 days, as measured by a distant observer, a signal is sent to the probe to
move to a greater distance. How long will have passed on the probe?
Time slows down for the probe, so it will be less time. We can compute the
corresponding rate pretty quickly by comparing the formula to the relevant formula for the
wavelength, so we have
  t 1

2GM
437
 t 0   7.00 d 
 5.75 d .
2
cr
532

(c) To what radius would the neutron star have to be compressed to cause it to fall
inside its Schwarzschild radius and become a black hole?
This is easily computed using the formula
11
3
2
30
2GM 2  6.673 10 m / kg / s  2.348 10 kg 
RS  2 
 3487 m  3.487 km .
2
c
 2.998 108 m/s 
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
Energy and momentum in relativity: E   mc 2
Orbits:
v
GM
r
Harmonic Oscillator Energy: En    n  12  , n  0,1, 2,
Gravitational time dilation:
Red Shift:


  0 1 


p   mv
  t 1
2GM 

c2r 
1/2
2GM
c2r
Schwarzschild radius: RS 
1/2
,
 2GM 
f  f 0 1  2 
cr 

Isotope Masses
2GM
c2