PH2400 - Spring 2001 - Exam II - Answers
Multiple Choice
1. a, c, and d
2. b
3. c
4. c
5. a
Short Answer
6. sqrt[(x2-x1)2 - c2
(t2-t1)2]. The space-time
interval is the same for all observers. Hence one can
simply compute the space-time interval for the Earth observer.
7. All have quantized energy levels. Looking at the shape
of the wavefunctions, the lowest energy state has 0 nodes
(excluding the walls for the "particle in a box"), the
next highest state 1 node, etc. For the atom, the energy
levels get closer together with increasing n, for the harmonic
oscillator they are uniformly spaced, and for the "particle in
a box" they get farther apart.
A rather subtle similarity, you may not have noticed, is that
for all, as you increase n, the particle is more likely to
be found farther from the center.
8. The ultraviolet catastrophy refers to the fact that the
classical theory for black body radiation predicts an intensity
which becomes infite for short wavelengths, and a total emitted
power which is infinite.
9. We know from the uncertainty principle, plus some example
solutions, that the ground state energy (the lowest possible)
does not have zero kinetic energy, and hence one would say
that the system is moving.
10. Tunneling refers to the process whereby a particle can
go from one region to another even if there is a classically
forbidden barrier in between. That is, the particles energy
is too small to make it over the barrier. If the tunneling
is rapid, that means the barrier is not very high and the
particle almost has enough energy to make it over.
Problems
11. 32.4 ns.
12. 0.909c.
13. A2 = 45/(64 b9).
14. 3/10
15. 1.83 GeV.
bonus: 0.767 c
Explanations
Multiple choice
1. They always see the same speed of light, so c & d are the same,
plus they see the same relative speed for the other observer, a.
2. Time to go 5 km as seen by policeman = 5km/(0.8c) = 20.83 us.
gamma = 1/sqrt(1-v2/c2) = 1/0.6
Time on car's clock (which is proper time, and hence shortest) =
tpolice/gamma = 12.5 us
Alternately, the distance traveled by the car as seen by the driver is
5 km/gamma = 3 km. The driver sees the road go by at 0.8c so
it takes a time 3 km/(0.8c) = 12.5 us
3. The stopping potential does not depend on intensity, so c.
4. emission happens when energy of electron is lowered, i.e. n gets
smaller. Smallest n can get is 1, which gives the biggest change in
energy. Biggest change in energy gives smallest wavelength. Hence,
delta-E = 13.6 eV (1 - 1/32) = 12.089 eV. Then
lambda = hc/delta-E = 1240 eV.nm/12.089 eV = 102.6 eV. Ans is c.
5. Wavelength is h/p. Situation with smallest momentum, p, has
longest wavelength. That is a.
Problems
11. K- sees proper time = shortest time. Experimenter
sees gamma(1.237 x 10-8) s. To find gamma,
E = 493.7 + 800 MeV = gamma(493.7 MeV) so gamma = 2.62.
Hence, measure 2.62 x 1.237 x 10-8s = 32.4 ns.
12. Earth observer sees
ux = (ux' + v)/(1 + ux'v/c2)
where v = 0.950c, ux' = -0.3c. Substituting in, get 0.909c.
13. Need integral from -b to b of psi squared = 1. On doing this
integral, get
2 A2 b9 (1 - 2/5 + 1/9) = 1
so need A2 = 45/(64 b9).
14. Can do integral, but easier to look at figure. Each "hump"
has 1/5th the area, and a half-hump has 1/10th the area.
Total shaded area = 3/10 of the total area.
15. lambda = h/p = hc/pc = 0.3 fm so pc = hc/0.3 fm = 4.13 GeV
and E2 = (pc)2 + (mc2)2
= (4.13 GeV)2 + (3.76 GeV)2 = 31.19 (GeV)2
so E = 5.585 GeV.
but also, E = 3.76 GeV + K so K = 5.585 - 3.76 = 1.83 GeV
bonus: with E = 5.585 GeV = gamma(3.76 GeV) get gamma = 1.557
Solve for v to get 0.767 c.
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