Giving or receiving aid is cause for dismissal from the University. You may have one two-sided, 8-1/2x11-inch page of notes. Calculators may be used, but not computers. Show all work for Short and Long Problems on the exam booklet. No other materials may be used during the exam. The exam is 2 hours long.

g = [1/(Ö[(1-v²/c²)])] x¢ = g(x - vt) t¢ = g(t - vx/c²) f = dp/dt

E = gm c² = Ö[( m² c⁴ + c² p²)] K = E - mc² p = gm v c p/ E = v/c

K_max = hf - F 2dsinq = nl l¢- l = (h/mc) (1-cosf) E_n = -Z² hcR/n²

p = h/l = (^h/_2p) k E = hf = (^h/_2p)w Dx Dp ³ (^h/_2p)/2 DE Dt ³ (^h/_2p)/2

c = 3.00 ×10⁸ m/sec e = 1.60×10^-19 C h = 6.63×10^-34 J·sec
1 eV = 1.60×10^-19 J (^h/_2p) = h/2p = 1.06×10^-34 J·sec
hc = 1240 eV·nm (^h/_2p) c = 197 eV·nm R = 0.0110 nm^-1 hcR = 13.6 eV

m_e = 0.511 Mev/c² = 9.11 ×10^-31 kg
m_p = 938 Mev/c² = 1.673 ×10^-27 kg
m_n = 940 Mev/c² = 1.675 ×10^-27 kg

I. Six Multiple Choice Problems (30 points total, each 5 points, no part marks)

Indicate the correct answer.

1. What is the rest mass of a photon?
(a) hf
(b) hc/l
(c) 0
(d) ¥
(e) none of the above

2. An electron has velocity v = (Ö3/2) c. What is its total energy?
(a) 3 mc²
(b) 2 mc²
(c) (3/2) mc²
(d) mc²
(e) none of the above

3. Suppose a particle of rest mass 0.5 MeV/c² is put in a small box of size » 0.1 nm. What is its approximate kinetic energy?

(a) 0.01 eV
(b) 10 eV
(c) 10,000 eV = 10 keV
(d) 10,000,000 eV = 10 Mev
(e) 10,000,000,000 eV = 10 GeV

4. Suppose a photon hits a particle of mass m initially at rest. The photon's final momentum is in the opposite direction from its initial momentum. Assume total momentum and energy are conserved during the collision. What is l_final - l_initial?
(a) 0
(b) h/(2mc)
(c) h/mc
(d) 2 h/mc
(e) none of the above

5. What is the minimum wavelength photon that can be emitted from an excited He⁺ ion (to ±1 nm accuracy, note that He has two protons in its nucleus)?
(a) 3475 nm
(b) 879 nm
(c) 155 nm
(d) 23 nm
(e) none of the above

6. What is the wavelength of electrons with momentum 124 eV/c?
(a) 0.01 nm
(b) 0.1 nm
(c) 1 nm
(d) 10 nm
(e) none of the above

II. Two Short Problems (30 points total, each 15 points)

7. (a) At fantastic taxpayer expense, NASA launches an astronaut off to Alpha Centari. The rocket travels there at a velocity of 0.5 c relative to the earth. After reaching Alpha Centari, the astronaut immediately turns around and travels back, again at a velocity of 0.5 c relative to the earth. According to clocks on the spaceship, 13.5 years pass during the mission. How many years pass on the clocks back on earth? (7 points)

(b) Suppose that NASA communicates with the spaceship during the flight using radio waves sent from the earth at frequency f = 100.0 MHz. Find the frequencies that the astronaut must set his radio to during the outgoing and return trips. (8 points)

8. Two photons of momenta of equal magnitude but opposite directions collide, and turn into one p particle of mass 140 MeV/c².

(a) What is the total momentum of the p particle? (3 marks)

(b) What are the energies and wavelengths of the two photons? (7 marks)

(c) The p particle has a lifetime of 2.2×10^-6 sec. What is the minimum possible uncertainty in a measurement of its rest mass? (5 marks)

III. Two Longer Problems (40 points total, each 20 points)

9. Electrons are reflected from a crystal with spacing of its atomic planes of 1.7 Å. The m = 1 Bragg diffraction peak is observed at a Bragg scattering angle of 30^°.

(a) What is the kinetic energy of each electron? (6 marks)

(b) If the electron beam of (a) has a momentum uncertainty of 1%, estimate the minimum uncertainty in position of any one electron. (6 marks)

(c) What energy neutrons lead to the same Bragg scattering as in (a), i.e. a m = 1 peak at a Bragg scattering angle of 30^°? (5 marks)

(d) If photons of the same momentum as the electrons of (a) are Bragg-scattered from the same crystal planes, find the angle at which the m = 1 diffraction peak is observed. (3 marks)

10. This problem concerns a particle of mass m, confined inside a one-dimensional box of length L. You should consider the particle to be non-relativistic (i.e. v << c).

(a) Use the uncertainty principle to estimate the minimum kinetic energy possible for the particle in the box. (your answer is a formula) (4 marks)

(b) Sketch the shape of the space-dependent part (y(x)) of the ground-state (lowest-energy) wavefunction, and write down the precise ground-state kinetic energy (again as a formula). (4 marks)

(c) Suppose the particle has a wavefunction y(x) µ sin(3 px/L). Find its momentum and kinetic energy (as a formula). (4 marks)

(d) If the particle is a proton and if L = 1×10^-14 m, find actual numbers for (a), (b) and (c). (4 marks)

(e) In a few words, explain why (a) < (b) < (c). (4 marks)

1. (c) Photons are massless.

2. (b) g = 1/(1-v²/c²)^1/2 = 1/(1-3/4)^1/2 = 1/(1/2) = 2 so E = gmc² = 2 mc².

3. (b) Recall that the kinetic energy of an electron in a hydrogen atom - which is a 0.5 MeV/c² particle in a box - is about 13.6 eV. This already gives the approximate answer. From uncertainty principle, p » (^h/_2p)/a for each component of momentum for a particle in a box of diameter a. So K = p²/2m » (^h/_2p)² /(2m a²) where a is the box size, maybe multiply by 3 for three space dimensions, to get roughly K » (^h/_2p)² / (m a²) = ((^h/_2p) c)² / (mc² a²). Plug in (^h/_2p) c = 200 eV·nm, mc² = 500,000 eV, a = 0.1 nm, to find K » 10 eV as the best answer.

4. (d) This is Compton scattering (elastic scattering of a massless particle such as a photon from an electron or other massive particle); for reversal of photon momentum, q = 180^° and cosq = -1, thus l¢- l = 2 h/mc.

5. (d) The minimum wavelength is the maximum energy, which occurs for highly excited states (n®¥) near E = 0 making a transition all the way to the n = 1 state. Recall that He⁺ is a one-electron ion with Z = 2. The energy of such transitions is thus Z² ×13.6 eV = 4×13.6 eV = 54.4 eV, so l = hc/E = 1240 eV·nm/54.4 eV = 22.8 nm.

6. (d) Use l = h/p = hc/(cp) = 1240 eV·nm/124 eV = 10 nm.

Recall that you must use three inertial frames to understand why this is the appropriate way to analyze the problem - earth, departing spacecraft, and returning spacecraft. You cannot use the spacecraft as one inertial frame to analyze how the clocks are observed to run back on earth.

(b) While the spacecraft travels away from the earth, the transmission frequency is Doppler shifted to a lower value,
f = f_source (1-v/c)^1/2/(1+v/c)^1/2.
Plug in v/c = 0.5 and f_source = 100 MHz to find f = 57.7 MHz.

On the way back, the transmissions are shifted to higher frequency,
f = f_source (1+v/c)^1/2/(1-v/c)^1/2 = 173.2 MHz.

(b) The two photons have equal and opposite momenta so they have equal energies. Since their total energy must add up to 140 MeV (the total energy of the p particle created during the collsion), each photon must have 70 MeV.

Their wavelengths are also equal and are given by l = hc/E = 1240 eV·nm/70×10⁶ eV = 1.77×10^-5 nm = 1.77×10^-14 m.

(c) Use DE Dt ³ (^h/_2p)/2, or DE = (^h/_2p) c /(2 c Dt). Plug in (^h/_2p) c = 197 eV·nm, 2cDt = 2×3×10¹⁷ nm/sec×2.2×10^-6 sec = 13.2×10¹¹ nm, so DE = 197 / 13.2×10¹¹ eV = 1.5×10^-10 eV. The mass uncertainty is just this result divided by c², or Dm = 1.5×10^-10 eV/c².

(b) Since p = 7300 eV/c, the momentum uncertainty is Dp = 0.01 p = 73 eV/c. Thus
Dx ³ (^h/_2p)/(2 Dp) = (^h/_2p) c / (2 cDp) = 197 eV·nm/(2 ×73 eV) = 1.34 nm. The electrons can be located with as good as 1.34 nm spatial precision.

(c) The neutrons must have the same wavelength as in (a), and therefore the same momentum p = 7300 eV/c. This means that their kinetic energy must be K = p²/(2m) = (pc)²/(2mc²). We plug in pc = 730 eV and mc² = 940,000,000 eV to find K = (7300)² / (2×940,000,000) eV = 0.028 eV.

(There was a slight ambiguity in the way this question was asked: I asked for the `energy' when I should have said `kinetic energy'. Almost everyone interpreted the question this way. But, a few students correctly argued that if the neutrons are to have the same wavelength as the photons of (a), then they should have the same frequency of their matter waves. Since E = hf holds for the total energy E = Ö[(p² c² + m² c⁴)] this implies that they should have a total energy which is the same as that of (a). This gives E = 7300 eV. The only trouble is that this is less than mc² for a neutron, and therefore impossible! But this line of reasoning still shows good understanding of the material, so the grader of that problem (Jerry Gladky) gave full marks for E = 7300 eV.)

(d) Since the Bragg condition depends only on wavelength of the particle, which is in turn dependent only on momentum (l = h/p) the angle at which the m = 1 peak is seen remains 30^°.

(b) The ground state, or lowest-energy, wavefunction for the particle in the box is

The kinetic energy of this state is

(c) This is the n = 3 state, which has k = 3p/L and therefore momentum p = (^h/_2p) k = 3p(^h/_2p)/L = 3h/(2L). The kinetic energy is E₃ from the formula sheet, or you could compute it:

(d) The answers to (a), (b) and (c) are

If we plug in L = 1×10^-15 m = 10×10^-6 nm, (^h/_2p) c = 197 eV·nm and mc² = 938,000,000 eV we have

The results for (b) and (c) are just p² and 9p² times this, or 2,040,000 eV = 2.04 MeV
and 18,400,000 eV = 18.4 MeV.

( If you multiplied all these results by 3 since I mentioned that this was a real proton and therefore it should be in a 3-d box, that would be fine. In fact, these energies roughly describe the kinetic energy of protons inside nuclei - a big nucleus has a diameter around 5×10^-15 m. )

(e) The result from the uncertainty principle (a) must be below the ground state energy (b), and since (c) has nodes, it must have a higher energy than the nodeless ground state. So (a) < (b) < (c).

You could also simply say that (c) has a shorter wavelength than (b) and therefore greater momentum and energy.