15.6 Compton Scattering

1. AIts wavelength is shorter, since it has given energy to the electron
   A student who thinks less energy means a shorter wavelength picks this. E = hc/λ: energy and wavelength are inversely related, so the photon that has lost energy has the longer wavelength.
2. BIts wavelength is longer, since it has given up energy to the electron Correct
   The electron recoils, gaining kinetic energy, so the photon that emerges has less energy. Since E = hc/λ, less energy means a longer wavelength; this is the Compton effect.
3. CIts wavelength is unchanged, since it simply bounces off the electron
   A student who pictures the photon bouncing off like a ball off a wall picks this. The electron is free to recoil and takes energy from the photon, so the photon's wavelength increases.
4. DIts speed is now less than c, since it has given energy to the electron
   A student who thinks a photon loses energy by slowing down picks this. Every photon moves at c in a vacuum; the energy loss shows up as a lower frequency and a longer wavelength.

CED 15.6.A.1 · Read this in Fix

1. AEnergy alone, since a massless photon carries no momentum
   A student who thinks a photon has no momentum because it has no mass picks this. A photon has momentum p = h/λ, and conservation of momentum is needed to explain how the change in wavelength depends on the angle.
2. BMomentum alone, since the photon loses energy in the collision
   A student who confuses the photon losing energy with the system losing energy picks this. The energy the photon loses is gained by the electron, so the total energy of the system is conserved.
3. CThe photon's own energy, since it bounces off unchanged
   A student who pictures the photon bouncing off like a ball off a wall picks this. The free electron recoils and takes energy from the photon; it is the total for photon and electron that is conserved, not the photon's own energy.
4. DEnergy and momentum, since the electron gains what is lost Correct
   Compton scattering is explained by treating the photon as a particle and applying conservation of energy and conservation of momentum to its collision with the electron. The electron gains the energy and momentum that the photon loses.

CED 15.6.A.2.i · Read this in Fix

1. A5.05 × 10⁻¹¹ m Correct
   Δλ = (h/(m_e c))(1 − cos θ) = (2.43 × 10⁻¹² m)(1 − cos 37°) = (2.43 × 10⁻¹² m)(0.20) = 4.9 × 10⁻¹³ m. The scattered wavelength is longer: 5.00 × 10⁻¹¹ m + 0.049 × 10⁻¹¹ m = 5.05 × 10⁻¹¹ m.
2. B4.95 × 10⁻¹¹ m
   A student who thinks the photon's wavelength gets shorter as it loses energy picks this, subtracting the shift. Less energy means a longer wavelength, so Δλ is added: 5.05 × 10⁻¹¹ m.
3. C5.15 × 10⁻¹¹ m
   A student who uses sin 37° = 0.60 in place of 1 − cos 37° = 0.20 picks this. The Compton shift depends on 1 − cos θ, which is zero for an undeflected photon.
4. D5.44 × 10⁻¹¹ m
   A student who uses 1 + cos 37° = 1.80 picks this. The shift is (h/(m_e c))(1 − cos θ); with a plus sign an undeflected photon would have the largest shift, which is impossible.

Working h/(m_e c) = 6.63 × 10⁻³⁴/[(9.11 × 10⁻³¹)(3.00 × 10⁸)] = 2.43 × 10⁻¹² m. Δλ = (2.43 × 10⁻¹² m)(1 − cos 37°) = (2.43 × 10⁻¹² m)(0.201) = 4.89 × 10⁻¹³ m. λ′ = 5.00 × 10⁻¹¹ m + 0.0489 × 10⁻¹¹ m = 5.05 × 10⁻¹¹ m. (Subtracting: 4.95; sin 37°: 5.15; 1 + cos 37°: 5.44, all × 10⁻¹¹ m.)

CED 15.6.A.3 · Read this in Fix

1. Aλ₁ > λ₂ > λ₃
   A student who thinks a photon's wavelength gets shorter as it loses energy picks this. The photons scattered through larger angles do lose more energy, but less energy means a longer wavelength, so λ₃ is the longest.
2. Bλ₁ = λ₂ = λ₃
   A student who thinks the photons bounce off without losing energy picks this. The electrons recoil, and the amount of energy the photon loses, and so its change in wavelength, depends on how much its direction changes.
3. Cλ₃ > λ₂ > λ₁ Correct
   The change in wavelength grows with the change in the photon's direction: Δλ = (h/(m_e c))(1 − cos θ), with 1 − cos θ = 0.13 at 30°, 1.00 at 90° and 1.87 at 150°. Detector 3, at the largest angle, records the longest wavelength.
4. Dλ₂ > λ₁ = λ₃
   A student who uses sin θ in place of 1 − cos θ picks this: sin 30° = sin 150° = 0.5 and sin 90° = 1. The shift depends on 1 − cos θ, which increases steadily from 0° to 180°.

Working Δλ ∝ 1 − cos θ: 30° → 0.13, 90° → 1.00, 150° → 1.87, so λ₃ > λ₂ > λ₁. (sin θ: 0.5, 1, 0.5 gives λ₂ > λ₁ = λ₃.)

CED 15.6.A.1 · Read this in Fix

1. AScattered X-rays had longer wavelengths, by amounts set by the scattering angle Correct
   In the classical wave model, electrons shaken by the incident wave re-radiate at the same frequency, so the scattered X-rays would have the incident wavelength. The observed increase in wavelength, depending on the angle, is what collisions between particles of light and electrons predict.
2. BScattered X-rays had the same wavelength as the incident X-rays at every angle
   A student who pictures photons bouncing off electrons like balls off a wall picks this. That was not observed, and it is what the classical wave model predicts; the evidence for photons is that the wavelength increases.
3. CThe wavelength shift was larger when the incident X-rays were more intense
   A student who expects a stronger beam to have a larger effect picks this. In the photon model each shift comes from one photon–electron collision, so it depends on the angle, not on the intensity; that is what was observed.
4. DThe scattered X-rays traveled more slowly than the incident X-rays did
   A student who thinks a photon loses energy by slowing down picks this. Scattered X-rays still travel at c; their lower energy shows up as a longer wavelength.

CED 15.6.A.2 · Read this in Fix

1. A1.25
   A student who takes momentum to be proportional to wavelength picks this. From λ = h/p, p = h/λ, so a longer wavelength means a smaller momentum: 1/1.25 = 0.80.
2. B0.80 Correct
   λ = h/p, so p = h/λ: momentum is inversely proportional to wavelength. p_scattered/p_incident = λ_incident/λ_scattered = 1/1.25 = 0.80.
3. C1.00
   A student who thinks a photon's momentum is fixed because its speed is always c picks this. A photon's momentum is h/λ, which changes when its wavelength changes.
4. D0.89
   A student who uses K = p²/(2m) for the photon picks this: the energy ratio is 1/1.25 = 0.80 and √0.80 = 0.89. For a photon p = h/λ, so the momentum ratio is 0.80, the same as the energy ratio.

Working p = h/λ ⇒ p′/p = λ/λ′ = 1/1.25 = 0.80. (p ∝ λ: 1.25; p fixed: 1.00; p ∝ √E with E ∝ 1/λ: √(1/1.25) = 0.89.)

CED 15.6.A.2.ii · Read this in Fix

1. AArrow P
   A student who thinks the electron is pushed straight along the incident photon's direction picks this. The scattered photon carries momentum perpendicular to the incident direction, so the electron must carry an equal and opposite perpendicular component.
2. BArrow Q
   A student who thinks the photon bounces off without losing energy picks this: equal momenta for the incident and scattered photons would put the electron at exactly 45°. The scattered photon has less momentum than the incident one, so the angle is less than 45°.
3. CArrow R
   A student who adds the two photons' momenta picks this. Conservation gives p⃗_electron = p⃗_incident − p⃗_scattered: the scattered photon's momentum is subtracted, so the electron's perpendicular component points away from the scattered photon's side.
4. DArrow S Correct
   Momentum is conserved: p⃗_incident = p⃗_scattered + p⃗_electron, so p⃗_electron = p⃗_incident − p⃗_scattered. Its component along the incident direction equals the incident photon's momentum, and its component perpendicular to it is opposite to the scattered photon's and equal in size. The scattered photon has a longer wavelength and so less momentum (p = h/λ), so the electron moves below the incident direction at less than 45°: arrow S.

Working Take x along the incident photon, y along the scattered photon. Before: (p_i, 0). After: photon (0, p_f) + electron (p_ex, p_ey). So p_ex = p_i, p_ey = −p_f. Since λ_f > λ_i, p_f < p_i, so the electron's direction is below +x at tan⁻¹(p_f/p_i) < 45°: arrow S. (Only x: P; p_f = p_i: Q at 45°; adding: R above +x.)

CED 15.6.A.2.i · Read this in Fix

1. A√(E²+E′²)/c Correct
   A photon's momentum is p = h/λ = hf/c = E/c. The momentum of the photon–electron system is conserved: before, it is E/c in the +x-direction; after, the photon carries E′/c in the +y-direction, so the electron has components E/c in the +x-direction and E′/c in the −y-direction. Its magnitude is √((E/c)² + (E′/c)²) = √(E²+E′²)/c.
2. B(E−E′)/c
   A student who conserves momentum as a scalar picks this, writing E/c = E′/c + p_e. The photon's momenta before and after are in different directions, so each component is conserved separately: the electron has E/c along +x and E′/c along −y.
3. CE/c
   A student who thinks the electron always moves along the incident photon's direction picks this, conserving the x-component alone. The scattered photon has momentum in the +y-direction, so the electron must also have a y-component, E′/c in the −y-direction.
4. D√2·E/c
   A student who thinks a photon's momentum cannot change, since it always moves at c, gives the scattered photon momentum E/c and picks this: √((E/c)² + (E/c)²) = √2·E/c. The scattered photon has less energy, E′ < E, so its momentum E′/c is smaller than E/c.

Working Photon momentum: E = hf and λ = h/p give p = h/λ = hf/c = E/c. Momentum of the photon–electron system is conserved. Before: E/c in +x. After: photon E′/c in +y. Electron: p_x = E/c, p_y = −E′/c, so |p_e| = √((E/c)² + (E′/c)²) = √(E²+E′²)/c. Errors: magnitudes conserved as scalars, E/c = E′/c + p_e → (E−E′)/c; x-component only (electron along +x) → E/c; scattered photon keeps momentum E/c → √2·E/c.

CED 15.6.A.2.i · Read this in Fix

1. Ah/λ + h/λ′ Correct
   The photon's momentum is h/λ in the +x-direction before and h/λ′ in the −x-direction after. The momentum of the photon–electron system is conserved: h/λ = −h/λ′ + p_e, so p_e = h/λ + h/λ′. The photon's momentum reverses, so the electron must take up more than the photon's initial momentum.
2. Bh/λ − h/λ′
   A student who conserves momentum as a scalar picks this, writing h/λ = h/λ′ + p_e. The scattered photon moves in the −x-direction, so its momentum counts as −h/λ′: p_e = h/λ + h/λ′.
3. Ch/(λ′ − λ)
   A student who finds the photon's change in momentum from the change in wavelength alone picks this. Momentum is inversely proportional to wavelength, so each momentum must be found separately, with its direction: the change is h/λ − (−h/λ′) = h/λ + h/λ′.
4. Dλ/h + λ′/h
   A student who takes a photon's momentum to be proportional to its wavelength, p = λ/h, picks this. From λ = h/p, p = h/λ: the longer the wavelength, the smaller the momentum.

Working Photon momentum from λ = h/p: p = h/λ before, h/λ′ after. Momentum of the photon–electron system is conserved along x: h/λ = −h/λ′ + p_e, so p_e = h/λ + h/λ′ (in the +x-direction). Errors: magnitudes conserved as scalars, h/λ = h/λ′ + p_e → h/λ − h/λ′; momentum change from Δλ → h/(λ′ − λ); p ∝ λ, p = λ/h → λ/h + λ′/h.

CED 15.6.A.2.ii · Read this in Fix

1. A3.21 × 10⁻¹³ J
   A student who uses sin 60° = 0.866 in place of 1 − cos 60° = 0.500 picks this: the larger shift gives λ′ = 2.60 × 10⁻¹² m. The Compton shift depends on 1 − cos θ, which is greatest, 2h/(m_e c), for a photon sent straight back (θ = 180°); sin θ would wrongly give no shift there.
2. B3.50 × 10⁻¹³ J
   A student who uses 1 + cos 60° = 1.50 picks this: λ′ = 4.14 × 10⁻¹² m. The shift is (h/(m_e c))(1 − cos θ); with a plus sign an undeflected photon would have the largest shift.
3. C2.82 × 10⁻¹³ J Correct
   Δλ = (h/(m_e c))(1 − cos 60°) = (2.43 × 10⁻¹² m)(0.500) = 1.21 × 10⁻¹² m, so λ′ = 1.71 × 10⁻¹² m. Energy is conserved, so the electron gains what the photon loses: K = hc/λ − hc/λ′ = 3.98 × 10⁻¹³ J − 1.16 × 10⁻¹³ J = 2.82 × 10⁻¹³ J.
4. D1.64 × 10⁻¹³ J
   A student who takes the energy transferred to be hc/Δλ = (1.989 × 10⁻²⁵ J·m)/(1.21 × 10⁻¹² m) picks this. Photon energy is inversely proportional to wavelength, so find the photon's energy before and after, 3.98 × 10⁻¹³ J and 1.16 × 10⁻¹³ J, and subtract.

Working h/(m_e c) = 6.63 × 10⁻³⁴/[(9.11 × 10⁻³¹)(3.00 × 10⁸)] = 2.43 × 10⁻¹² m. Δλ = (2.43 × 10⁻¹² m)(1 − cos 60°) = 1.21 × 10⁻¹² m, so λ′ = 5.00 × 10⁻¹³ m + 1.21 × 10⁻¹² m = 1.71 × 10⁻¹² m. Energy is conserved: K = hc/λ − hc/λ′ = (1.989 × 10⁻²⁵ J·m)(1/(5.00 × 10⁻¹³ m) − 1/(1.71 × 10⁻¹² m)) = 3.98 × 10⁻¹³ J − 1.16 × 10⁻¹³ J = 2.82 × 10⁻¹³ J. (sin 60°: 3.21 × 10⁻¹³ J; 1 + cos 60°: 3.50 × 10⁻¹³ J; hc/Δλ: 1.64 × 10⁻¹³ J.)

CED 15.6.A.2.i · Read this in Fix