These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data and assumptions for every question: h = 6.63 × 10⁻³⁴ J·s; c = 3.00 × 10⁸ m/s; m_e = 9.11 × 10⁻³¹ kg; h/(m_e c) = 2.43 × 10⁻¹² m = 2.43 pm; 1 eV = 1.60 × 10⁻¹⁹ J. Each electron is free and at rest before the collision. All data are invented for practice. A scientific calculator is assumed. Round only at the end.
Question 1 (multiple choice · foundation)
An X-ray photon scatters off a free electron. Which property of the photon is the same after the collision as before it?
- (A) Its speed
- (B) Its frequency
- (C) Its wavelength
- (D) The magnitude of its momentum
Answer and explanation
Answer: (A). A photon in a vacuum always travels at c. It gives some energy to the electron, so its energy hf goes down.
- (B) Lower energy means lower frequency, since E = hf.
- (C) Lower frequency at the same speed means a longer wavelength, since c = fλ.
- (D) p = h/λ, so a longer wavelength means a smaller momentum.
Question 2 (multiple choice · core)
A beam of X-rays of one wavelength strikes a target. Detectors at 45°, 90° and 135° to the original beam measure the wavelength of the scattered X-rays: λ₄₅, λ₉₀ and λ₁₃₅. Which ranking is correct?
- (A) λ₁₃₅ > λ₉₀ > λ₄₅
- (B) λ₄₅ > λ₉₀ > λ₁₃₅
- (C) λ₉₀ > λ₁₃₅ > λ₄₅
- (D) λ₄₅ = λ₉₀ = λ₁₃₅
Answer and explanation
Answer: (A). Δλ = (2.43 pm)(1 − cos θ) increases steadily from 0° to 180°: the shifts are 0.71 pm, 2.43 pm and 4.14 pm. The larger the angle, the more energy the photon gives up and the longer its new wavelength.
- (B) reverses the trend, as if photons gained energy when deflected more.
- (C) thinks the shift peaks at 90°, perhaps from sin θ, which does peak at 90°. The equation uses 1 − cos θ.
- (D) is the classical wave prediction, which experiments rule out.
Question 3 (multiple choice · core)
An X-ray photon is scattered through 120°. What is the change in its wavelength?
- (A) 3.64 pm
- (B) 1.21 pm
- (C) 4.85 pm
- (D) 2.43 pm
Answer and explanation
Answer: (A). cos 120° = −0.5, so 1 − cos θ = 1.5 and Δλ = 1.5 × 2.43 pm = 3.64 pm.
- (B) uses cos 120° = +0.5, giving 1 − 0.5 = 0.5. Cosines of angles above 90° are negative.
- (C) is the maximum shift, which happens only at 180°.
- (D) is the shift at 90°.
Question 4 (multiple choice · core)
Photon X has wavelength 20 pm and photon Y has wavelength 80 pm. Each scatters off a free electron through the same angle. Which statement is correct?
- (A) Both have the same Δλ, and X loses a larger fraction of its energy.
- (B) X has the larger Δλ, because it has more energy.
- (C) Both have the same Δλ, and both lose the same fraction of their energy.
- (D) Both have the same Δλ, and Y loses a larger fraction of its energy.
Answer and explanation
Answer: (A). Δλ depends only on θ, so it is the same for both. The fraction of energy lost is 1 − λ/λ′. The same Δλ is a bigger part of 20 pm than of 80 pm, so X loses more of its energy. At 90°, for example, X loses about 11% and Y about 3%.
- (B) thinks Δλ depends on the starting wavelength or energy. It does not.
- (C) mixes up an equal absolute change with an equal fractional change.
- (D) reverses the fractions.
Question 5 (calculation · core)
An X-ray photon of wavelength 40.0 pm scatters off a free electron at rest and is detected at 120° to its original direction.
(a) Calculate the wavelength of the scattered photon. (b) Calculate the energy of the photon before and after the collision, in joules. (c) Calculate the kinetic energy of the electron, in joules and in keV. (d) State the change in the magnitude of the photon’s momentum.
Worked solution and suggested Marlbridge rubric
- (a) Δλ = 2.43 pm × (1 − cos 120°) = 2.43 × 1.5 = 3.64 pm. λ′ = 40.0 + 3.64 = 43.6 pm.
- (b) E = hc/λ = (6.63 × 10⁻³⁴)(3.00 × 10⁸) ÷ (40.0 × 10⁻¹²) = 4.97 × 10⁻¹⁵ J. E′ = (1.989 × 10⁻²⁵) ÷ (43.64 × 10⁻¹²) = 4.56 × 10⁻¹⁵ J.
- (c) K_e = E − E′ = 4.9725 × 10⁻¹⁵ − 4.5579 × 10⁻¹⁵ = 4.15 × 10⁻¹⁶ J. In eV: 4.146 × 10⁻¹⁶ ÷ 1.60 × 10⁻¹⁹ = 2.59 × 10³ eV = 2.59 keV.
- (d) p = h/λ = 1.658 × 10⁻²³ kg·m/s; p′ = h/λ′ = 1.519 × 10⁻²³ kg·m/s. The magnitude decreases by 1.38 × 10⁻²⁴ kg·m/s.
| Point | What earns it |
|---|---|
| 1 | λ′ = 43.6 pm, using 1 − cos 120° = 1.5 |
| 1 | E and E′ both correct |
| 1 | K_e from energy conservation, 4.15 × 10⁻¹⁶ J and 2.59 keV |
| 1 | Change in momentum magnitude, 1.38 × 10⁻²⁴ kg·m/s, as a decrease |
Common error: rounding E and E′ to 2 significant figures before subtracting, which can give K_e = 0 or 1 × 10⁻¹⁶ J. Answers within about 1% of these, from using hc = 1240 eV·nm, are acceptable.
Question 6 (constructed response · core)
An X-ray photon of wavelength λ travels in the +x direction and hits a free electron at rest. The photon bounces straight back (θ = 180°) with wavelength λ′.
(a) Draw a momentum diagram showing the momentum vectors of the photon before the collision, and of the photon and the electron after it. (b) Starting from conservation of momentum, derive an expression for the magnitude of the electron’s momentum p_e in terms of λ, λ′ and h. (c) A student says: “The electron ends up with more momentum than the photon had at the start, so momentum is not conserved.” Evaluate this claim. (d) For λ = 25.0 pm, calculate λ′ and p_e.
Worked solution and suggested Marlbridge rubric
(a) Before: one arrow along +x, labelled h/λ. After: a slightly shorter arrow along −x, labelled h/λ′ (longer wavelength, so smaller momentum), and an arrow along +x for the electron, labelled p_e. The electron’s arrow is longer than the incoming photon’s arrow.
(b) Total x-momentum before = total after: h/λ = −h/λ′ + p_e. So p_e = h/λ + h/λ′.
(c) The student is right that p_e > h/λ, but wrong about conservation. Momentum is a vector. After the collision the photon moves in the −x direction, so its momentum is negative. The electron must carry the original momentum plus enough to cancel the photon’s backward momentum. The total (p_e − h/λ′) still equals h/λ.
(d) λ′ = 25.0 pm + 2 × 2.43 pm = 29.9 pm (29.85 pm). p_e = 6.63 × 10⁻³⁴ ÷ 25.0 × 10⁻¹² + 6.63 × 10⁻³⁴ ÷ 29.85 × 10⁻¹² = 2.652 × 10⁻²³ + 2.221 × 10⁻²³ = 4.87 × 10⁻²³ kg·m/s, about 1.84 times the incoming photon’s momentum.
| Point | What earns it |
|---|---|
| 1 | Diagram: photon after along −x and shorter than before; electron along +x |
| 1 | Starts from Σp before = Σp after with the photon’s final momentum negative |
| 1 | p_e = h/λ + h/λ′ |
| 1 | Rejects the claim by treating momentum as a vector (the backward photon momentum is negative) |
| 1 | λ′ = 29.9 pm and p_e = 4.87 × 10⁻²³ kg·m/s |
Do not award the derivation points for p_e = h/λ − h/λ′, which ignores the change of direction.
Question 7 (constructed response · stretch)
A student scatters X-rays of a single (unknown) wavelength from a target and measures the wavelength λ′ of the scattered X-rays at five angles.
| θ | 30° | 60° | 90° | 120° | 150° |
|---|---|---|---|---|---|
| λ′ (pm) | 63.3 | 64.2 | 65.4 | 66.6 | 67.5 |
(a) Explain what quantity to plot against λ′ to get a straight line. State what the slope and the vertical intercept represent. (b) Calculate the values for your horizontal axis, plot the graph and draw a best-fit line. (c) Use your line to find the incident wavelength and the electron mass, taking h and c as known. (d) Explain how these data support the photon model and not a classical wave model.
Worked solution and suggested Marlbridge rubric
(a) λ′ = λ + (h/m_e c)(1 − cos θ). Plot λ′ (vertical) against (1 − cos θ) (horizontal). The slope is h/(m_e c) and the vertical intercept is λ, the incident wavelength.
(b) 1 − cos θ = 0.134, 0.500, 1.000, 1.500 and 1.866. The points lie close to a straight line.
(c) A least-squares line has slope 2.42 pm and intercept 63.0 pm. So the incident wavelength is about 63.0 pm. From the slope, m_e = h ÷ (c × slope) = 6.63 × 10⁻³⁴ ÷ [(3.00 × 10⁸)(2.42 × 10⁻¹²)] = 9.14 × 10⁻³¹ kg, within about 0.3% of the accepted 9.11 × 10⁻³¹ kg. A hand-drawn line giving a slope between about 2.3 and 2.5 pm (m_e from about 8.8 to 9.6 × 10⁻³¹ kg) is consistent with the data.
(d) A classical wave model predicts that the scattered wave has the same wavelength as the incident wave at every angle: a horizontal line at λ′ = λ. The data show λ′ > λ, with the shift increasing with angle in the way predicted by treating each X-ray as a photon of momentum h/λ that collides with an electron and conserves energy and momentum. The slope even gives a sensible electron mass, which a wave model cannot explain.
| Point | What earns it |
|---|---|
| 1 | Axes: λ′ against (1 − cos θ), with slope = h/(m_e c) and intercept = λ |
| 1 | Correct (1 − cos θ) values, plotted with labelled axes, units and a best-fit line |
| 1 | Incident wavelength ≈ 63.0 pm from the intercept |
| 1 | m_e from the slope, in the range 8.8–9.6 × 10⁻³¹ kg |
| 1 | States the wave-model prediction (no change in wavelength, at any angle) |
| 1 | Links the angle-dependent increase to energy and momentum transfer in a photon–electron collision |
Carry forward the student’s own slope into the value of m_e.
How did you do?
- Q1 or Q4 wrong: re-read “The momentum of a photon” and “Comparing scattering situations” in the study guide.
- Q2 or Q3 wrong: revisit “The Compton shift equation” and its table.
- Q5 wrong: redo Worked example 1, keeping unrounded energies.
- Q6 wrong: redo Worked example 2 and check the signs of your momentum components.
- Q7 incomplete: compare the wave and particle predictions in “A photon that bounces”.
Then tick off the topic checklist.