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 for every question: h = 6.63 × 10⁻³⁴ J·s; c = 3.00 × 10⁸ m/s; e = 1.60 × 10⁻¹⁹ C; 1 eV = 1.60 × 10⁻¹⁹ J; electron mass mₑ = 9.11 × 10⁻³¹ kg; proton and neutron mass 1.67 × 10⁻²⁷ kg. All wavelengths of light are vacuum wavelengths unless stated. All speeds are much less than c. A scientific calculator is assumed. Round only at the end.
Question 1 (multiple choice · foundation)
Lamp X gives out blue light of wavelength 450 nm. Lamp Y gives out red light of wavelength 675 nm. What is the ratio (energy of one photon from X) ÷ (energy of one photon from Y)?
- (A) 1.5
- (B) 0.67
- (C) 2.25
- (D) 1.0, because all photons travel at the same speed
Answer and explanation
Answer: (A). E = hc/λ, so E ∝ 1/λ. E_X/E_Y = λ_Y/λ_X = 675 nm ÷ 450 nm = 1.5. (Check: E_X = 2.76 eV and E_Y = 1.84 eV.)
- (B) treats E as proportional to λ. Shorter wavelength means higher frequency and more energy, not less.
- (C) squares the ratio. Photon energy depends on 1/λ, not 1/λ².
- (D) confuses speed with energy. All photons in a vacuum share the speed c, but their energies depend on frequency.
Question 2 (multiple choice · core)
Orange light of wavelength 600 nm in air passes into water, which has an index of refraction of 1.33. Which row describes the light in the water?
- (A) Speed 2.26 × 10⁸ m/s, wavelength 451 nm, photon energy unchanged
- (B) Speed 2.26 × 10⁸ m/s, wavelength 600 nm, photon energy smaller
- (C) Speed 3.00 × 10⁸ m/s, wavelength 451 nm, photon energy larger
- (D) Speed 3.99 × 10⁸ m/s, wavelength 798 nm, photon energy unchanged
Answer and explanation
Answer: (A). v = c/n = (3.00 × 10⁸ m/s) ÷ 1.33 = 2.26 × 10⁸ m/s. The frequency is set by the source and does not change, so E = hf does not change. From v = fλ, the wavelength falls by the same factor as the speed: 600 nm ÷ 1.33 = 451 nm. (Treating air as n ≈ 1.)
- (B) keeps the wavelength fixed, so it must change the frequency, which would change the energy. In fact the frequency, not the wavelength, stays fixed.
- (C) keeps the vacuum speed in the water. Photons in a medium travel at c/n.
- (D) multiplies by n instead of dividing. A larger index means a slower photon.
Question 3 (multiple choice · core)
An electron and a proton move at the same speed. Which statement about their de Broglie wavelengths is correct?
- (A) The electron’s wavelength is about 1830 times the proton’s.
- (B) The electron’s wavelength is about 43 times the proton’s.
- (C) The proton’s wavelength is about 1830 times the electron’s.
- (D) Their wavelengths are equal, because they move at the same speed.
Answer and explanation
Answer: (A). λ = h/(mv). With the same v, λ ∝ 1/m, so λ_e/λ_p = mₚ/mₑ = 1.67 × 10⁻²⁷ ÷ 9.11 × 10⁻³¹ = 1.83 × 10³. The lighter particle has the smaller momentum and so the longer wavelength.
- (B) uses √(mₚ/mₑ) = 42.8. That ratio applies when the two particles have the same kinetic energy, not the same speed.
- (C) has the ratio upside down: it gives the heavier particle the longer wavelength.
- (D) forgets that λ depends on momentum, which includes the mass.
Question 4 (multiple choice · core)
A very dim beam of electrons is aimed at two narrow slits so that only one electron passes through the apparatus at a time. A screen records where each electron arrives. Which description of the results is correct?
- (A) Each electron is recorded at a single point; after many electrons, the points form bright and dark fringes.
- (B) Each electron is spread out across the screen as a faint copy of the full interference pattern.
- (C) The electrons form two bright bands, one behind each slit, because each electron is a particle.
- (D) No pattern forms, because interference needs two electrons to pass through the slits together.
Answer and explanation
Answer: (A). Each detection happens at one point, which is particle-like behaviour. The wave model gives the probability of arriving at each point, so many detections build up the interference fringes, which is wave-like behaviour.
- (B) is not what detectors record. An electron is never detected as a spread-out smear; it arrives whole at one place.
- (C) is the classical particle prediction. It fails because the electron’s wave passes through both slits and interferes.
- (D) is contradicted by experiment: the fringes still appear when the electrons pass one at a time. Each electron’s wave interferes with itself.
Question 5 (calculation · core)
A sterilising lamp gives out 2.0 W of ultraviolet light, all of wavelength 250 nm.
(a) Calculate the energy of one photon, in joules and in electron-volts. (b) Calculate the number of photons the lamp emits each second. (c) An infrared heater gives out 2.0 W of light of wavelength 1000 nm. How many times more photons per second does it emit than the ultraviolet lamp? Explain without a full calculation.
Worked solution
- (a) E = hc/λ = (6.63 × 10⁻³⁴ J·s)(3.00 × 10⁸ m/s) ÷ (250 × 10⁻⁹ m) = 7.96 × 10⁻¹⁹ J.
- In eV: 7.956 × 10⁻¹⁹ J ÷ 1.60 × 10⁻¹⁹ J/eV = 4.97 eV.
- (b) N/t = P/E = 2.0 J/s ÷ 7.956 × 10⁻¹⁹ J = 2.5 × 10¹⁸ photons per second.
- (c) The infrared wavelength is 4 times longer, so each infrared photon carries 1/4 of the energy (1.24 eV). To deliver the same 2.0 W, the heater must emit 4 times as many photons per second (1.0 × 10¹⁹ per second).
Suggested mark points (4): 1 for E = 7.96 × 10⁻¹⁹ J; 1 for 4.97 eV; 1 for 2.5 × 10¹⁸ photons per second; 1 for the factor of 4 with the reason (E ∝ 1/λ at equal power).
Common error: writing f = cλ instead of f = c/λ, which gives an energy near 5 × 10⁻³² J. Always check that a visible or ultraviolet photon has a few eV.
Question 6 (constructed response · core)
A particle of mass m and charge q starts from rest and is accelerated through a potential difference ΔV.
(a) Derive an expression for its de Broglie wavelength λ in terms of m, q, ΔV and physical constants. (b) The potential difference is increased to 4ΔV. By what factor does λ change? Justify using your expression. (c) An electron and a proton are each accelerated from rest through the same potential difference. Determine the ratio λ_electron/λ_proton. (d) Calculate λ for an electron accelerated through 50 V, and say whether a crystal with atomic spacing about 2 × 10⁻¹⁰ m could diffract it.
Worked solution and suggested Marlbridge rubric
(a) The work done by the field gives the kinetic energy: K = qΔV. Also K = ½mv² = p²/(2m), so p = √(2mK) = √(2mqΔV). Then λ = h/p = h/√(2mqΔV).
(b) λ ∝ 1/√ΔV. Multiplying ΔV by 4 multiplies λ by 1/√4, so λ halves.
(c) The two particles have charges of equal size, so with the same ΔV, λ ∝ 1/√m. λ_e/λ_p = √(mₚ/mₑ) = √(1.67 × 10⁻²⁷ ÷ 9.11 × 10⁻³¹) = 42.8.
(d) λ = (6.63 × 10⁻³⁴) ÷ √(2 × 9.11 × 10⁻³¹ × 1.60 × 10⁻¹⁹ × 50) = 1.7 × 10⁻¹⁰ m. This is comparable to the atomic spacing, so yes, the crystal can diffract the electrons.
| Point | What earns it |
|---|---|
| 1 | Uses energy conservation K = qΔV for a particle starting from rest |
| 1 | Correct momentum p = √(2mqΔV) (or v = √(2qΔV/m) then p = mv) |
| 1 | Final expression λ = h/√(2mqΔV) |
| 1 | (b) Factor 1/2, justified by λ ∝ 1/√ΔV from the expression |
| 1 | (c) Ratio 42.8 (accept 43), electron longer |
| 1 | (d) λ ≈ 1.7 × 10⁻¹⁰ m and a conclusion based on λ being comparable to the spacing |
Do not award (b) for “λ quarters” (that treats λ ∝ 1/ΔV). Accept a ratio in (c) written as proton/electron = 0.0234 if the student states clearly which particle has the longer wavelength.
Question 7 (constructed response · stretch)
A research team compares two experiments with fictional data.
- Experiment 1: a beam of neutrons moving at 2.0 × 10³ m/s is aimed at a crystal whose rows of atoms are about 2 × 10⁻¹⁰ m apart.
- Experiment 2: a dust grain of mass 1.0 × 10⁻¹⁵ kg and diameter about 1 × 10⁻⁶ m drifts at 1.0 × 10⁻³ m/s towards a pair of slits.
(a) Calculate the de Broglie wavelength in each experiment. (b) Predict, with reasons, in which experiment a diffraction or interference pattern could be observed. (c) A student says: “Neutrons have no charge, so they cannot be diffracted. Only charged particles and light have waves.” Evaluate this claim. (d) The neutrons in Experiment 1 have a kinetic energy of about 0.02 eV. Explain, using the de Broglie relation, why physicists use slow neutrons rather than fast ones to study crystals.
Worked solution and suggested Marlbridge rubric
(a) Neutron: λ = h/(mv) = (6.63 × 10⁻³⁴) ÷ (1.67 × 10⁻²⁷ × 2.0 × 10³) = 1.99 × 10⁻¹⁰ m. Dust grain: λ = (6.63 × 10⁻³⁴) ÷ (1.0 × 10⁻¹⁵ × 1.0 × 10⁻³) = 6.6 × 10⁻¹⁶ m.
(b) Experiment 1. The neutron wavelength is about the same as the spacing of the rows of atoms, so the crystal acts as a grating and strong diffraction is expected. The dust grain’s wavelength is about 10⁻⁹ of its own size, and far smaller than any slit it could pass through, so no pattern can be seen.
(c) The claim is wrong. The de Broglie wavelength λ = h/p depends only on momentum, not on charge. Any particle with momentum has a wavelength, so neutral neutrons can be diffracted (as the numbers in (a) and (b) show). Photons are also neutral, and they diffract.
(d) Kinetic energy 0.02 eV means a small momentum, so a long wavelength. At this energy λ is close to the atomic spacing. Much faster neutrons would have larger momentum and a much shorter wavelength than the spacing, so the diffraction angles would be too small to measure.
| Point | What earns it |
|---|---|
| 1 | Neutron λ ≈ 2.0 × 10⁻¹⁰ m |
| 1 | Dust grain λ ≈ 6.6 × 10⁻¹⁶ m |
| 1 | Chooses Experiment 1 because λ is comparable to the atomic spacing |
| 1 | Explains no pattern for the dust grain by comparing its λ with its size or any slit width |
| 1 | Rejects the claim: λ = h/p does not involve charge, with neutrons or photons as evidence |
| 1 | (d) Links low energy to low momentum to a wavelength that matches the spacing |
Accept, in (d), reasoning that compares the wavelengths of slow and fast neutrons using λ ∝ 1/v. Carry forward an error from (a) into (b) once.
How did you do?
- Q1 or Q5 wrong: re-read “Light as photons” in the study guide and Worked example 1. Remember E ∝ 1/λ.
- Q2 wrong: revisit “How photons travel”: frequency and energy stay fixed in a medium.
- Q3 or Q6 wrong: work through “Matter waves” and Worked example 2; write λ = h/p before every ratio.
- Q4 wrong: look again at Figure 1 and “Evidence for both models”.
- Q7 incomplete: revisit “When do matter waves matter?”
Then tick off the topic checklist.