These are original Marlbridge practice questions, not past exam questions; each combines two or more topics. The rubrics are a suggested Marlbridge rubric, not official scoring. Use the data in the calculator note. Unnamed atoms and metals are invented.
Big ideas of the unit
- Quantum theory was needed. Classical physics could not explain spectra, blackbody radiation or the photoelectric effect (Topic 15.1).
- Light comes in photons with E = hf, p = h/λ, no mass and no charge. Brighter light means more photons.
- Matter has waves. λ = h/p. Quantum effects matter when λ is comparable to the size of the system.
- Atoms have structure: a small, massive nucleus (Z protons, A − Z neutrons) with electrons around it. Bohr orbits must hold a whole number of electron wavelengths, so energy is quantized (Topic 15.2).
- Spectra are fingerprints. An atom absorbs or emits a photon only if its energy equals a gap between levels (Topic 15.3).
- Hot objects glow with a continuous spectrum. λ_max = b/T and P = σAT⁴; Planck’s quantized light explained the curve (Topic 15.4).
- Two collisions prove photons. In the photoelectric effect one photon is absorbed by one electron (Topic 15.5); in Compton scattering a photon bounces off an electron, conserving energy and momentum (Topic 15.6).
- Nuclei change by the same conservation laws. Mass converts to energy (E = mc²), decay is random with a fixed half-life (Topic 15.7), and each decay type conserves A, charge and lepton number (Topic 15.8).
Key relationships and methods
| Idea | Relationship or method | Watch out for |
|---|---|---|
| Photon (15.1) | E = hf = hc/λ; p = h/λ; v = c/n in a medium | Using λ in a medium in E = hc/λ |
| Matter wave (15.1) | λ = h/p = h/√(2mK) | K in eV, not J |
| Bohr orbits (15.2) | ke²/r² = mv²/r; 2πr = nλ; r = n²r₁ | Speed falls as r grows |
| Transitions (15.3) | E_photon = |ΔE|; λ (nm) = 1240 ÷ E (eV) | Exact gaps only; ionization takes any E above the binding energy |
| Blackbody (15.4) | λ_max = b/T; P = σAT⁴ | Kelvin; T⁴ |
| Photoelectric (15.5) | K_max = hf − φ = eV_s | Intensity changes the number of electrons, not K_max |
| Compton (15.6) | Δλ = (h/mₑc)(1 − cos θ); conserve p in x and y | Δλ depends only on θ |
| Mass-energy (15.7) | E = Δmc²; 1 u ↔ 931 MeV | Mass is not conserved; nucleon number is |
| Decay rate (15.7) | N = N₀e^(−λt); λ = ln 2 / t½ | Expected values only |
| Decay types (15.8) | α: A − 4, Z − 2; β⁻: Z + 1 with ν̄; β⁺: Z − 1 with ν; γ: no change | Lepton number |
| Two products from rest | Equal and opposite p; K ∝ 1/m | The lighter product gets more energy |
Question 1 (multiple choice · mixed)
Topics 15.1, 15.5, 15.6. Which pair of observations both give evidence that light is a stream of photons?
- (A) Light spreads out after a narrow slit; the maximum kinetic energy of photoelectrons does not depend on intensity.
- (B) The maximum kinetic energy of photoelectrons does not depend on intensity; X-rays scattered by electrons have a longer wavelength that grows with the scattering angle.
- (C) Light travels more slowly in glass than in air; scattered X-rays have a longer wavelength than the incident X-rays.
- (D) A thin oil film shows coloured fringes; light can be polarized.
Answer and explanation
Answer: (B). Both need a photon of energy hf (and, for Compton scattering, momentum h/λ) meeting one electron. A wave model predicts faster electrons in brighter light and no wavelength change on scattering.
- (A) Diffraction at a slit is wave behaviour.
- (C) A lower speed in glass fits a wave model; only the Compton half is photon evidence.
- (D) Interference and polarization are both wave behaviour.
Question 2 (multiple choice · mixed)
Topics 15.3, 15.4. Light from the hot, dense surface of a star passes through the cooler, thinner gas around it. Its peak intensity per unit wavelength is at 500 nm. Which row is correct?
- (A) A continuous spectrum with no lines, because the gas re-emits what it absorbs; surface temperature about 5.8 × 10³ K
- (B) Bright lines on a dark background; surface temperature about 5.8 × 10³ K
- (C) A continuous spectrum with dark lines; surface temperature about 1.7 × 10⁻⁴ K
- (D) A continuous spectrum with dark lines; surface temperature about 5.8 × 10³ K
Answer and explanation
Answer: (D). The surface acts as a blackbody, so its spectrum is continuous, with T = b/λ_max = 2.898 × 10⁻³ ÷ 500 × 10⁻⁹ = 5796 K. Atoms in the cool gas absorb photons that match their energy gaps, leaving dark lines.
- (A) The gas re-emits in all directions, so little continues towards Earth.
- (B) is the emission spectrum of a hot, thin gas against a dark background.
- (C) inverts Wien’s law (λ_max ÷ b).
Question 3 (multiple choice · mixed)
Topics 15.2, 15.8. A neutral atom of carbon-14 (¹⁴₆C) undergoes beta-minus decay. Just after the decay, before it gains or loses any electrons, what remains?
- (A) ¹⁴₅B⁻: 5 protons, 9 neutrons, 6 electrons
- (B) ¹⁴₇N: 7 protons, 7 neutrons, 7 electrons
- (C) ¹⁴₇N⁺: 7 protons, 7 neutrons, 6 electrons
- (D) ¹⁴₆C⁺: 6 protons, 8 neutrons, 5 electrons
Answer and explanation
Answer: (C). In the nucleus a neutron becomes a proton, emitting an electron and an antineutrino. Z rises to 7 (nitrogen) and A stays 14, so there are 7 neutrons. The 6 original electrons remain, so the charge is +7e − 6e = +e: a positive ion.
- (A) moves Z the wrong way. Beta-minus raises Z.
- (B) assumes the atom stays neutral. No electron was added.
- (D) takes the beta particle from the outer electrons. It is created in the nucleus, and the element changes.
Question 4 (constructed response · mixed)
Topics 15.1, 15.2, 15.3. In the Bohr model of hydrogen, r_n = n²r₁ with r₁ = 5.29 × 10⁻¹¹ m, and E_n = −13.6 eV ÷ n².
(a) Calculate the wavelength of the photon emitted when the atom drops from n = 3 to n = 2. Name the part of the spectrum. (b) Using the standing-wave condition, calculate the electron’s de Broglie wavelength in the n = 2 orbit, and hence its speed. (c) The atom (mass 1.67 × 10⁻²⁷ kg) was at rest. Calculate the photon’s momentum and the atom’s recoil speed. (d) A photon of wavelength 300 nm hits an atom in the n = 2 state. Show that the atom is ionized and calculate the de Broglie wavelength of the freed electron.
Worked solution and suggested Marlbridge rubric
(a) E₃ = −1.51 eV and E₂ = −3.40 eV, so E = 1.89 eV. λ = 1240 ÷ 1.889 = 656 nm, visible (red).
(b) Two wavelengths fit around the orbit: λ = 2π(4r₁) ÷ 2 = 4πr₁ = 6.65 × 10⁻¹⁰ m. v = h/(mₑλ) = 6.63 × 10⁻³⁴ ÷ (9.11 × 10⁻³¹ × 6.648 × 10⁻¹⁰) = 1.09 × 10⁶ m/s, half the n = 1 speed, as v ∝ 1/n requires.
(c) p = h/λ = 6.63 × 10⁻³⁴ ÷ 656 × 10⁻⁹ = 1.01 × 10⁻²⁷ kg·m/s. By momentum conservation the atom recoils the opposite way at v = p/m = 0.605 m/s, with negligible kinetic energy.
(d) E = 1240 ÷ 300 = 4.13 eV, more than the 3.40 eV binding energy of n = 2, so the atom is ionized. K = 0.73 eV = 1.17 × 10⁻¹⁹ J. p = √(2mₑK) = 4.62 × 10⁻²⁵ kg·m/s, so λ = h/p = 1.43 × 10⁻⁹ m.
| Point | What earns it |
|---|---|
| 1 | Energy gap 1.89 eV |
| 1 | 656 nm, visible |
| 1 | λ = 2πr₂/2 with r₂ = 4r₁ |
| 1 | v = 1.09 × 10⁶ m/s |
| 1 | Photon momentum and recoil speed 0.605 m/s, opposite to the photon |
| 1 | 4.13 eV > 3.40 eV, so ionized, with K = 0.73 eV |
| 1 | K converted to joules and λ = 1.43 × 10⁻⁹ m |
Total: 7 points. Carry an error from (a) into (c) once.
Question 5 (constructed response · mixed)
Topics 15.1, 15.4, 15.5. A lamp filament, modelled as a blackbody, has surface area 2.0 × 10⁻⁵ m² and runs at 2900 K.
(a) Calculate its peak wavelength and the total power it radiates. (b) A filter passes only 450 nm light. 2.0 mW of it falls on a metal with work function 2.10 eV. Calculate the photons per second reaching the metal, K_max in eV, and the stopping potential. (c) The filament is turned up to 3300 K. By what factor does its total power rise? A student says: “The filament is hotter, so the electrons from behind the filter will be faster.” Evaluate the claim. (d) The filter is removed. Explain why K_max now increases, even though the filament’s peak wavelength is far longer than the metal’s threshold wavelength.
Worked solution and suggested Marlbridge rubric
(a) λ_max = 2.898 × 10⁻³ ÷ 2900 = 999 nm (infrared). P = σAT⁴ = (5.67 × 10⁻⁸)(2.0 × 10⁻⁵)(2900)⁴ = 80.2 W.
(b) E = hc/λ = 4.42 × 10⁻¹⁹ J = 2.76 eV. Photons per second = 2.0 × 10⁻³ ÷ 4.42 × 10⁻¹⁹ = 4.5 × 10¹⁵. K_max = 2.76 − 2.10 = 0.66 eV, so V_s = 0.66 V.
(c) P ∝ T⁴: (3300/2900)⁴ = 1.68. The claim is wrong. The filter fixes the photon energy at 2.76 eV, so K_max stays 0.66 eV. The hotter filament emits more 450 nm photons per second, so more electrons are emitted: the current rises, not the speed.
(d) The spectrum is continuous: the filament also emits some violet and near-ultraviolet light, shorter than 450 nm. K_max is set by the most energetic photons present: 400 nm photons (3.10 eV) give 1.0 eV. Photons longer than λ₀ = 1240 ÷ 2.10 = 590 nm, most of the output, eject nothing.
| Point | What earns it |
|---|---|
| 1 | λ_max = 999 nm |
| 1 | P = 80.2 W |
| 1 | 4.5 × 10¹⁵ photons per second |
| 1 | K_max = 0.66 eV and V_s = 0.66 V |
| 1 | Power factor 1.68 |
| 1 | Claim wrong: the filter fixes photon energy; more photons give more current only |
| 1 | (d) Continuous spectrum; K_max set by the most energetic photons |
Total: 7 points.
Question 6 (constructed response · mixed)
Topics 15.1, 15.5, 15.6. An X-ray photon of wavelength 62.0 pm travels along +x and scatters off a free electron at rest. The electron recoils with kinetic energy 0.800 keV. The scattered photon leaves above the x-axis.
(a) Calculate the energy of the incoming photon in keV, and the wavelength of the scattered photon. (b) Calculate the angle θ through which the photon is scattered. (c) Use conservation of momentum to find the magnitude and direction of the electron’s momentum. Check your answer against the kinetic energy. (d) Calculate the electron’s de Broglie wavelength. Explain why the electron can be treated as free, when in the photoelectric effect the work function matters.
Worked solution and suggested Marlbridge rubric
(a) E = 1240 eV·nm ÷ 0.0620 nm = 20.0 keV. By energy conservation, E′ = 20.0 − 0.800 = 19.2 keV and λ′ = 1240 ÷ 19 200 = 0.06458 nm = 64.6 pm.
(b) Δλ = 64.58 − 62.0 = 2.58 pm. So 1 − cos θ = 2.583 ÷ 2.43 = 1.063, cos θ = −0.063 and θ = 93.6°. Keep unrounded values: Δλ is a small difference.
(c) p = h/λ = 1.069 × 10⁻²³ kg·m/s; p′ = h/λ′ = 1.027 × 10⁻²³ kg·m/s.
- x: p_ex = p − p′cos θ = 1.069 × 10⁻²³ + 0.065 × 10⁻²³ = 1.134 × 10⁻²³ kg·m/s.
- y: p_ey = 0 − p′sin θ = −1.025 × 10⁻²³ kg·m/s.
- p_e = √(1.134² + 1.025²) × 10⁻²³ = 1.53 × 10⁻²³ kg·m/s, at 42.1° below the +x axis.
Check: p_e²/(2mₑ) = 1.28 × 10⁻¹⁶ J = 0.80 keV.
(d) λ = h/p_e = 43.4 pm. The 20 keV photon has thousands of times the few eV that bind an outer electron, so the binding is negligible. Photoelectric photons have only a few eV, so there the work function is a large share.
| Point | What earns it |
|---|---|
| 1 | E = 20.0 keV and E′ = 19.2 keV by energy conservation |
| 1 | λ′ = 64.6 pm |
| 1 | θ = 93.6° (accept 93°–94°) from the Compton equation |
| 1 | Photon momenta and both components, y-component negative |
| 1 | 1.53 × 10⁻²³ kg·m/s at 42° below the +x axis |
| 1 | λ = 43.4 pm |
| 1 | Photon energy ≫ binding energy, unlike the photoelectric case |
Total: 7 points.
Question 7 (constructed response · mixed)
Topics 15.1, 15.7, 15.8. Nucleus X (A = 214, Z = 84), at rest, alpha decays to nucleus D, which then beta-minus decays to nucleus E. Atomic masses: X 213.99520 u; D 209.98420 u; ⁴He 4.00260 u. X has a half-life of 0.16 ms. Take each mass as proportional to its nucleon number when sharing energy.
(a) Write both decay equations, giving A and Z for D and E. Show that lepton number is conserved in the second. (b) Calculate the energy released in the alpha decay, and the kinetic energy of the alpha particle. (c) Sometimes D is left in an excited state 0.80 MeV above its ground state. Calculate the alpha particle’s kinetic energy in this case, and the wavelength of the gamma photon D then emits. (d) A sample starts with 6.0 × 10⁹ nuclei of X. Calculate the expected number of alpha particles emitted in the first 0.40 ms.
Worked solution and suggested Marlbridge rubric
(a) ²¹⁴₈₄X → ²¹⁰₈₂D + ⁴₂He. Then ²¹⁰₈₂D → ²¹⁰₈₃E + ⁰₋₁e + ν̄. Lepton number: 0 before; (+1) + (−1) = 0 after.
(b) Δm = 213.99520 − 209.98420 − 4.00260 = 0.00840 u (electron masses cancel). Q = 0.00840 × 931 = 7.82 MeV. Equal and opposite momenta, so K ∝ 1/m: K_α = 7.82 × 210/214 = 7.67 MeV (D gets 0.15 MeV).
(c) Only 7.82 − 0.80 = 7.02 MeV is shared, so K_α = 7.02 × 210/214 = 6.89 MeV. Gamma: λ = 1240 eV·nm ÷ 0.80 × 10⁶ eV = 1.55 × 10⁻³ nm = 1.55 pm. A and Z of D do not change.
(d) 0.40 ms is 2.5 half-lives. Remaining = 6.0 × 10⁹ × (½)^2.5 = 1.06 × 10⁹. Decayed, so alpha particles emitted: 4.9 × 10⁹.
| Point | What earns it |
|---|---|
| 1 | Alpha equation with D: A = 210, Z = 82 |
| 1 | Beta-minus equation with E: A = 210, Z = 83, electron and antineutrino |
| 1 | Lepton number 0 = (+1) + (−1) |
| 1 | Q = 7.82 MeV from the mass decrease |
| 1 | K_α = 7.67 MeV, using equal momenta |
| 1 | K_α = 6.89 MeV in the excited case |
| 1 | Gamma wavelength 1.55 pm |
| 1 | 4.9 × 10⁹ decays |
Total: 8 points. No lepton point for “electron and neutrino”.
How did you do?
Questions 1–3 score 1 point each; Questions 4–7 total 29. There is no pass mark: use the checklists that match the points you lost.
- photons and matter waves (Q1, 4(b)–(d), 6(d)): Topic 15.1 checklist
- atoms, ions and Bohr orbits (Q3, 4(b)): Topic 15.2 checklist
- energy levels and spectra (Q2, 4(a), 4(d)): Topic 15.3 checklist
- blackbody laws (Q2, 5(a), 5(c)): Topic 15.4 checklist
- photoelectric effect (Q1, 5(b)–(d)): Topic 15.5 checklist
- Compton scattering (Q1, 6): Topic 15.6 checklist
- mass-energy and half-life (Q7(b)–(d)): Topic 15.7 checklist
- decay equations (Q3, 7(a), 7(c)): Topic 15.8 checklist
For a topic-by-topic check, take the Unit 15 diagnostic.