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. Conventions for every question: lepton number is +1 for electrons and neutrinos, −1 for positrons and antineutrinos, and 0 for nuclei, alpha particles and photons. When sharing energy between two products, take each mass as proportional to its nucleon number. The chain in Question 6 and the energies in Question 7 are invented for practice. Round only at the end.
Question 1 (multiple choice · foundation)
Polonium-210 decays: ²¹⁰₈₄Po → ²⁰⁶₈₂Pb + X. What is X?
- (A) An alpha particle, ⁴₂He
- (B) A neutral helium-4 atom
- (C) A positron and a neutrino
- (D) A gamma photon
Answer and explanation
Answer: (A). Nucleon number: 210 = 206 + 4, so X has A = 4. Charge: 84 = 82 + 2, so X has charge +2. A particle with A = 4 and charge +2 is a helium-4 nucleus: an alpha particle.
- (B) has the right nucleon number but no net charge, so charge would not balance (84 on the left, 82 on the right). An alpha particle has no electrons.
- (C) cannot change A from 210 to 206; positrons and neutrinos have no nucleons.
- (D) changes neither A nor Z, so the product would still be polonium-210.
Question 2 (multiple choice · core)
Sodium-22 (²²₁₁Na) undergoes beta-plus decay. Which equation is correct?
- (A) ²²₁₁Na → ²²₁₀Ne + ⁰₊₁e + ν
- (B) ²²₁₁Na → ²²₁₂Mg + ⁰₋₁e + ν̄
- (C) ²²₁₁Na → ²²₁₀Ne + ⁰₊₁e + ν̄
- (D) ²²₁₁Na → ¹⁸₉F + ⁴₂He
Answer and explanation
Answer: (A). In beta-plus decay a proton becomes a neutron, so Z falls by 1 (to neon) and A stays 22. A positron is emitted with a neutrino. Lepton number: 0 = (−1) + (+1).
- (B) is a balanced equation, but for beta-minus decay. The question asks for beta-plus.
- (C) pairs the positron with an antineutrino. The lepton number after would be −2, not 0.
- (D) balances A (22 = 18 + 4) and Z (11 = 9 + 2), but it describes alpha decay, not beta-plus decay: no positron is emitted.
Question 3 (multiple choice · core)
Why are neutrinos so difficult to detect?
- (A) They have no charge and negligible mass, and interact only through the weak force and gravity, so they rarely interact with matter.
- (B) They are stopped by the first few millimetres of any material they hit, so they never reach a detector.
- (C) They have a large charge and are repelled by the nuclei in a detector.
- (D) They are absorbed by the nucleus that emits them, so they never leave it.
Answer and explanation
Answer: (A). A neutrino feels neither the electric force nor the strong force. With almost no interaction, most neutrinos pass straight through a detector, or even the whole Earth.
- (B) describes the opposite behaviour, which is closer to how alpha particles behave in matter.
- (C) is wrong because neutrinos are uncharged.
- (D) contradicts the decay equations: the neutrino is emitted and carries energy away.
Question 4 (multiple choice · core)
After an alpha decay, a thorium-234 nucleus is left in an excited state, ²³⁴₉₀Th*. It then emits a gamma photon. Which describes the nucleus after the photon is emitted?
- (A) ²³⁴₉₀Th, with less energy than before
- (B) ²³⁴₉₁Pa, with less energy than before
- (C) ²³⁰₈₈Ra, with less energy than before
- (D) ²³⁴₉₀Th, with more energy than before
Answer and explanation
Answer: (A). A photon has no charge and no nucleons, so A and Z do not change. The nucleus is still thorium-234, now in a lower energy state, because the photon carried energy away.
- (B) changes Z by +1, which is what beta-minus decay does.
- (C) applies a second alpha decay.
- (D) has the energy change backwards. Emitting a photon lowers the energy of the nucleus.
Question 5 (constructed response · core)
(a) Complete each equation and name the type of decay.
- ³²₁₅P → ? + ⁰₋₁e + ?
- ¹⁵₈O → ¹⁵₇N + ? + ?
- ²⁴¹₉₅Am → ? + ⁴₂He
(b) Strontium-90 (Z = 38) decays to yttrium-90 (Z = 39). Name the decay, list the particles emitted, and show that lepton number is conserved.
Worked solution and suggested Marlbridge rubric
(a)
- ³²₁₅P → ³²₁₆S + ⁰₋₁e + ν̄. Beta-minus. A: 32 = 32. Charge: 15 = 16 − 1.
- ¹⁵₈O → ¹⁵₇N + ⁰₊₁e + ν. Beta-plus. Z falls by 1 with A unchanged, so a positron is emitted, and a neutrino keeps lepton number at 0.
- ²⁴¹₉₅Am → ²³⁷₉₃Np + ⁴₂He. Alpha. A: 241 − 4 = 237. Z: 95 − 2 = 93 (neptunium).
(b) A stays 90 and Z rises by 1, so a neutron has become a proton: beta-minus decay. An electron and an antineutrino are emitted. Lepton number: 0 before; (+1) + (−1) = 0 after.
| Point | What earns it |
|---|---|
| 1 | Equation 1: ³²₁₆S and ν̄, named beta-minus |
| 1 | Equation 2: positron and neutrino, named beta-plus |
| 1 | Equation 3: ²³⁷₉₃Np (accept “element with Z = 93”), named alpha |
| 1 | (b) Identified as beta-minus, with electron and antineutrino |
| 1 | (b) Lepton number shown as 0 before and (+1) + (−1) = 0 after |
Do not award the equation 2 point for “positron and antineutrino”: that breaks lepton number conservation.
Question 6 (constructed response · core)
A nucleus with Z = 90 and N = 140 undergoes four decays in this order: alpha, beta-minus, beta-minus, alpha.
(a) On a grid of proton number Z (vertical) against neutron number N (horizontal), mark the starting nucleus and draw an arrow for each decay. (b) Give Z, N and A of the final nucleus. (c) A student says: “After the first three decays, Z is back to 90, so the nucleus is the same nucleus as at the start.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) Start at (N, Z) = (140, 90).
- Alpha: two left, two down → (138, 88).
- Beta-minus: one left, one up → (137, 89).
- Beta-minus: one left, one up → (136, 90).
- Alpha: two left, two down → (134, 88).
(b) Final nucleus: Z = 88, N = 134, A = 222.
(c) The claim is partly right. After three decays Z = 90 again, so it is the same element. But A = 136 + 90 = 226, not 230. It is a different isotope of the same element, with 4 fewer neutrons. A nucleus is identified by both Z and A.
| Point | What earns it |
|---|---|
| 1 | Axes labelled with Z up and N across, starting point at (140, 90) |
| 1 | Both alpha arrows drawn two left and two down |
| 1 | Both beta-minus arrows drawn one left and one up |
| 1 | Final Z = 88, N = 134 and A = 222 |
| 1 | Agrees it is the same element, because Z is the same |
| 1 | Rejects “same nucleus”, because A has fallen from 230 to 226 (a different isotope) |
Allow carried-forward errors from a wrong arrow in (a) into (b) once.
Question 7 (constructed response · stretch)
A nucleus with nucleon number 212, at rest, undergoes alpha decay. The alpha particle has 6.00 MeV of kinetic energy.
(a) Calculate the kinetic energy of the daughter nucleus, and the total energy released by the decay.
A different source undergoes beta-minus decay. A student measures the kinetic energies of many emitted electrons and finds every value from nearly zero up to a maximum.
(b) Explain why every alpha particle from the first decay (to the same final state) has the same kinetic energy. (c) Explain why the electrons from the beta-minus decay do not all have the same kinetic energy, and what this tells you about the decay.
Worked solution and suggested Marlbridge rubric
(a) The daughter has A = 208. Total momentum is zero, so the alpha particle and the daughter have equal and opposite momenta. With equal p, K = p²/2m is inversely proportional to mass: K_d = 6.00 MeV × 4/208 = 0.115 MeV. Energy released = 6.00 + 0.115 = 6.12 MeV (6.115 MeV).
(b) Only two objects share the released energy. Momentum conservation fixes the ratio of their kinetic energies, and energy conservation fixes their total. So there is only one possible way to share the energy, and every alpha particle gets the same amount.
(c) In beta-minus decay three particles leave: the daughter nucleus, the electron and an antineutrino. Energy and momentum can be shared among three particles in many different ways, so the electron can get anything from almost nothing up to nearly all of the available energy. The spread shows that a third particle, which the detector did not record, carries away the rest of the energy. That particle is the antineutrino, which barely interacts with matter.
| Point | What earns it |
|---|---|
| 1 | Equal and opposite momenta, so K inversely proportional to mass |
| 1 | K_d = 0.115 MeV |
| 1 | Energy released = 6.12 MeV |
| 1 | (b) Two products: momentum and energy conservation give one unique share |
| 1 | (c) Three products, so the sharing can vary |
| 1 | (c) Links the spread to an undetected antineutrino, with a reason it is not detected (no charge, negligible mass, weak interaction) |
A remark that the maximum electron energy is close to the energy released (when the antineutrino gets almost none) is good physics but earns no extra mark.
How did you do?
- Q1, Q4 or Q5 wrong: re-read “The four types of decay” in the study guide and check each equation for A and Z.
- Q2 wrong: revisit “Conservation rules for every decay” and the lepton number column of the particle table.
- Q3 wrong: reread “The particles involved”.
- Q6 wrong: work through Figure 1 and Worked example 2 again.
- Q7 incomplete: redo Worked example 3, including its interpretation paragraph.
Then tick off the topic checklist.