AQA A-Level Physics Paper 1, November 2020: Question 10

1 mark · Medium difficulty · Multiple Choice

Identify the resulting nuclide after thorium-232 undergoes one alpha emission followed by two beta-minus emissions.

Practise this question

Question

A multiple choice question numbered 10 asking which nuclide is formed when the radioactive nuclide 232/90 Th decays by one alpha emission followed by two beta-minus emissions. Four options are provided: A (238/92 U), B (230/90 Th), C (228/90 Th), and D (228/88 Rn), each with a selection box next to it.
Question text

10 The radioactive nuclide 232 Th decays by one α emission followed by two β− emissions.

Which nuclide is formed as a result of these decays?

[1 mark]

238 U

A 92

230 Th

B 90

228Th

C 90

228 Rn

D 88

Mark scheme

Show the mark scheme The mark scheme table showing question number 10 corresponding to the correct answer C.

10 C

How to answer it

Radioactive Decay & Successive Emissions

What this question tests

This question assesses your understanding of how alpha (α) and beta-minus (β⁻) radioactive decays affect the nucleon number ($A$) and proton number ($Z$) of a nuclide, and how to track these changes sequentially through a multi-step decay chain.

Question Part 10

Determining the Resultant Nuclide

Path: Th-232 → One Alpha Emission → Two Beta-Minus Emissions

✅ Correct Answer: C

Option C ( ^228_90 Th ) is the correct nuclide formed after the sequence of decays.

💡 Key Knowledge

  • Alpha decay (α): Emits a helium nucleus ( ^4_2 He ). Nucleon number decreases by 4; proton number decreases by 2.
  • Beta-minus decay (β⁻): Emits an electron and an antineutrino. Nucleon number remains unchanged; proton number increases by 1.

🧠 Exam Technique

Break the multi-step process down step-by-step. Keep a running tally of both the top number (nucleon number) and bottom number (proton number) after each individual decay event.

❌ Common Errors

  • Forgetting that two beta-minus emissions occur, only accounting for one.
  • Incorrectly changing the nucleon number during a beta-minus decay.
  • Failing to realise that losing 2 protons from alpha decay and then gaining 2 protons back from two beta decays brings the proton number back to the original value (90).

📐 Step-by-Step Breakdown

  1. Starting Nuclide: Thorium-232 is written as ^232_90 Th (Nucleon number = 232, Proton number = 90).
  2. Step 1: One Alpha (α) Emission
    • Nucleon number: 232 - 4 = 228
    • Proton number: 90 - 2 = 88
    • Intermediate nuclide: ^228_88 Ra
  3. Step 2: Two Beta-Minus (β⁻) Emissions
    • Nucleon number changes: 228 + 0 + 0 = 228 (remains unchanged)
    • Proton number changes: 88 + 1 + 1 = 90 (increases by 2 overall)
    • Final nuclide: ^228_90 Th
Mark Scheme Allocation: 1 mark total for selecting option C.

Topics

Physics · 3.8 Nuclear physics (A-level only)

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.