AQA A-Level Physics Paper 1, June 2023: Question 9

1 mark · Medium difficulty · Multiple Choice

Identify which combination of neutron number N and proton number Z gives a nucleus with the greatest specific charge.

Practise this question

Question

A multiple-choice question numbered 0 9 asking which combination of N neutrons and Z protons gives a nucleus with the greatest specific charge, worth 1 mark. A table presents four options labelled A to D with values for N and Z: A has N=6, Z=5; B has N=8, Z=7; C has N=16, Z=13; D has N=20, Z=17.
Question text

09 A nucleus contains N neutrons and Z protons.

Which combination of N and Z gives a nucleus with the greatest specific charge?

[1 mark]

N Z

A 6 5

B 8 7

C 16 13

D 20 17

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer for question 09 is B, corresponding to N = 8 and Z = 7.

09 B 8 7

How to answer it

Specific Charge of a Nucleus

📌 What this question tests

This question assesses your understanding of specific charge (charge-to-mass ratio), specifically how to define it for a nucleus in terms of proton number ($Z$) and nucleon number ($N + Z$), and how to compare ratios across multiple choice options efficiently.

Question 09 Analysis

Evaluating Specific Charge Combinations

✅ Correct Answer: Option B

N = 8, Z = 7

This combination yields the highest numerical ratio of charge-to-mass when evaluated across the given choices.

💡 Key Knowledge

  • Specific Charge Definition: Specific charge = Total charge / Total mass.
  • Nuclear Charge: Due entirely to protons, so total charge is proportional to Z .
  • Nuclear Mass: Proportional to total nucleon number ( N + Z ), assuming proton and neutron masses are approximately equal.
  • Therefore, maximize the ratio: Z / (N + Z) .

🧠 Exam Technique

  • Don't waste time multiplying by exact values for elementary charge and atomic mass units. They cancel out when comparing ratios!
  • Quickly calculate Z / (N + Z) or Z / A as a decimal for each option to find the maximum.

❌ Common Errors

  • Dividing Z by N instead of total mass ( N + Z ).
  • Confusing nucleon number ( A = N + Z ) with neutron number ( N ) when estimating total mass.

📐 Step-by-Step Calculation / Ratio Check

  1. Option A: N = 6, Z = 5 → Nucleon number A = 11 . Ratio 5 / 11 = 0.4545
  2. Option B: N = 8, Z = 7 → Nucleon number A = 15 . Ratio 7 / 15 = 0.4667 (Highest)
  3. Option C: N = 16, Z = 13 → Nucleon number A = 29 . Ratio 13 / 29 = 0.4483
  4. Option D: N = 20, Z = 17 → Nucleon number A = 37 . Ratio 17 / 37 = 0.4595
🎯 Mark Scheme Allocation: 1 mark awarded for selecting B.

Topics

Physics · 3.2 Particles and radiation

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.