AQA A-Level Physics Paper 2, June 2025: Question 25

1 mark · Medium difficulty · Multiple Choice

Calculate the initial distance from a gamma radiation source given the initial and new count rates when the distance is increased by 10 cm.

Practise this question

Question

Question 25 asks: A radiation meter is placed at a distance d from a source of gamma radiation. The corrected count rate is 800 counts per second. When the radiation meter is moved a further 10 cm from the source the corrected count rate is 600 counts per second. What is d? Four multiple choice options are given: A: 8.7 cm, B: 13 cm, C: 30 cm, D: 64 cm.

Mark scheme

Show the mark scheme Mark scheme table row showing question number 25, correct answer D (64 cm), and assessment objective AO2.

How to answer it

Gamma Radiation & The Inverse-Square Law

📌 What this question tests

This question assesses your understanding and algebraic application of the inverse-square law for a point source of gamma radiation ( I ∝ 1/r² ). You must be able to set up a ratio connecting count rate and distance, handle algebraic rearrangement cleanly, and work in consistent distance units without unnecessary unit conversions.

Question 25 (Multiple Choice)

AQA A-Level Physics • Nuclear Physics • 1 Mark • AO2

✅ Correct Answer

D — 64 cm

Mark Scheme: Award 1 mark for option D.

💡 Key Knowledge

  • Gamma radiation spreads out spherically from an isotropic point source with negligible absorption in air over short distances.
  • Intensity (and therefore corrected count rate, C) obeys the inverse-square law:
    C = k / r² or C₁r₁² = C₂r₂²
  • Corrected count rate: The background radiation has already been subtracted, meaning the measured rate is directly proportional to intensity.

📐 Step-by-Step Calculation

  1. Identify given values:
    Initial distance: r₁ = d
    Initial count rate: C₁ = 800 s⁻¹
    New distance: r₂ = d + 10 cm
    New count rate: C₂ = 600 s⁻¹
  2. Set up the ratio equation:
    Since C₁ × r₁² = C₂ × r₂² :
    800 × d² = 600 × (d + 10)²
  3. Simplify the ratio:
    Divide both sides by 200:
    4d² = 3(d + 10)²
    (d + 10)² / d² = 4 / 3
  4. Take the square root of both sides:
    (d + 10) / d = √(4 / 3)
    1 + (10 / d) = 2 / √3 ≈ 1.1547
  5. Solve for d:
    10 / d = 1.1547 - 1 = 0.1547
    d = 10 / 0.1547 ≈ 64.6 cm ≈ 64 cm (to 2 s.f., matching D).

🧠 Exam Technique & Shortcuts

  • Keep distances in cm: Because all answer choices are in centimetres, converting to metres adds needless steps and risks power-of-ten mistakes.
  • Quick estimation trick: The count rate drops by only 25% (from 800 to 600). Since C ∝ 1/r² , a small fractional drop in count rate means the distance increased by only a small fraction:
    r₂/r₁ = √(800/600) ≈ 1.155
    An increase of only 15.5% corresponds to 10 cm. Therefore, the original distance must be relatively large ( 10 / 0.155 ≈ 64 cm ). This instantly rules out small values like 8.7 cm or 13 cm!

❌ Common Errors to Avoid

  • Linear approximation trap: Assuming count rate is inversely proportional to distance ( C ∝ 1/r ). This gives 800d = 600(d + 10) ⇒ d = 30 cm (distractor C).
  • Misinterpreting "a further 10 cm": Setting r₂ = 10 cm instead of d + 10 cm .
  • Expanding quadratics unnecessarily: Expanding 3(d + 10)² = 3d² + 60d + 300 and solving d² - 60d - 300 = 0 via the quadratic formula works, but wastes precious exam time compared to taking square roots directly.

Topics

Physics · Required Practicals · 3.8 Nuclear physics (A-level only) · A-Level practicals (7–12)

Question and mark scheme from the AQA A-Level Physics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.