AQA A-Level Physics Paper 2, June 2025: Question 26

1 mark · Medium difficulty · Multiple Choice

Determine the energy of the emitted gamma photon that has the shortest wavelength following the alpha decay of americium-241 to neptunium-237.

Practise this question

Question

Energy level diagram showing americium-241 at 5.546 MeV decaying via alpha emission to two excited states of neptunium-237 at 0.103 MeV and 0.060 MeV, above the ground state at 0 MeV. The multiple-choice options for the energy of the emitted gamma photon with shortest wavelength are: A 5.546 MeV, B 5.486 MeV, C 0.103 MeV, and D 0.043 MeV.

Mark scheme

Show the mark scheme Mark scheme table showing question 26 with correct answer C (0.103 MeV), testing assessment objective AO2.

How to answer it

Nuclear Energy Levels & Gamma Ray Photon Emission

📌 What this question tests
  • Nuclear de-excitation: Understanding that after alpha decay to an excited state, a daughter nucleus de-excites by emitting gamma photons.
  • Photon energy & wavelength relationship: Using the inverse proportionality E = hc / λ to recognise that the shortest wavelength corresponds to the maximum photon energy.
  • Energy level diagrams: Reading transitions between nuclear levels correctly and distinguishing between alpha decay energy and gamma emission transitions.
Question 26 Analysis

Identifying the Shortest Wavelength Gamma Photon

AQA A-Level Physics • Nuclear Physics • Multiple Choice (1 Mark)

✅ Correct Answer

C: 0.103 MeV

Awarded 1 mark (AO2) for selecting Option C.

💡 Key Knowledge

  • Photon energy equation: E = hf = hc / λ .
  • Because λ = hc / E , the shortest wavelength corresponds strictly to the largest photon energy (ΔE).
  • Gamma photons originate only from nuclear de-excitation transitions within the daughter nucleus (²³⁷₉₃Np), not the alpha decay step from ²⁴¹₉₅Am.

📐 Step-by-Step Calculation & Transition Analysis

  1. Relate wavelength to energy:
    λ ∝ 1 / E → Shortest wavelength requires the maximum transition energy.
  2. Identify all possible gamma transitions in ²³⁷₉₃Np:
    • From 0.103 MeV to 0 MeV : ΔE₁ = 0.103 - 0 = 0.103 MeV
    • From 0.103 MeV to 0.060 MeV : ΔE₂ = 0.103 - 0.060 = 0.043 MeV
    • From 0.060 MeV to 0 MeV : ΔE₃ = 0.060 - 0 = 0.060 MeV
  3. Select the maximum energy:
    The largest transition is from the top excited state directly to the ground state:
    E_max = 0.103 MeV

🧠 Exam Technique & Decoding Distractors

  • Option A (5.546 MeV): Trap. This is the total energy level of the parent nucleus ²⁴¹Am, not a gamma photon.
  • Option B (5.486 MeV): Trap. This is 5.546 - 0.060 MeV , which is related to the kinetic energy released during alpha decay, not gamma emission.
  • Option D (0.043 MeV): Trap. This is the smallest energy transition ( 0.103 - 0.060 MeV ). It corresponds to the longest wavelength gamma photon, confusing shortest with longest.

❌ Common Student Errors

  • Confusing shortest wavelength with smallest energy: Inversely proportional quantities trip students up under time pressure. Remember: high frequency/energy = short wavelength!
  • Including the alpha transition: Forgetting that the question specifically asks for the emitted gamma photon, which only occurs between states of ²³⁷Np.

Topics

Physics · 3.8 Nuclear physics (A-level only) · 3.2 Particles and radiation

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.