AQA AS Level Physics Paper 2, November 2021: Question 9

1 mark · Medium difficulty · Multiple Choice

Calculate the maximum wavelength of the photons produced when a muon and an antimuon annihilate into the minimum number of photons.

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Question

Multiple choice question 09 asking for the maximum wavelength of photons produced when a muon and an antimuon annihilate to produce the minimum number of photons. Four options are given: A (5.9 x 10^-15 m), B (1.2 x 10^-14 m), C (5.9 x 10^-9 m), and D (1.2 x 10^-8 m).
Question text

09 A muon and an antimuon annihilate to produce the minimum number of photons.

What is the maximum wavelength of the photons?

[1 mark]

A 5.9 × 10−15 m

B 1.2 × 10−14 m

C 5.9 × 10−9 m

D 1.2 × 10−8 m

Mark scheme

Show the mark scheme Mark scheme for question 09 indicating the correct answer is B, corresponding to 1.2 x 10^-14 m.

09 B 1.2 × 10−14 m

How to answer it

Muon-Antimuon Annihilation & Photon Wavelength

AQA AS Level Physics • Particles & Radiation

What this question tests

This question assesses your understanding of particle-antiparticle annihilation, conservation of energy and momentum, rest mass energy of fundamental particles (specifically muons), and the relationship between photon energy, wavelength, and frequency.

Question 09: Maximum Wavelength of Annihilation Photons

Exam breakdown and expert tutor guidance

✅ Correct Answer

Option B ( 1.2 × 10⁻¹⁴ m )

Mark: 1 / 1 available for selecting B.

💡 Key Knowledge

  • Annihilation & Minimum Photons: Charge and momentum conservation dictate that a particle-antiparticle pair must annihilate into at least two photons.
  • Rest Mass Energy: Look up the rest mass of a muon in your AQA data booklet ( 105.66 MeV or find it via specific charge/mass values).
  • Energy-Wavelength Link: Maximum wavelength ( λ_max ) corresponds to minimum energy ( E_min ) per photon using E = hc / λ .

🧠 Exam Technique

Don't panic over unfamiliar particles! Whether it's an electron-positron or muon-antimuon pair, the physical principles remain identical. Always set the total initial rest energy equal to the energy of the resulting photons.

❌ Common Errors

  • Assuming only one photon is produced (violates conservation of momentum).
  • Forgetting to convert the rest mass energy from MeV to Joules using the elementary charge 1.60 × 10⁻¹⁹ J .
  • Using only the mass of one muon instead of the combined system energy of the muon and antimuon pair.

📐 Step-by-Step Calculation

  1. Identify minimum number of photons: To conserve momentum in the centre of mass frame, a pair must produce at least 2 photons travelling in opposite directions.
  2. Find total rest mass energy: A single muon has a rest mass energy of approximately 105.6 MeV (or calculate via m = 1.88 × 10⁻²⁸ kg ).
    Total energy for muon + antimuon = 2 × 105.6 MeV = 211.2 MeV .
  3. Convert energy to Joules: E_total = 211.2 × 10⁶ eV × 1.60 × 10⁻¹⁹ J/eV = 3.379 × 10⁻¹¹ J .
  4. Energy per photon: Since 2 identical photons share this energy equally (assuming particles were stationary prior to annihilation):
    E_photon = (3.379 × 10⁻¹¹ J) / 2 = 1.690 × 10⁻¹¹ J .
  5. Calculate maximum wavelength: Using λ = hc / E :
    λ = (6.63 × 10⁻³⁴ Js × 3.00 × 10⁸ ms⁻¹) / (1.690 × 10⁻¹¹ J)
    λ = 1.176 × 10⁻¹⁴ m ≈ 1.2 × 10⁻¹⁴ m (matching Option B).

Topics

Physics · 3.2 Particles and radiation

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.