AQA A-Level Physics Paper 2, June 2025: Question 28
1 mark · Medium difficulty · Multiple Choice
Calculate the kinetic energy of an alpha particle emitted during the decay of polonium-210 to lead-206 using nuclear masses in atomic mass units.
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AQA A-Level Physics • Nuclear Physics
Alpha Decay Energy Release & Mass Defect
What this question tests
This question evaluates your ability to apply Einstein's mass-energy equivalence to a radioactive decay reaction. Specifically, it assesses:
- Calculating the mass defect (Δm) of an alpha decay reaction from rest masses given in atomic mass units (u).
- Converting mass defect in u directly into released kinetic energy in MeV using the data sheet equivalence ( 1 u = 931.5 MeV ).
- Understanding conservation of momentum and energy approximations (why negligible recoil speed implies the alpha particle carries essentially all the released kinetic energy).
- Spotting metric prefix traps ( meV vs MeV ).
Question 28 Walkthrough
Calculation of Alpha Particle Kinetic Energy
✅ Correct Answer
C — 4.7 MeV
The mass defect is 0.005 u . Multiplying by 931.5 MeV u⁻¹ yields 4.66 MeV ≈ 4.7 MeV .
💡 Key Knowledge
- Nuclear Equation: ²¹⁰₈₄Po → ²⁰⁶₈₂Pb + ⁴₂α
- Mass-Energy Conversion: On the AQA data sheet, 1 u = 931.5 MeV . You do not need to convert to kg and J first!
- Recoil Assumption: Since the daughter lead nucleus is ~51 times heavier than the alpha particle and recoil speed is stated as negligible, virtually 100% of the decay energy goes to the kinetic energy of the alpha particle.
📐 Step-by-Step Calculation
Step 1: Calculate total initial mass (Reactants) Mass of ²¹⁰Po nucleus = 209.983 u
Step 2: Calculate total final mass (Products) Mass of ²⁰⁶Pb + Mass of α = 205.975 u + 4.003 u = 209.978 u
Step 3: Calculate the mass defect (Δm) Δm = Initial Mass − Final Mass
Δm = 209.983 u − 209.978 u = 0.005 u
Δm = 209.983 u − 209.978 u = 0.005 u
Step 4: Convert mass defect to energy in MeV Using the data sheet equivalence factor 1 u = 931.5 MeV :
E = Δm × 931.5 MeV u⁻¹
E = 0.005 × 931.5 = 4.6575 MeV
Rounding to 2 significant figures: 4.7 MeV
E = Δm × 931.5 MeV u⁻¹
E = 0.005 × 931.5 = 4.6575 MeV
Rounding to 2 significant figures: 4.7 MeV
🧠 Exam Technique & Shortcuts
- Use 931.5 MeV: Avoid converting u → kg → J → eV → MeV . That 4-step route wastes 2–3 minutes in a multiple-choice section and introduces rounding errors.
- Inspect orders of magnitude: Typical alpha particles emitted from radioactive decays have energies between 4 MeV and 9 MeV . Even without calculation, GeV (option D) is astronomical for alpha decay, and meV (option A) is far too tiny.
❌ Common Traps & Distractors
- Prefix Blindness (Option A: 5.0 meV): Lowercase m means milli (10⁻³ eV), while uppercase M means mega (10⁶ eV). A difference of 9 orders of magnitude!
- Rough Rounding Error: Rounding 0.005 × 931.5 very crudely as ~5 and picking A without reading the prefix.
- Unit Mix-up (Option D: 3.7 GeV): Using speed of light squared ( c² ) incorrectly on values already expressed in mass units.
Topics
Physics · 3.8 Nuclear physics (A-level only)
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.