AQA A-Level Physics Paper 2, June 2025: Question 3

📐 Step-by-Step Calculation

1. Method 1 (via atomic mass unit u):
   Mass of 1 atom of ²³⁸Pu ≈ 238 × 1.661 × 10⁻²⁷ kg = 3.953 × 10⁻²⁵ kg
   Total mass = 3.953 × 10⁻²⁵ × (3.23 × 10²⁴)
   Mass = 1.28 kg
2. Method 2 (via moles & molar mass):
   Number of moles n = N / N_A = (3.23 × 10²⁴) / (6.02 × 10²³) = 5.365 mol
   Total mass = 5.365 mol × 238 g mol⁻¹ = 1277 g = 1.28 kg

✅ Correct Answer & Marking

• Mark 1: Finding mass of one atom ( 3.953 × 10⁻²⁵ kg ) OR number of moles ( 5.37 mol ).
• Mark 2: Final answer of 1.28 kg (accepts 1.29 kg if using proton/neutron masses).

❌ Common Errors

• Forgetting to convert grams to kilograms at the end when using the molar mass route ( 238 g mol⁻¹ ).
• Using standard proton mass instead of the unified atomic mass unit u = 1.661 × 10⁻²⁷ kg .

🧠 Exam Technique

• Check the unit requested on the answer line! The unit line states kg, so leaving an answer of 1280 g loses the second mark without clear conversion.

📐 Step-by-Step Calculation

1. Initial Activity (A₀):
   A₀ = λN₀ = (2.51 × 10⁻¹⁰ s⁻¹) × (3.23 × 10²⁴)
   A₀ = 8.11 × 10¹⁴ Bq (or s⁻¹)
2. Required Total Thermal Power:
   Minimum electrical power required = 29.6 W with efficiency η = 4.58% = 0.0458 .
   P_thermal = 29.6 / 0.0458 = 646.3 W
3. Minimum Activity Needed (A):
   Each decay yields E_alpha = 8.94 × 10⁻¹³ J .
   A = P_thermal / E_alpha = 646.3 / (8.94 × 10⁻¹³) = 7.23 × 10¹⁴ Bq
4. Decay Time in Seconds:
   Using A = A₀ e^(−λt) ⇒ t = −ln(A / A₀) / λ
   t = −ln(7.23 × 10¹⁴ / 8.11 × 10¹⁴) / (2.51 × 10⁻¹⁰)
   t = −ln(0.8915) / (2.51 × 10⁻¹⁰) = 4.58 × 10⁸ s
5. Convert Seconds to Earth Years:
   t = (4.58 × 10⁸ s) / (365 × 24 × 3600 s yr⁻¹)
   Time = 14.5 Earth years (accepts 14 to 15 years)

✅ Mark Scheme Breakdown (5 Marks)

• Mark 1: Initial activity A₀ = 8.11 × 10¹⁴ Bq .
• Mark 2: Dividing output power by efficiency ( 29.6 / 0.0458 = 646 W ) OR multiplying initial power/activity by 0.0458.
• Mark 3: Correct relationship between power and activity using alpha particle energy ( P = A × 8.94 × 10⁻¹³ ).
• Mark 4: Correct logarithmic decay formula applied: t = ln(A/A₀) / −λ .
• Mark 5: Correct time unit conversion giving 14.5 Earth years (allow 14 or 15).

❌ Common Traps & Misconceptions

• Efficiency blunder: Multiplying electrical power by 0.0458 instead of dividing to find total thermal power needed. Missing efficiency gives an absurd answer of 404 years.
• Sign errors in logarithms: Forgetting the negative sign when using ln(A/A₀) = −λt , leading to negative time.
• Unit conversion omission: Leaving time in seconds ( 4.58 × 10⁸ s ) instead of Earth years.

💡 Key Knowledge

• Power is rate of energy transfer: P = (ΔN/Δt) × E = A × E .
• Efficiency relation: P_elec = Efficiency × P_thermal .
• Since A ∝ N ∝ P , you can directly use power ratio: t = −ln(P / P₀) / λ , where initial electrical power P₀ = A₀ × E_alpha × 0.0458 = 33.2 W .

📐 Step-by-Step Calculation

1. Energy to warm ice from −25 °C to 0.0 °C:
   Q₁ = mcΔθ = 1.0 × 10⁻³ kg × 2100 J kg⁻¹ K⁻¹ × 25 K = 52.5 J
2. Energy to melt ice at 0.0 °C:
   Q₂ = mL_f = 1.0 × 10⁻³ kg × (334 × 10³ J kg⁻¹) = 334 J
3. Energy to vaporise liquid water at 0.0 °C:
   Q₃ = mL_v = 1.0 × 10⁻³ kg × (2500 × 10³ J kg⁻¹) = 2500 J
4. Total Energy Required:
   Q_total = 52.5 + 334 + 2500 = 2886.5 J ≈ 2890 J (or 2.9 kJ)
5. Energy Supplied by Laser:
   E_laser = P × t = 1.0 × 10³ W × 3.0 s = 3000 J (3.0 kJ)
6. Conclusion:
   Since 2887 J < 3000 J , the laser is able to vaporise the ice.

✅ Mark Scheme Breakdown (4 Marks)

• Mark 1: Valid Q = mcΔθ calculation ( 52.5 J ).
• Mark 2: Valid Q = mL calculation for fusion ( 334 J ) or vaporisation ( 2500 J ).
• Mark 3: Full calculation combining all three heat terms ( 2887 J ) AND laser energy calculation ( E = 1000 × 3.0 = 3000 J ).
• Mark 4: Correct comparison ( 2900 J < 3000 J ) and explicit affirmative conclusion ("Yes").

❌ Common Errors

• Mass unit trap: Using m = 1.0 instead of converting grams to kilograms ( 1.0 × 10⁻³ kg ).
• Prefix neglect: Overlooking the 'k' in kJ kg⁻¹ (e.g. using 334 instead of 334,000).
• Missing the prompt's instruction: The question explicitly says that direct vaporisation energy equals energy to melt + energy to vaporise melted ice. Forgetting either L_f or L_v loses marks.

🧠 Exam Technique: Alternative Comparisons