## ERQ · 12 marks · Topics: E.3 Radioactive decay + B.1 Thermal energy transfers
**Stem.** An industrial sterilization facility uses a sealed cobalt-60 source housed inside a cylindrical lead shielding block. Cobalt-60 decays by beta-minus emission followed by gamma emission, with a half-life of 5.27 years. The source initially contains 2.0 g of pure Co-60 (molar mass 60 g mol⁻¹). The lead shielding has a mass of 5.0 kg and a specific heat capacity of 130 J kg⁻¹ K⁻¹. On average, 3.0 MeV of energy is released per decay, and 85% of this energy is absorbed by the lead. The facility manager wants to estimate the thermal effect on the shielding over two half-lives of operation (t = 10.54 years ≈ 3.32 × 10⁸ s). Assume the Avogadro constant N_A = 6.02 × 10²³ mol⁻¹ and 1 eV = 1.60 × 10⁻¹⁹ J.
### Part (a) State [2 marks] · AO1 · Topic: E.3
State what is meant by the *activity* of a radioactive source, and write down the relationship between the decay constant λ and the half-life T₁/₂.
### Part (b)(i) Calculate [3 marks] · AO2 · Topic: E.3
Calculate the initial activity A₀ of the 2.0 g Co-60 sample, in Bq.
### Part (b)(ii) Determine [2 marks] · AO2/AO3 · Topic: E.3
Determine the total number of Co-60 nuclei that decay between t = 0 and t = 10.54 years (two half-lives).
### Part (c) Calculate [3 marks] · AO3 · Topic: E.3 + B.1
Using your answer to (b)(ii) and the energy data in the stem, calculate the temperature rise of the lead shielding over the 10.54-year period. Assume no heat loss to the surroundings.
### Part (d) Suggest [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR
Suggest **two** reasons why the actual temperature rise of the lead shielding measured at the facility would be significantly smaller than the value calculated in (c).
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## Mark Scheme
### Part (a) [2 marks]
- M1: Activity = number of nuclear decays (disintegrations) per unit time / rate of decay [ECF: no]
- M2: λ = ln 2 / T₁/₂ (or equivalent, e.g. λ = 0.693/T₁/₂) [ECF: no]
### Part (b)(i) [3 marks]
- M1: N₀ = (m/M)·N_A = (2.0/60) × 6.02 × 10²³ = 2.01 × 10²² nuclei; and λ = ln2 / (5.27 × 3.156 × 10⁷) = 4.17 × 10⁻⁹ s⁻¹ [ECF: no]
- M2: A₀ = λN₀ with correct substitution = (4.17 × 10⁻⁹)(2.01 × 10²²) [ECF: yes from M1]
- M3: A₀ ≈ 8.4 × 10¹³ Bq (accept 8.3–8.5 × 10¹³ Bq, 2 s.f., correct unit) [ECF: yes]
### Part (b)(ii) [2 marks]
- M1: Recognise after 2 half-lives N remaining = N₀/4, so number decayed = (3/4)N₀ [ECF: no]
- M2: Decays = 0.75 × 2.01 × 10²² ≈ 1.5 × 10²² (accept 1.50–1.51 × 10²²) [ECF: yes from b(i) M1]
### Part (c) [3 marks]
- M1: Total energy absorbed = 0.85 × N_decay × 3.0 × 10⁶ × 1.60 × 10⁻¹⁹ J [ECF: yes from b(ii)]
- M2: E_absorbed = 0.85 × 1.51 × 10²² × 4.80 × 10⁻¹³ ≈ 6.2 × 10⁹ J [ECF: yes]
- M3: ΔT = E/(mc) = 6.2 × 10⁹ / (5.0 × 130) ≈ 9.5 × 10⁶ K (accept 9–10 × 10⁶ K) [ECF: yes]
### Part (d) [2 marks]
Award 1 mark for each distinct valid reason (max 2):
- Heat is continuously lost from the lead to the surroundings by conduction/convection/radiation over the 10-year period, so steady-state temperature is reached far below ΔT_calc [ECF: no]
- Gamma photons have significant penetration — a large fraction of the 3.0 MeV escapes the lead rather than being absorbed (the 85% absorption figure is optimistic) [ECF: no]
- Neutrinos emitted in β⁻ decay carry away energy that is not deposited in the lead [ECF: no]
- The lead would melt (m.p. 600 K) long before reaching ~10⁷ K, so the model breaks down — phase change absorbs energy [ECF: no]
- The calculation ignores that activity decreases over time — but energy still deposited is over-counted if all decays assumed at initial rate (only credit if linked to thermal loss being the dominant effect)
### Marker notes
- Alternative method for (b)(ii): integrate A(t) from 0 to 2T₁/₂, giving N₀(1 − e^(−2ln2)) = 0.75 N₀ — full credit.
- (b)(i): allow use of T₁/₂ = 5.3 yr giving A₀ ≈ 8.3 × 10¹³ Bq.
- (c): if student uses A₀ × t (treating activity as constant) instead of total decays from (b)(ii), cap at M1 only — this is the physics error E.3 is testing.
- (d) discriminator: must reference a *physical mechanism*, not just "the model is wrong". Two distinct mechanisms required; do not award two marks for two phrasings of "heat loss".