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## ERQ · 13 marks · Topics: E.3 Radioactive decay + A.3 Work, energy and power · Archetype: modeling_and_assumptions
**Integration:** primary=E.3 Radioactive decay, secondary=A.3 Work, energy and power (strength: supporting)
**Stem.** A radiotherapy unit at a regional hospital uses a sealed cobalt-60 source to treat deep-seated tumours. Cobalt-60 decays by beta emission to an excited nickel-60 nucleus, which then emits two gamma photons of average total energy 1.25 MeV per decay. The half-life of Co-60 is 5.27 years. When commissioned, the source had an activity of A₀ = 8.5 × 10¹⁴ Bq. The medical physicist models the source as a constant-power gamma emitter for the purpose of planning a patient's 7-day treatment course, during which the patient receives a continuous fractionated exposure. All emitted gamma energy is assumed to be absorbed by tissue inside the treatment field.
### Part (a) State [2 marks] · AO1 · Topic: E.3
State the relationship between the activity A of a radioactive sample and the number N of undecayed nuclei present, and define the decay constant λ.
### Part (b)(i) Show that [3 marks] · AO2 · Topic: E.3
Using the relationship dN/dt = −λN, show that the decay constant of Co-60 is λ ≈ 4.17 × 10⁻⁹ s⁻¹.
### Part (b)(ii) Calculate [2 marks] · AO2 · Topic: E.3
Calculate the activity of the source after a 7-day treatment course, taking the initial activity as 8.5 × 10¹⁴ Bq.
### Part (c) Determine [3 marks] · AO2 · Topic: E.3 + A.3
Each decay deposits, on average, 1.25 MeV of gamma energy in the patient. Treating the activity as effectively constant over the 7 days, determine the total energy, in joules, delivered to the patient and hence the mean power deposited.
### Part (d) Evaluate [3 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR · Topic: E.3 + A.3
Evaluate the assumption, made in part (c), that the treatment power remains constant. Refer to the clinical implications of activity decay for dose uniformity over both the 7-day treatment window and the multi-year operational lifetime of the source.
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## Mark Scheme
### Part (a) [2 marks]
- M1: Activity A = λN (rate of decay is proportional to number of undecayed nuclei present) [no ECF]
- M2: λ is the probability per unit time that a given nucleus will decay (accept: constant of proportionality between A and N, with units s⁻¹) [no ECF]
### Part (b)(i) [3 marks] — Show that
- M1: Identifies λ = ln 2 / T₁/₂ from integration of dN/dt = −λN OR uses N = N₀ e^(−λt) with half-life condition [no ECF]
- M2: Converts T₁/₂ = 5.27 yr × 3.156 × 10⁷ s/yr = 1.663 × 10⁸ s [no ECF]
- M3: λ = 0.693 / 1.663 × 10⁸ = 4.17 × 10⁻⁹ s⁻¹ (accept 4.16–4.18 × 10⁻⁹ s⁻¹) [no ECF]
*Show-that target given to 3 sig figs; subsequent parts MUST use λ = 4.17 × 10⁻⁹ s⁻¹.*
### Part (b)(ii) [2 marks] — Calculate
- M1: t = 7 × 86 400 = 6.05 × 10⁵ s; λt = 4.17 × 10⁻⁹ × 6.05 × 10⁵ = 2.52 × 10⁻³ [ECF from (b)(i) λ value]
- M2: A = 8.5 × 10¹⁴ × e^(−2.52 × 10⁻³) = 8.48 × 10¹⁴ Bq (accept 8.47–8.49 × 10¹⁴ Bq; accept 8.5 × 10¹⁴ Bq if student recognises decay is negligible) [ECF from M1]
### Part (c) [3 marks] — Determine
- M1: Total number of decays N_d = A × t = 8.5 × 10¹⁴ × 6.05 × 10⁵ = 5.14 × 10²⁰ decays [ECF from (b)(ii) for A]
- M2: Energy per decay = 1.25 × 10⁶ × 1.60 × 10⁻¹⁹ = 2.00 × 10⁻¹³ J; total energy E = 5.14 × 10²⁰ × 2.00 × 10⁻¹³ = 1.03 × 10⁸ J [ECF from M1]
- M3: Mean power P = E/t = 1.03 × 10⁸ / 6.05 × 10⁵ = 170 W (accept 1.7 × 10² W) [ECF from M2]
### Part (d) [3 marks] — Evaluate (position + supporting + limiting per §4.4.1)
- M1 (Position): The constant-power assumption is justified for the 7-day treatment window but fails over the operational lifetime of the source.
- M2 (Supporting consideration): Since λt ≈ 2.5 × 10⁻³ ≪ 1 [ECF from (b)(ii) λt value], activity falls by only ~0.25% over one week — negligible compared to typical clinical dose tolerances (2–5%) — so the power delivered on day 7 is essentially identical to day 1, validating the model for one treatment course.
- M3 (Limiting consideration): However, A(t) = A₀ e^(−λt) is exponential, so over months and years the power falls substantially (≈12% per year, 50% after one half-life of 5.27 years); the physicist must therefore progressively lengthen exposure times or replace the source to maintain dose uniformity across patients treated throughout the source's lifetime.
### Marker notes
- Alternative method accepted for (c): computing instantaneous power P = A × E_γ = 8.5 × 10¹⁴ × 2.00 × 10⁻¹³ = 170 W directly, then E = P × t = 1.03 × 10⁸ J — full marks if both quantities stated.
- (d) accept any of: ignores attenuation/scattering in tissue, neglects beta-particle energy (absorbed in source capsule, so model is acceptable here), assumes 100% absorption (in reality only fraction deposited in target volume), neglects geometric inverse-square divergence, ignores source self-absorption increasing with age.
- Show-that target in (b)(i): λ = 4.17 × 10⁻⁹ s⁻¹ given to 3 sig figs; student-derived 4.16–4.18 × 10⁻⁹ s⁻¹ acceptable.
- Sig-fig penalty applied globally per §4.2, not per part.
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