Generated ERQ

✓ passed E.3 Radioactive decay × B.1 Thermal energy transfers 12 marks SL 2 passes 69.29s $0.4517
``` ## ERQ · 12 marks · Topics: E.3 Radioactive decay + B.1 Thermal energy transfers · Archetype: theory_application **Integration:** primary=E.3 Radioactive decay, secondary=B.1 Thermal energy transfers (strength: supporting) **Stem.** A school physics class investigates the thermal effect of radioactive decay using a sealed cobalt-60 source of initial activity 1.76 × 10¹¹ Bq, embedded in a thermally insulated copper block surrounded by 0.250 kg of water. Each decay of a cobalt-60 nucleus releases a total energy of 2.82 MeV, distributed between a beta particle, an antineutrino, and two gamma photons. The half-life of cobalt-60 is 5.27 years. The students assume that all of the decay energy is absorbed by the water, and that the container is a perfect calorimeter. The specific heat capacity of water is 4180 J kg⁻¹ K⁻¹. The initial water temperature is 20.0 °C. ### Part (a) State [2 marks] · AO1 · Topic: E.3 Radioactive decay State what is meant by the *activity* of a radioactive sample, and state one assumption made when defining the half-life of a radioactive nuclide. ### Part (b) Calculate [3 marks] · AO2 · Topic: E.3 Radioactive decay Calculate the initial power, in watts, released by the cobalt-60 source, assuming all decay energy is converted to thermal energy in the water. ### Part (c) Determine [4 marks] · AO2 · Topic: E.3 Radioactive decay + B.1 Thermal energy transfers The students wish to raise the temperature of the water from 20.0 °C to 21.0 °C. Determine the time, in hours, required to achieve this temperature rise. Assume the activity of the source remains constant over this interval. ### Part (d) Evaluate [3 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR One student claims: *"Because the half-life of cobalt-60 is so long, the assumption of constant activity is justified, and the measured temperature rise should match the predicted value."* Evaluate this claim. --- ## Mark Scheme ### Part (a) [2 marks] — State - M1: activity = number of nuclear decays (disintegrations) per unit time [no ECF] - M2: half-life assumes a large number of nuclei (so that statistical decay constant applies) / OR decay is a random spontaneous process / OR decay constant is independent of external conditions [no ECF] ### Part (b) [3 marks] — Calculate - M1: energy per decay E = 2.82 × 10⁶ × 1.60 × 10⁻¹⁹ = 4.51 × 10⁻¹³ J [no ECF] - M2: substitution P = A × E = 1.76 × 10¹¹ × 4.51 × 10⁻¹³ [ECF from M1] - M3: P = 0.0794 W (accept 0.079–0.080 W) [ECF from M1, M2] ### Part (c) [4 marks] — Determine - M1: Q = mcΔT = 0.250 × 4180 × 1.00 = 1045 J [no ECF] - M2: recognises required time t = Q / P using P from (b) [ECF from (b)] - M3: t = 1045 / 0.0794 = 1.316 × 10⁴ s [ECF from (b), M1] - M4: t = 1.316 × 10⁴ / 3600 = 3.66 h (accept 3.6–3.7 h) [ECF from M3] ### Part (d) [3 marks] — Evaluate (position + supporting + limiting per §4.4.1) - M1: **Position** — the claim is partially correct: the constant-activity assumption is justified, *but* the predicted temperature rise will not be matched in practice [no ECF] - M2: **Supporting consideration** — the elapsed time (~3.7 h) is many orders of magnitude smaller than the half-life of 5.27 years, so the fractional decrease in activity over the experiment is negligible (< 0.01%), validating the constant-activity assumption - M3: **Limiting consideration** — however, antineutrinos escape the apparatus carrying energy that is not deposited as heat (and/or some gamma photons penetrate the calorimeter walls without being absorbed), so the *measured* ΔT will be *less* than the predicted value ### Marker notes - Alternative method accepted for (c): direct one-line calculation t = mcΔT / (A × E) provided all substitutions visible — award M1 for Q, M2 for correct combined expression, M3 for t in seconds, M4 for conversion to hours. - (b) accept use of 2.82 MeV directly with conversion factor 1.60 × 10⁻¹³ J MeV⁻¹ giving 4.51 × 10⁻¹³ J. - (d) M3: accept any ONE valid limiting factor — antineutrino escape, incomplete gamma absorption, heat loss to copper block/surroundings, or non-ideal calorimeter. Do not award M3 for restating that activity decreases (that contradicts M2). - (d) if student argues claim is entirely wrong without acknowledging the validity of the constant-activity part, award maximum [2/3] (M1 forfeited, M3 only if limiting factor identified). - ECF chain: (b) → (c) M2,M3,M4 fully ECF-eligible. ```