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✓ passed E.3 Radioactive decay × A.1 Kinematics 10 marks SL 3 passes 107.22s $0.7233
## ERQ · 10 marks · Topics: E.3 Radioactive decay + A.1 Kinematics · Archetype: data_response **Integration:** primary=E.3 Radioactive decay, secondary=A.1 Kinematics (strength: supporting) **Stem.** A laboratory uses a small sealed source of polonium-210, which decays by alpha emission with a half-life of 138 days. A student investigates the source using a vacuum chamber in which a silicon detector is placed 4.50 cm from the source along the axis of an evacuated tube. The detector records both the count rate (in counts per second) and the time-of-flight of individual alpha particles between the source and the detector. The measured activity averaged over the first hour is 4.20 × 10⁴ Bq. A representative time-of-flight measurement for a single alpha particle is 2.93 × 10⁻⁹ s. The mass of an alpha particle is 6.64 × 10⁻²⁷ kg. | Quantity | Measured value | |---|---| | Source–detector distance, d | 4.50 × 10⁻² m | | Mean activity (first hour), A₀ | 4.20 × 10⁴ Bq | | Alpha time-of-flight, t | 2.93 × 10⁻⁹ s | | Half-life of Po-210, T½ | 138 days | | Alpha mass, mα | 6.64 × 10⁻²⁷ kg | ### Part (a) Define [2 marks] · AO1 · Topic: E.3 Define the terms *activity* and *half-life* of a radioactive source. ### Part (b)(i) Calculate [3 marks] · AO2 · Topic: E.3 Calculate the decay constant of polonium-210, in s⁻¹, and hence determine the number of undecayed Po-210 nuclei in the source at the start of the measurement. ### Part (b)(ii) Determine [2 marks] · AO2 · Topic: A.1 (+ E.3) Using the time-of-flight data, determine the kinetic energy of an emitted alpha particle, in MeV. ### Part (c) Suggest [2 marks] · AO3 · Topic: E.3 + A.1 The student finds that the count rate measured by the detector remains essentially constant (within experimental scatter) over the full 48-hour run. Suggest why the count rate does not show a measurable decrease, despite radioactive decay occurring throughout the run. ### Part (d) Evaluate [1 mark] · AO3 · ASSUMPTIONS DISCRIMINATOR In part (b)(ii) the student assumed the alpha particle travels in a straight line at constant speed from source to detector. Evaluate this assumption for the experimental setup described. --- ## Mark Scheme ### Part (a) [2 marks] - M1: Activity = number of nuclear disintegrations (decays) per unit time ✓ [no ECF] - M2: Half-life = time taken for the activity (or number of undecayed nuclei) to fall to half its initial value ✓ [no ECF] ### Part (b)(i) [3 marks] — Calculate - M1: λ = ln2 / T½ with T½ converted to seconds (138 × 24 × 3600 = 1.192 × 10⁷ s) ✓ [no ECF] - M2: λ = 0.693 / 1.192 × 10⁷ = 5.81 × 10⁻⁸ s⁻¹ ✓ [ECF from unit conversion] - M3: N = A/λ = 4.20 × 10⁴ / 5.81 × 10⁻⁸ = 7.23 × 10¹¹ nuclei ✓ [ECF from M2] ### Part (b)(ii) [2 marks] — Determine - M1: v = d/t = 4.50 × 10⁻² / 2.93 × 10⁻⁹ = 1.54 × 10⁷ m s⁻¹ ✓ [no ECF] - M2: KE = ½mv² = ½(6.64 × 10⁻²⁷)(1.54 × 10⁷)² = 7.86 × 10⁻¹³ J ≈ 4.91 MeV ✓ [ECF from M1] ### Part (c) [2 marks] — Suggest (proposal + reasoning per §4.4.1) - M1: Over 48 hours the fraction of nuclei decayed is very small because 48 h ≪ T½ (138 days), so the activity decreases by only ≈ 1% ✓ - M2: This expected change is smaller than (or comparable to) the random statistical/Poisson scatter in the count rate, so no trend is resolvable ✓ ### Part (d) [1 mark] — Evaluate (position + supporting + limiting per §4.4.1) - M1: Assumption is reasonable — in the evacuated chamber there are no air molecules to scatter or decelerate the alpha particle so it travels at essentially constant speed; however the assumption would fail if significant external fields or residual gas were present, though over only 4.50 cm and given the alpha's high KE (~5 MeV) gravitational deflection and any residual-gas energy loss are negligible ✓ ### Marker notes - (b)(i) Accept T½ in days throughout if λ then converted: λ = 5.02 × 10⁻³ day⁻¹ → must convert A to day⁻¹ (3.63 × 10⁹ day⁻¹) before dividing; final N must still be ≈ 7.2 × 10¹¹. - (b)(ii) Accept KE in range 4.85–4.95 MeV (consistent with Po-210 alpha energy of 5.3 MeV — the slight discrepancy is the recoil energy of the daughter nucleus, not penalised). - (c) Accept quantitative estimate: ΔA/A₀ = 1 − exp(−λ × 48 × 3600) ≈ 1.0 × 10⁻² (≈ 1%) as equivalent to M1. - (d) Accept any of the following as the limiting consideration: residual gas in imperfect vacuum, gravitational deflection, stray electric/magnetic fields, alpha straggling at low energies — provided each is paired with a justification that it is negligible in this setup.