Generated ERQ

✓ passed A.2 Forces and momentum × E.3 Radioactive decay 12 marks HL 3 passes 127.54s $0.8199
``` ## ERQ · 12 marks · Topics: A.2 Forces and momentum + E.3 Radioactive decay · Archetype: theory_application **Integration:** primary=A.2 Forces and momentum, secondary=E.3 Radioactive decay (strength: co_equal) **Stem.** A laboratory source contains a small mass of ²²⁶Ra, which alpha-decays to ²²²Rn with Q-value 4.87 MeV and half-life 1.60 × 10³ years. A silicon surface-barrier detector counts the emitted alpha particles, while a separate recoil-ion detector records the daughter nuclei. At the start of an experiment the measured activity of the source is 3.70 × 10⁵ Bq. Atomic masses: m(²²⁶Ra) = 226.025 u, m(²²²Rn) = 222.018 u, m(⁴He) = 4.0026 u; 1 u = 1.66 × 10⁻²⁷ kg = 931.5 MeV c⁻². The parent nuclei may be assumed stationary in the laboratory frame. ### Part (a) State [2 marks] · AO1 · Topic: A.2 State the law of conservation of linear momentum and state the condition under which it applies. ### Part (b)(i) Show that [3 marks] · AO2 · Topic: E.3 + A.2 Using the decay equation ²²⁶Ra → ²²²Rn + ⁴He and the atomic masses above, show that the Q-value of the decay is approximately 4.87 MeV. ### Part (b)(ii) Calculate [3 marks] · AO2 · Topic: A.2 Assuming the alpha particle carries kinetic energy equal to 4.78 MeV, calculate the recoil speed of the ²²²Rn nucleus immediately after the decay. Use non-relativistic mechanics. ### Part (c) Determine [3 marks] · AO3 · Topic: E.3 + A.2 The detector resolves the alpha energy to ±15 keV. The decay actually has two alpha branches: α₁ (to the ²²²Rn ground state, Q₁ = 4.87 MeV) and α₂ (to an excited state, Q₂ = 4.68 MeV). After 4.80 × 10³ years of operation, determine the activity of the source AND the number of α₁ events detected during a 1.00 hour counting interval, given that 94.5% of decays follow the α₁ branch and the detector subtends a solid-angle fraction of 1.20 × 10⁻³ of 4π sr. ### Part (d) Suggest [1 mark] · AO3 · ASSUMPTIONS DISCRIMINATOR Suggest one physical reason why the recoil ²²²Rn nucleus may NOT, in practice, be detected with the kinetic energy predicted in part (b)(ii), even when the alpha energy is measured exactly. --- ## Mark Scheme ### Part (a) [2 marks] — State - M1: The total linear momentum of a system is constant / remains the same before and after an interaction [no ECF] - M2: Provided no (net) external force acts on the system / system is isolated [no ECF] ### Part (b)(i) [3 marks] — Show that - M1: Mass defect Δm = 226.025 − (222.018 + 4.0026) = 4.4 × 10⁻³ u (accept 0.0044 u) [no ECF] - M2: Q = Δm × 931.5 MeV — substitution shown [no ECF] - M3: Q ≈ 4.87 MeV (student value 4.84–4.90 MeV acceptable; target given to 3 sf) [no ECF] ### Part (b)(ii) [3 marks] — Calculate - M1: Apply conservation of momentum: p(Rn) = p(α), and use E_k(α) = p²/(2m_α) to obtain p(α) = √(2 m_α E_k(α)); substitution with m_α = 4.0026 × 1.66 × 10⁻²⁷ kg and E_k = 4.78 × 1.60 × 10⁻¹³ J [no ECF] - M2: p(α) ≈ 1.01 × 10⁻¹⁹ kg m s⁻¹ (accept 1.00–1.02 × 10⁻¹⁹) [ECF from M1] - M3: v(Rn) = p / m(Rn) ≈ 1.01 × 10⁻¹⁹ / (222.018 × 1.66 × 10⁻²⁷) ≈ 2.74 × 10⁵ m s⁻¹ [ECF from M2] ### Part (c) [3 marks] — Determine (E.3 + A.2 co-equal) - M1: Decay constant λ = ln2 / T₁/₂ = ln2 / (1.60 × 10³ × 3.156 × 10⁷) = 1.37 × 10⁻¹¹ s⁻¹; OR equivalently use t/T₁/₂ = 3.00 [no ECF] - M2: A = A₀ e^(−λt) = 3.70 × 10⁵ × 2^(−3.00) = 4.63 × 10⁴ Bq (accept 4.6 × 10⁴ Bq) [no ECF] - M3: N(α₁ detected) = A × 0.945 × (1.20 × 10⁻³) × 3600 ≈ 1.89 × 10⁵ counts (accept 1.8–1.9 × 10⁵) [ECF from M2] ### Part (d) [1 mark] — Suggest (proposal + physics justification per §4.4.1) - M1: ONE valid proposal with physics-based reason, e.g.: the recoil nucleus loses kinetic energy by ionising/scattering within the source material (self-absorption) before reaching the recoil detector, because its range in solid matter is only tens of nm at ~100 keV; OR the parent ²²⁶Ra is bound in a crystal lattice, so momentum is partly transferred to neighbouring atoms rather than to a free ²²²Rn nucleus; OR a fraction of decays populate the excited ²²²Rn* state (α₂ branch), so the recoil energy is shared with a subsequent γ photon and is lower than the value calculated for the ground-state branch. ### Marker notes - (b)(i) Show-that target: 4.87 MeV given to 3 sf; student-derived 4.84–4.90 MeV acceptable per §4.5 - (b)(ii) Alternative method: use m_α v_α = m_Rn v_Rn with v_α from ½ m_α v_α² = E_k directly → v_Rn = (m_α/m_Rn) v_α; full marks if executed correctly - (c) Accept use of (½)³ = 0.125 in place of exponential since t = 3 T₁/₂ exactly - (d) Accept any ONE physically valid limitation with mechanism stated; bare "relativistic effects" without quantitative or causal reasoning is NOT sufficient (γ ≈ 1.00004 at v_α ~ 1.5 × 10⁷ m s⁻¹, so relativity is negligible here and should be rejected as the primary limitation) - ECF chain: (b)(ii) M2→M3; (c) M2→M3; (b)(i) is no-ECF (Show-that) ```