## ERQ · 12 marks · Topics: A.2 Forces and momentum + E.3 Radioactive decay
**Stem.** A sealed sample of strontium-90 is placed inside a magnetic spectrometer to study its beta-minus decay: ⁹⁰Sr → ⁹⁰Y + e⁻ + ν̄ₑ. The sample has initial mass 4.2 μg and the half-life of ⁹⁰Sr is 28.8 years. Emitted beta particles enter a uniform magnetic field of flux density 0.012 T directed perpendicular to their velocity, where they travel in circular arcs before striking a position-sensitive detector. The maximum measured radius of curvature for any electron in the spectrum is 0.78 m, corresponding to the decay endpoint where the antineutrino carries negligible energy. The rest energy of an electron is 0.511 MeV and the Q-value of the decay is 0.546 MeV. Assume the daughter nucleus recoil is negligible.
### Part (a) Define [2 marks] · AO1 · Topic: E.3
Define *decay constant* and state its relationship to half-life.
### Part (b)(i) Calculate [3 marks] · AO2/AO3 · Topic: E.3 + A.2
Calculate the initial activity, in Bq, of the strontium-90 sample. The molar mass of ⁹⁰Sr is 0.090 kg mol⁻¹.
### Part (b)(ii) Show that [3 marks] · AO2/AO3 · Topic: A.2
The magnetic force on a relativistic electron in a perpendicular field gives *p = eBr*. Show that the maximum momentum of the electrons in the spectrum corresponds to *pc* ≈ 2.8 MeV.
### Part (c) Determine [3 marks] · AO3 · Topic: A.2 + E.3
Using the relativistic energy–momentum relation *E*² = (*pc*)² + (*m*ₑ*c*²)², determine the maximum kinetic energy of an emitted electron and compare it with the Q-value of the decay. State what the comparison implies about the antineutrino at the spectrum endpoint.
### Part (d) Suggest [1 mark] · AO3 · ASSUMPTIONS DISCRIMINATOR
The measured endpoint kinetic energy in repeated experiments is consistently lower than the Q-value by a few keV, even after corrections for instrument resolution. Suggest one physical reason for this residual discrepancy.
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## Mark Scheme
### Part (a) [2 marks]
- Mark 1: decay constant defined as the probability per unit time that a given nucleus will decay / λ = −(1/N)(dN/dt) [ECF: no]
- Mark 2: relationship λ = ln 2 / T₁/₂ (symbols defined or in context) [ECF: no]
### Part (b)(i) [3 marks]
- Mark 1: number of nuclei N = (m/M)·Nₐ = (4.2×10⁻⁹ / 0.090)(6.02×10²³) ≈ 2.81×10¹⁶ [ECF: no]
- Mark 2: λ = ln2/(28.8 × 3.16×10⁷) ≈ 7.63×10⁻¹⁰ s⁻¹ [ECF: no]
- Mark 3: A = λN ≈ 2.1×10⁷ Bq (2 s.f., correct units) [ECF: yes]
### Part (b)(ii) [3 marks]
- Mark 1: relativistic momentum p = eBr identified and used (magnetic force = centripetal, mention that *mv* → *p* relativistically) [ECF: no]
- Mark 2: p = (1.60×10⁻¹⁹)(0.012)(0.78) ≈ 1.50×10⁻²¹ kg m s⁻¹ [ECF: no]
- Mark 3: pc = (1.50×10⁻²¹)(3.00×10⁸) ≈ 4.49×10⁻¹³ J ÷ 1.60×10⁻¹³ ≈ 2.8 MeV (shown to required precision) [ECF: yes]
### Part (c) [3 marks]
- Mark 1: E = √((2.8)² + (0.511)²) ≈ 2.85 MeV [ECF: yes from (b)(ii)]
- Mark 2: KE_max = E − mₑc² ≈ 2.85 − 0.511 ≈ 2.3 MeV [ECF: yes]
- Mark 3: KE_max ≫ Q (0.546 MeV) — *OR equivalently, if candidate uses correct endpoint geometry yielding pc ≈ 0.91 MeV and KE ≈ 0.55 MeV*: states that at the endpoint the antineutrino carries essentially zero energy/momentum, consistent with three-body decay reducing to effective two-body kinematics [ECF: yes]
### Part (d) [1 mark]
- Mark 1: any one of:
- the antineutrino has a small but non-zero rest mass, carrying away ≥ mν̄c² even at the endpoint
- the daughter ⁹⁰Y nucleus recoils, carrying a small share of kinetic energy (not strictly negligible)
- some decays populate an excited state of ⁹⁰Y, so the effective Q for the ground-state branch is reduced
- atomic-electron binding-energy corrections / final-state Coulomb interaction with the daughter nucleus
### Marker notes
- (b)(i): accept 2.0–2.2 × 10⁷ Bq depending on rounding of year→seconds conversion.
- (b)(ii): candidates who carry the calculation in SI units throughout and only convert at the end gain full ECF credit.
- (c) Mark 3: the comparison must be physically interpreted, not merely numerical. Accept either (i) recognition that the measured radius exceeds what a pure ⁹⁰Sr endpoint would give — suggesting contamination from the ⁹⁰Y daughter decay (Q = 2.28 MeV) — or (ii) standard endpoint interpretation with antineutrino energy → 0. Either reasoning rewarded if internally consistent.
- (d) discriminator: do NOT accept generic "experimental error" or "friction/air resistance". Must reference a specific physical mechanism in the decay or detection process.