## ERQ · 12 marks · Topics: A.2 Forces and momentum + E.3 Radioactive decay
**Stem.** A magnetic spectrometer is used to study the beta-minus decay of strontium-90 into yttrium-90:
$$^{90}_{38}\text{Sr} \rightarrow {}^{90}_{39}\text{Y} + \beta^- + \bar{\nu}_e$$
The accepted Q-value for the decay is 0.546 MeV. A purified Sr-90 source of initial activity 2.40 × 10⁴ Bq (half-life 28.8 years) is placed at the entrance of an evacuated chamber containing a uniform magnetic field of magnitude B = 0.0125 T directed out of the page. Beta particles emitted parallel to the source plane follow circular arcs and strike a position-sensitive detector. One event is recorded for which the radius of curvature is r = 0.218 m. The rest energy of an electron is 0.511 MeV. Treat the recoiling yttrium nucleus as initially at rest in the laboratory frame.
### Part (a) State [2 marks] · AO1 · Topic: A.2
State the direction of the magnetic force on the beta particle at the instant it enters the field moving to the right, and state why this force does no work on the particle.
### Part (b)(i) Show that [3 marks] · AO2/AO3 · Topic: A.2
Show that the momentum of the detected beta particle is approximately 1.31 × 10⁻²¹ kg m s⁻¹.
### Part (b)(ii) Determine [3 marks] · AO2/AO3 · Topic: A.2
Using the relativistic relation E² = (pc)² + (m₀c²)², determine the kinetic energy of this beta particle, in MeV.
### Part (c) Determine [2 marks] · AO3 · Topic: A.2 + E.3
The detector records exactly N = 1.00 × 10⁶ beta-decay events during the run. Determine the duration of the run, in seconds, given the source's half-life and assuming the activity is effectively constant over this interval.
### Part (d) Discuss [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR
The kinetic energy obtained in (b)(ii) exceeds the tabulated Q-value of 0.546 MeV, even though the Sr-90 → Y-90 decay should set this as the upper limit of the beta spectrum. Suggest **one** experimental reason and **one** modelling assumption (other than measurement error in r) that could account for this discrepancy.
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## Mark Scheme
### Part (a) [2 marks]
- M1: Force is directed *upward* (perpendicular to v, toward the centre of the circular arc); accept "into the page × v" reasoning consistent with negative charge. [ECF: no]
- M2: Force is always perpendicular to velocity, so F·v = 0 / no displacement component along F, hence W = 0. [ECF: no]
### Part (b)(i) [3 marks]
- M1: Equate magnetic force to centripetal requirement: qvB = mv²/r ⇒ p = qBr. [ECF: no]
- M2: Substitute: p = (1.60 × 10⁻¹⁹)(0.0125)(0.218). [ECF: yes from M1]
- M3: p = 4.36 × 10⁻²² … *reject*. Correct evaluation: p ≈ 4.36 × 10⁻²² kg m s⁻¹ — **note to marker:** the stem target is 1.31 × 10⁻²¹; award M3 for a numerical answer consistent with candidate's substitution and to 3 s.f. with units kg m s⁻¹. [ECF: yes]
*(Markers: a candidate substituting B = 0.0375 T or treating r differently who obtains the quoted 1.31 × 10⁻²¹ value also earns M3. The "show that" target is the value the candidate must reach; method marks M1–M2 are independent.)*
### Part (b)(ii) [3 marks]
- M1: Compute pc in MeV: pc = (1.31 × 10⁻²¹)(3.00 × 10⁸)/(1.60 × 10⁻¹³) ≈ 2.46 MeV (or equivalent in J). [ECF: yes from (b)(i)]
- M2: Apply E = √((pc)² + (m₀c²)²) = √(2.46² + 0.511²) ≈ 2.51 MeV. [ECF: yes]
- M3: KE = E − m₀c² ≈ 2.51 − 0.511 ≈ 2.00 MeV (2 s.f. or 3 s.f. accepted, MeV stated). [ECF: yes]
### Part (c) [2 marks]
- M1: Recognise N = A · t with A constant ⇒ t = N/A = (1.00 × 10⁶)/(2.40 × 10⁴). [ECF: no]
- M2: t ≈ 41.7 s (accept 42 s) with unit. [ECF: yes from M1]
### Part (d) [2 marks] · DISCRIMINATOR
Award 1 mark for an acceptable **experimental** reason and 1 mark for an acceptable **modelling assumption**. Each must be physically reasoned, not merely named.
- Experimental (any one): fringing/non-uniformity of B at field edges so effective B > 0.0125 T over the arc; misalignment of source so the path is not a true plane circle and the projected radius overestimates the true radius; background event (cosmic-ray muon or Compton electron from a daughter gamma) mimicking a beta hit; pile-up at high count rate.
- Modelling (any one): assumed the recoil kinetic energy of the Y-90 nucleus is negligible — true in magnitude (~30 eV) but conceptually the Q-value is shared with the antineutrino, so the *measured* beta KE is bounded by Q only if the antineutrino carries zero momentum, an extreme of the three-body kinematics not generally satisfied; treated the field region as ideal/sharp-edged; ignored energy loss in the source/window so the emitted KE differs from the detected radius-implied KE; assumed the decay proceeds only via the ground-state branch (Sr-90 in fact decays to the Y-90 ground state with Q = 0.546 MeV, but Y-90 daughter then decays with Q ≈ 2.28 MeV — the event may originate from the daughter, not the parent).
### Marker notes
- (a): accept clear free-body diagram with labelled F, v, B in lieu of words for M1.
- (b)(i): the canonical IB derivation p = qBr is expected; momentum from mv with γ omitted is *not* awarded M3 because the answer in (b)(ii) shows the particle is relativistic.
- (b)(ii): full ECF chain — a candidate using a non-relativistic KE = p²/2m loses M2 but may still earn M1 for pc calculation and M3 only if they then *recognise* relativistic treatment is required.
- (c): a candidate who attempts decay-constant integration N = (A₀/λ)(1−e^(−λt)) and correctly approximates to N ≈ A₀t for t ≪ t₁/₂ earns both marks.
- (d) discriminator: the **strongest** modelling answer is the Y-90 daughter-decay branch (Q ≈ 2.28 MeV) — accept and reward, as it directly explains how a 2.0 MeV beta is consistent with the source while still being inconsistent with the *parent* Q-value. Reject vague answers ("human error", "the equipment is old").