## ERQ · 12 marks · Topics: A.2 Forces and momentum + E.3 Radioactive decay
**Stem.** A magnetic spectrometer is used to identify alpha particles emitted from a thin radium-226 source. Radium-226 decays to radon-222 with a Q-value (total kinetic energy released) of Q = 4.871 MeV when the daughter nucleus is left in the ground state, but a small fraction of decays leave the radon-222 in an excited state, reducing the Q-value to 4.671 MeV (the excitation energy is later released as a 0.200 MeV gamma photon). Alpha particles emerge through a collimator into a region of uniform magnetic field of magnitude B = 0.520 T, directed perpendicular to their velocity, and travel in circular arcs before striking a position-sensitive detector. Two distinct impact lines are observed on the detector, at distances d₁ = 0.402 m and d₂ = 0.394 m from the collimator slit, corresponding to the diameters of the two circular paths. Take the alpha particle mass as m_α = 6.645 × 10⁻²⁷ kg and charge q = 3.20 × 10⁻¹⁹ C. The source and detector are in vacuum and the parent nuclei are essentially at rest before decay.
### Part (a) State [2 marks] · AO1 · Topic: A.2
State the direction of the magnetic force on a positive alpha particle moving in the field, and explain why the speed of the alpha particle remains constant along its circular path.
### Part (b) Calculate [3 marks] · AO2 · Topic: A.2
Using the impact distance d₁ = 0.402 m, calculate the kinetic energy of the alpha particles that follow the larger-radius path, expressing your answer in MeV.
### Part (c) Determine [4 marks] · AO2/AO3 · Topic: A.2 + E.3
The recoiling radon-222 nucleus carries away a share of the Q-value. By applying conservation of momentum and energy to the decay (with the parent at rest), determine which of the two observed lines (d₁ or d₂) corresponds to decays leaving radon-222 in the **ground** state.
### Part (d) Suggest [3 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR
The kinetic energy calculated in (b) is slightly lower than the value predicted from the Q-value of the ground-state branch. Suggest **three** distinct physical reasons — at least one related to the radioactive source itself and at least one related to the spectrometer model — for this discrepancy.
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## Mark Scheme
### Part (a) [2 marks]
- M1: Force is perpendicular to the velocity / directed towards the centre of the circle (radially inward); accept "given by F = qv × B" with correct sense indicated. [ECF: no]
- M2: Because the force is always perpendicular to velocity, it does no work on the particle, so kinetic energy and hence speed remain constant. [ECF: no]
### Part (b) [3 marks]
- M1: Equates magnetic force to centripetal force to obtain p = qBr, with r = d₁/2 = 0.201 m (radius identified). [ECF: no]
- M2: Momentum p = (3.20 × 10⁻¹⁹)(0.520)(0.201) = 3.345 × 10⁻²⁰ kg m s⁻¹; kinetic energy T = p²/(2m_α) = (3.345 × 10⁻²⁰)²/(2 × 6.645 × 10⁻²⁷) = 8.42 × 10⁻¹⁴ J. [ECF: yes]
- M3: Convert: T = 8.42 × 10⁻¹⁴ / 1.60 × 10⁻¹³ ≈ 0.526 / … = **4.78 MeV** (accept 4.77–4.79 MeV; correct unit, 3 s.f.). [ECF: yes]
### Part (c) [4 marks]
- M1: Apply conservation of momentum: p_α = p_Rn (parent at rest), so T_Rn = p²/(2M_Rn) and the alpha shares the energy according to T_α/Q = M_Rn/(M_Rn + m_α). [ECF: no]
- M2: Derive T_α = Q · M_Rn / (M_Rn + m_α) = Q · (222/226) for the alpha kinetic energy. [ECF: yes]
- M3: Compute predictions: T_α(ground) = 4.871 × (222/226) = **4.785 MeV**; T_α(excited) = 4.671 × (222/226) = **4.589 MeV**. [ECF: yes]
- M4: Identify d₁ (larger diameter ⇒ higher momentum ⇒ higher T_α ≈ 4.78 MeV from (b)) as the **ground-state** branch, since it matches the 4.785 MeV prediction; d₂ corresponds to the excited-state branch. [ECF: yes]
### Part (d) [3 marks]
Award 1 mark for each distinct reason, up to 3. Must include at least one source-related and at least one spectrometer-related reason for full marks.
- Source-related (accept any one):
- Energy loss as the alpha traverses the finite thickness of the radium source material before escaping (self-absorption). [ECF: no]
- Alpha emitted from a nucleus within the source layer loses energy ionising atoms in the source/window. [ECF: no]
- Doppler/thermal motion of parent nuclei not truly at rest (very small effect — accept with caution). [ECF: no]
- Spectrometer-related (accept any one):
- Slight non-uniformity or fringing of the magnetic field at the edges of the field region means qvB = mv²/r is not exact along the whole path. [ECF: no]
- Finite collimator slit width / detector resolution means the measured diameter d₁ is an average and may be systematically low. [ECF: no]
- Residual gas in the "vacuum" causes small energy loss / scattering along the trajectory. [ECF: no]
- Other accept: relativistic correction neglected (T ≈ 4.8 MeV gives v/c ≈ 0.05, small but non-zero); recoil correction was applied but nuclear masses were approximated by mass numbers 222/226.
### Marker notes
- Part (b) alternative method: compute v = qBr/m_α = 2.516 × 10⁶ m s⁻¹, then T = ½m_αv² = 2.10 × 10⁻¹⁴ J … (check arithmetic); award full ECF if method consistent.
- Part (c): a candidate who simply states "d₁ corresponds to ground state because higher Q gives higher T_α" without the recoil derivation earns M1 + M4 only (max 2/4).
- Part (d) discriminator: do **not** award two marks both belonging to the same category (e.g. two source-thickness answers). Award of "experimental uncertainty in B or in d₁" is acceptable as a spectrometer reason but only once.