A-Level · AS · extended response
An unseen particle in a decay event
AS Level foundations · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Measured momentum-vector endpoints
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): py / MeV c⁻¹
Horizontal axis (x): px / MeV c⁻¹
Read the exact data as a table
| px / MeV c⁻¹ | py / MeV c⁻¹ |
|---|---|
| 0 | 0 |
| 3 | 0 |
| 0 | 4 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
A particle initially at rest decays into an electron, a positron and an unseen particle. The electron momentum components are (3,0) MeV/c and the positron components are (0,4) MeV/c; these supplied vectors are represented by points in momentum-component space. For this question the unseen particle is a photon. Use electron rest energy 0.511 MeV and E² = (pc)²+(mc²)². Supplied data (px / MeV c⁻¹; py / MeV c⁻¹): (0, 0), (3, 0), (0, 4).
- (a)
Infer the photon's momentum components and energy, then determine the parent's rest energy.
3 marks - (b)
Check charge and lepton-number conservation for the listed products, and compare leptons with a neutron's quark composition.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Infer the photon's momentum components and energy, then determine the parent's rest energy.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
The parent starts with zero momentum, so the unseen particle must cancel the measured vector sum: pγ = (−3,−4) MeV/c.
- 2
Its momentum magnitude is √(3²+4²) = 5 MeV/c. A photon is massless and has Eγ = pc = 5 MeV.
- 3
Electron energy is √(3²+0.511²)≈3.043 MeV and positron energy is √(4²+0.511²)≈4.033 MeV. Parent rest energy equals their sum plus 5, about 12.08 MeV.
(b)
Check charge and lepton-number conservation for the listed products, and compare leptons with a neutron's quark composition.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Electron charge is −e and lepton number +1; positron charge is +e and lepton number −1. The photon has neither charge nor lepton number, so both listed totals are zero.
- 2
A neutron is a baryon containing one up quark (+2/3 e) and two down quarks (−1/3 e each), giving zero net charge and baryon number one.
- 3
An electron is an elementary lepton, not a constituent quark. These conservation checks constrain a proposed process but do not alone establish that it occurs or determine all selection rules.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- The parent starts with zero momentum, so the unseen particle must cancel the measured vector sum: pγ = (−3,−4) MeV/c.
- Its momentum magnitude is √(3²+4²) = 5 MeV/c. A photon is massless and has Eγ = pc = 5 MeV.
- Electron energy is √(3²+0.511²)≈3.043 MeV and positron energy is √(4²+0.511²)≈4.033 MeV. Parent rest energy equals their sum plus 5, about 12.08 MeV.
- Electron charge is −e and lepton number +1; positron charge is +e and lepton number −1. The photon has neither charge nor lepton number, so both listed totals are zero.
- A neutron is a baryon containing one up quark (+2/3 e) and two down quarks (−1/3 e each), giving zero net charge and baryon number one.
- An electron is an elementary lepton, not a constituent quark. These conservation checks constrain a proposed process but do not alone establish that it occurs or determine all selection rules.
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