Exam-style question
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A neutron (n) decays into a proton (p), an electron (e⁻) and an electron antineutrino (ν̄ₑ). Which conservation law is satisfied by this process?.
Model answer
What a good answer should say
- The process conserves both lepton number and baryon number.
- The neutron and proton are baryons, so the baryon number remains +1 on both sides.
- The electron carries lepton number +1 and the antineutrino carries lepton number –1, giving a net lepton number of zero, which matches the initial state of zero lepton number.
Explanation
Why this works
The answer demonstrates understanding of lepton and baryon number conservation by correctly accounting for the particles involved and their quantum numbers.
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