A parent nucleus initially at rest undergoes β⁻-decay. In one event, the emitted electron has momentum magnitude 1.2 MeV·c⁻¹ and the antineutrino has momentum magnitude 0.9 MeV·c⁻¹. The angle between their momenta is 60°. If the daughter recoil momentum satisfies p(D)² = [N ÷ 100](MeV·c⁻¹)², find N.
Physics · Nuclear physics · radioactivity and law of decay
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Question

A parent nucleus initially at rest undergoes β⁻-decay. In one event, the emitted electron has momentum magnitude 1.2 MeV·c⁻¹ and the antineutrino has momentum magnitude 0.9 MeV·c⁻¹. The angle between their momenta is 60°. If the daughter recoil momentum satisfies p(D)² = [N ÷ 100](MeV·c⁻¹)², find N.

Physics · Nuclear physics · radioactivity and law of decay

G1 Physics · Accepted Answer

Momentum conservation for a parent initially at rest gives p⃗(D) = −[p⃗(e)+p⃗(ν̄)]. Therefore the daughter recoil momentum magnitude equals the magnitude of the vector sum of the electron and antineutrino momenta. Using the cosine rule: p(D)² = p(e)² + p(ν̄)² + 2p(e)p(ν̄)cos60°. Substitute: p(D)² = 1.2² + 0.9² + 2(1.2)(0.9)(0.5). = 1.44 + 0.81 + 1.08 = 3.33 (MeV·c⁻¹)². Write 3.33 = 333 ÷ 100. Hence N = 333.