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ID: past-exam-of-the-particle-and-nuclear-physics-course-of-the-university-of-cambridge/2022/section-b/4/d/ii/solution

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Past exam of the particle and nuclear physics course of the University of Cambridge / 2022 / Section B / 4 / d / ii / Solution by

Codex  0 2026-10-18

Let

K

be the common normalization converting the quoted

N_{i}

to photon numbers per parent decay. Apply nuclear level population balance to scheme A, starting with the top level:

Γ_{3} Γ_{2} + K N_{2} Γ_{1} + K (N_{3} + N_{5}) = K (N_{2} + N_{5} + N_{6}), = K (N_{3} + N_{4}), = K N_{1} .

(1)

For example, the

3 \to 2

transition contributes

K N_{2}

to the population of level 2; it must be subtracted from the total leaving level 2 to obtain its direct beta feeding. Similarly, a photon in the strong

1 \to 0

line need not imply a direct beta transition to level 1.

The three equations give

Γ_{1} = K (N_{1} - N_{3} - N_{5}), Γ_{2} = K (N_{3} + N_{4} - N_{2}), Γ_{3} = K (N_{2} + N_{5} + N_{6}) .

(2)

Adding them cancels the internal cascade counts:

1 - Γ_{0} = K (N_{1} + N_{4} + N_{6}), K = \frac{1 - Γ _{0}}{N _{1} + N _{4} + N _{6}} .

(3)

This also has a simple physical meaning: each decay that initially feeds an excited daughter eventually contributes exactly one of the three transitions into the ground state. The number of all emitted photons is not the correct normalization, because a single cascade can emit several.

Consequently the required formulae are

Γ_{1} Γ_{2} Γ_{3} = (1 - Γ_{0}) \frac{N _{1} - N _{3} - N _{5}}{N _{1} + N _{4} + N _{6}}, = (1 - Γ_{0}) \frac{N _{3} + N _{4} - N _{2}}{N _{1} + N _{4} + N _{6}}, = (1 - Γ_{0}) \frac{N _{2} + N _{5} + N _{6}}{N _{1} + N _{4} + N _{6}} .

(4)

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