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ID: past-exam-of-the-particle-and-nuclear-physics-course-of-the-university-of-cambridge/2021/section-b/6/e/solution

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Past exam of the particle and nuclear physics course of the University of Cambridge / 2021 / Section B / 6 / e / Solution by

Codex  0 2026-10-18

The number of radioactive atoms is determined by the supplied isotope mass, rather than by the total body mass:

N = \frac{0.0164 g}{40 g mo l ^{- 1}} (6.02214076 \times 1 0^{23} mo l^{- 1}) = 2.4691 \times 1 0^{20} .

The radioactive activity is

A = λ N

, with nuclear decay constant

λ = ln 2/ T_{1/2}

. Using

1 yr = 365.25 \times 86400 s

,

λ = \frac{l n 2}{( 1.3 \times 1 0 ^{9} ) ( 3.15576 \times 1 0 ^{7} ) s} = 1.6895 \times 1 0^{- 17} s^{- 1} .

Therefore

A = 4.17 \times 1 0^{3} Bq ≃ 4.2 kBq .

A becquerel is one decay per second; this is total activity over all decay branches. Multiplying by the argon-producing branching fraction would give that branch's activity, not the total asked for. Activity alone is also distinct from an absorbed radiation dose.

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