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

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

Codex  0 2026-10-18

The two processes are electron capture and beta-plus decay:

_{19}^{40} K + e^{-} \to_{18}^{40} Ar + ν_{e},_{19}^{40} K \to_{18}^{40} Ar + e^{+} + ν_{e} .

(1)

For neutral atomic masses, their available total energies obey

Q_{EC} = M_{a} (K) - M_{a} (Ar), Q_{β^{+}} = M_{a} (K) - M_{a} (Ar) - 2 m_{e} .

(2)

Hence electron capture has approximately $2 m_{e} c^{2} = 1.022 MeV$ more available energy. Electron-hole energies and nuclear recoil give smaller corrections for capture from a particular shell. With the physical mass difference,

Q_{EC} ≃ 1.505 MeV

and the positron kinetic-energy budget is only about

0.483 MeV

Electron capture dominates the argon-producing decays. Energetic advantage alone is not a universal theorem about branching fractions: electron density at the nucleus and nuclear transition matrix elements and selection rules also matter. In this case capture can populate the approximately

1.461 MeV

excited argon level, whereas positron emission cannot; this is an additional reason the physical capture branch is much larger.^[ref]

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