Past exam of the particle and nuclear physics course of the University of Cambridge / 2015 / Paper 1 / 4 / iii

Read the spin and parity from the actual diagram: N and Ne have $1/2^{-}$ ground states; O and F have $5/2^{+}$ ground states. The N→O beta-minus branch and Ne→F beta-plus branch therefore both have $|\Delta J|=2$ with a parity change: both are first-forbidden unique. The F→O beta-plus branch has $5/2^{+}\to 5/2^{+}$: it is allowed and can contain both Fermi and Gamow-Teller contributions.

The Ne ground-state branch is $0.2\%=0.002$, as explicitly printed on the diagram, not $0.2$ as a probability. The prose leaves off its percent sign. Interpret the supplied “lifetimes” as mean lifetimes $\tau$, so each partial decay rate is ${\lambda }_j=B_j/\tau$, with no extra $\ln 2$. ThenA decay width in energy units is ${\Gamma }_j=\hslash {\lambda }_j$, rather than a rate. Thus the corresponding partial energy widths, in the same order N, F, Ne, are $1.73\times 10^{-24}$, $6.93\times 10^{-24}$ and $8.78\times 10^{-24}\mathrm{MeV}$. If a different convention used the supplied numbers as half-lives, all these rates and widths would acquire $\ln 2$; their ratios would be unchanged. The literal mean-lifetime convention is used throughout.