= Solution
The state marked X in F is the $1/2^-$ analogue at $11.20\ \mathrm{MeV}$ excitation. Its absolute nuclear <energy> relative to O is $2.25+11.20=13.45\ \mathrm{MeV}$. Hence
$$
\boxed{Q_X=16.26-13.45-0.511=2.299\ \mathrm{MeV}.}
$$
It is a same-spin, same-parity <allowed beta decay> between <isobaric analog states>. The large analogue Fermi overlap makes it a \b[superallowed analogue transition], in the broad analogue-state terminology. Since $J=1/2$ rather than zero, this classification does not exclude a Gamow-Teller contribution.
For a transparent numerical estimate, take the Fermi part to dominate both this branch and the reference F-ground-state decay, neglect Coulomb and electron-mass corrections in their phase-space ratio, and use the <Sargent rule>. For $T_z=(N-Z)/2$, Ne has $T=T_z^{\rm magnitude}=3/2$ and goes from $T_z=-3/2$ to $-1/2$ in its analogue multiplet. The <isospin ladder strength in analogue beta decay> is therefore three. The F→O ground-state mirror transition in $T=1/2$ has Fermi strength one. Thus the <analogue beta-decay branching estimate> gives
$$
\lambda_X\simeq\frac3{95\ \mathrm s}\left(\frac{2.299}{1.739}\right)^5
=0.128\ \mathrm{s^{-1}},\qquad
\boxed{B_X\simeq(0.15\ \mathrm s)\lambda_X=0.0191\simeq1.9\%.}
$$
This percent-level result is an estimate under the stated assumptions, not a unique consequence of the level diagram. More generally, if $f$ denotes an allowed phase-space integral and $S=g_V^2|M_F|^2+g_A^2|M_{GT}|^2$, then
$$
B_X=\frac{\tau_{Ne}}{\tau_F}\frac{f_X}{f_F}\frac{S_X}{S_F}.
$$
The diagram fixes the Fermi strengths but not both axial strengths. Including them and proper <Positron> phase-space functions changes the numerical result. Assuming equal total allowed strengths instead of the Fermi-only ratio three would give approximately $0.64\%$; that alternative makes a different nuclear-structure assumption. The rate enhancement from removing the forbiddenness explains why the X branch can be appreciable despite its much smaller endpoint than the Ne ground-state branch.
Back to article page
