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

Beta decay converts a neutron to a proton or vice versa through the weak interaction, leaving the mass number fixed:Inside a nucleus the full initial and final nuclear masses determine whether either process is possible. Electron capture, $p+e^{-}\to n+{\nu }_e$, is the related bound-electron process. The emitted charged-lepton spectrum is continuous because energy is shared with the neutrino and nuclear recoil.

For allowed beta decay, the lepton wavefunctions' leading spatial terms are constant over the nucleus. A Fermi transition has $\Delta J=0$ and no parity change. A Gamow-Teller transition has $\Delta J=0,1$ and no parity change, excluding $0\to 0$. Thus a same-parity transition with nonzero initial and final spins and $\Delta J=0$ can contain both contributions. First-forbidden transitions change parity and permit $|\Delta J|=0,1,2$; the $|\Delta J|=2$ case is unique forbidden beta decay. Higher forbiddenness requires higher powers of the small lepton momentum times nuclear size. “Forbidden” denotes suppression of the leading allowed operator, not an impossible decay.^[ref]

Fermi's golden rule expresses a rate as a squared nuclear matrix element times leptonic phase space. Neglecting charged-lepton mass and Coulomb distortion, a constant allowed matrix element givesThis is the approximate Sargent rule, $\lambda \propto Q^5$ for comparable allowed strengths; it is not an exact law across arbitrary transitions. Finite electron mass, the Fermi function, forbidden shape factors and nuclear structure change the comparison. A microscopic first-forbidden unique operator introduces additional momentum factors.

Using bare nuclear masses $M_i,M_f$, the kinetic-energy release isThe daughter is different in the two cases: its charge is $Z+1$ for beta minus and $Z-1$ for beta plus. The relevant mass difference must exceed one electron mass. For neutral atomic masses, to leading order in small electronic binding corrections, the corresponding expressions are $Q_{{\beta }^{-}}=(M_i^{\mathrm{at}}-M_f^{\mathrm{at}})c^2$ and $Q_{{\beta }^{+}}=(M_i^{\mathrm{at}}-M_f^{\mathrm{at}}-2m_e)c^2$. Electron capture is energetically possible approximately when $M_i^{\mathrm{at}}>M_f^{\mathrm{at}}$, with available shell binding energy and daughter excitation included. Nuclear masses and atomic masses must not be mixed.

The mirror nuclei have interchanged neutron and proton numbers. Neglect their strong-interaction binding difference and describe them with the same uniformly charged sharp sphere of radius $R$. The nuclear Coulomb energy isChanging one neutron into a proton decreases the nucleon rest-mass sum by $m_n-m_p=1.294\mathrm{MeV}/c^2$, but raises the Coulomb energy. The diagram gives $(M_F-M_O)c^2=2.25\mathrm{MeV}$ for nuclear masses, soWith the standard electromagnetic conversion $\alpha \hslash c\simeq 1.44\mathrm{MeV}\mathrm{fm}$, the uniform-sphere Coulomb displacement energy estimate isThis $R$ is the uniform sphere's outer radius. If “charge radius” means the rms nuclear charge radius, the same distribution givesThe numerical result is an estimate within the charge-density model, not a model-independent radius measurement. A discrete-proton correction replaces $Z^2$ by $Z(Z-1)$, giving $\Delta [Z(Z-1)]=16$ and $R\simeq 3.90\mathrm{fm}$. The main answer above follows the literal continuum charged-sphere model. Strong isospin-breaking, nonuniform density and changes in the valence-nucleon distribution also limit the extraction.

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.

The bracketed diagram energies are nuclear, so subtract one electron rest energy from each parent-to-daughter energy difference. The three kinetic-energy endpoints areThe observed partial-rate ratios are ${\lambda }_N/{\lambda }_{Ne}=0.19672$ and ${\lambda }_N/{\lambda }_F=0.24918$.

To obtain the simple Sargent-law estimates, define an effective strength by $\lambda =C|M_{\mathrm{eff}}{|}^2{Q}^5$, using the same normalization $C$. ThenThus the two forbidden branches have effective strengths of the same broad order, whereas the N ground-state branch is highly suppressed relative to the allowed F branch after this approximate fifth-power kinematic scaling. The first comparison involves mirror transitions with related nuclear configurations; order-of-magnitude similarity is reasonable, without expecting exact equality from this estimate.

These effective strengths must not be confused with precisely extracted microscopic reduced nuclear matrix elements. Both N and Ne ground-state branches are first-forbidden unique, for which the first-forbidden unique beta-decay phase space contains an additional factor proportional to $p_e^2+E_{\nu }^2$. In the massless, no-Coulomb limit this gives $Q^7$, so using the same microscopic unique-forbidden normalization for those two branches instead yieldsComparing a unique-forbidden reduced operator directly with an allowed reduced operator also requires a convention for its powers of length and momentum. The number $8.05\times 10^{-5}$ is consequently an effective Sargent-normalized suppression, not a universal dimensionless ratio between different operator ranks. Electron-mass and opposite-sign Coulomb corrections matter, particularly for the low-energy F branch. The data justify approximate strength comparisons; they do not fix precision nuclear matrix elements without these extra ingredients.

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 $2.25+11.20=13.45\mathrm{MeV}$. HenceIt is a same-spin, same-parity allowed beta decay between isobaric analog states. The large analogue Fermi overlap makes it a 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 $T=T_z^{\mathrm{magnitude}}=3/2$ 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 givesThis 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 $S=g_V^2|M_F{|}^2+g_A^2|M_{GT}{|}^2$, thenThe 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.