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A first-forbidden unique beta decay has $|\Delta J|=2$ with a parity change. Its leading lepton shape factor contains two additional powers of momentum relative to an allowed beta decay. In the massless, no-Coulomb limit, ${\int }_0^Q{E}^2(Q-E{)}^2[E^2+(Q-E{)}^2]dE=2Q^7/105$. Thus its microscopic reduced nuclear matrix element multiplies a seventh-power phase-space factor, rather than the fifth power in the Sargent rule. A fifth-power comparison across different forbiddenness classes instead extracts convention-dependent effective strengths.

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 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.

Assume allowed beta decay, so the leading matrix element has negligible lepton-momentum dependence. Neglect nuclear recoil and neutrino mass. Let $Q$ be the kinetic energy available to the electron and antineutrino, $E_0=m_e+Q$, and $E=\sqrt{p^2+m_e^2}$. The neutrino energy and momentum magnitude are $k=E_0-E$.

Fermi's golden rule weights accessible final states. After angular integration, the momentum-space factors are $p^{2}dp$ and $k^{2}dk$. The energy-conserving Dirac delta function removes the neutrino integral:The constant includes the allowed nuclear matrix element, weak-coupling and angular factors. Including the usual Fermi function for electron-nucleus Coulomb distortion givesFor the kinetic-energy spectrum, $dp/dT=E/p$ instead yieldsThe two distributions differ by a Jacobian; forgetting it changes the shape.

A pure fifth-power law additionally neglects electron mass and Coulomb distortion and keeps the nuclear matrix element fixed. Then $E=p$, $E_0\simeq Q$, andTwo momentum powers come from each final particle and one from the remaining integration. This is the origin of the Sargent rule. Exact $Q^5$ dependence requires the relativistic-electron, constant-matrix-element limit. For $Q\ll m_e$, the same approximation without Coulomb distortion has $pE\simeq m_e\sqrt{2m_{e}T}$ and instead scales as $m_e^{3/2}{Q}^{7/2}$. Forbidden-transition factors, nuclear-structure changes and Coulomb distortion also modify the simple law.