Beta decay, phase space and the Kurie plot

Work in atomic masses at fixed A and the electron bookkeeping does itself. For β⁻ the parent atom carries Z electrons and the daughter atom needs Z + 1, which is exactly the electron that was emitted, so Q_β⁻ = [M(A, Z) − M(A, Z+1)]c² needs no correction at all. For β⁺ the daughter needs one electron fewer while a positron also leaves, so two electron masses must be created: Q_β⁺ = Q_EC − 2m_ec², a fixed toll of 1022.0 keV. Electron capture pays no toll, because the electron comes from the atom's own K or L shell; only its binding energy Bₙ is deducted, leaving the neutrino Q_EC − Bₙ. ⁶⁴Cu shows all three at once: Q_β⁻ = 579.4 keV to ⁶⁴Zn, Q_EC = 1675.0 keV to ⁶⁴Ni, and therefore Q_β⁺ = 653.0 keV — three open channels taking 38.5%, 43.9% and 17.6% of decays. ⁷Be shows the toll biting: Q_EC = 861.9 keV is positive, Q_β⁺ = −160.1 keV is not, so it captures and never emits a positron.

Fermi's golden rule, λ = (2π/ħ)|M_fi|²ρ(E_f), puts everything energy-dependent into ρ. Two light particles share the released energy, so normalise both in a box: the electron has V4πp²dp/(2πħ)³ states in dp and the antineutrino V4πq²dq/(2πħ)³ in dq. For a massless neutrino E_ν = qc, so q = (Q − Tₑ)/c and dq/dE_f = 1/c. Multiply the two densities and dλ/dp ∝ |M_fi|² F(Z, p) p² (Q − Tₑ)². Read that as a competition: p² wants the electron fast, (Q − Tₑ)² wants it slow, so the product peaks between and vanishes at both ends. For ³²P, with Q = 1711 keV, the peak sits near 690 keV and the mean at 695 keV. Nothing nuclear enters the shape — which is why a continuous spectrum set against a fixed nuclear mass difference could only mean a third particle was carrying the balance away unseen.

The golden rule wants the lepton wavefunction evaluated at the nucleus, and a charged particle leaving a charge Z_d does not have a plane wave's amplitude there. The correction is the Fermi function, the ratio of the Coulomb-distorted density at the origin to the plane-wave density; its non-relativistic form is F ≈ 2πη/(1 − e^(−2πη)) with η = Z_d α/β, positive for β⁻ and negative for β⁺. Take a 200 keV lepton and a daughter with Z_d = 16: pc = √(200² + 2 × 511.0 × 200) = 494.4 keV, E = 711.0 keV, β = 0.695, so η = ±0.168 and 2πη = ±1.055. Then F₋ = 1.055/(1 − e^(−1.055)) = 1.62 while F₊ = −1.055/(1 − e^(+1.055)) = 0.56, a factor of 2.9 between the two modes at the same energy. That is why β⁻ spectra pile up at low energy and β⁺ spectra are pushed towards the endpoint, and why F must come out before any claim about shape is made.

Expand the lepton plane waves across the nucleus: e^(ik⋅r) = 1 + ik⋅r + …, with k = p/ħ. A 1 MeV electron has pc = 1422 keV, so k = 0.0072 fm⁻¹, and a mass-32 nucleus has R = 1.2A¹⁄³ = 3.8 fm, giving kR = 0.0275 and (kR)² ≈ 8 × 10⁻⁴ in rate. Keep only the leading 1 and just two operators survive between nuclear states: the Fermi operator Σₖ t±(k), carrying no angular momentum, and the Gamow-Teller operator Σₖ σ(k)t±(k), carrying one unit. The selection rules follow at once. Fermi: ΔJ = 0, no parity change, connecting isobaric analogue states. Gamow-Teller: ΔJ = 0 or 1 with no parity change, but 0 → 0 excluded, because σ cannot connect two J = 0 states. So 0⁺ → 0⁺ is pure Fermi, ΔJ = 1 is pure Gamow-Teller, and everything between is a mixture. Kill the leading term with a parity change, ΔJ up to 2, and the next term ik⋅r must carry the transition: the rate falls by roughly that (kR)², first forbidden. ΔJ = 2 with no parity change needs one term more and pays (kR)⁴, second forbidden. That is the whole meaning of forbidden.

A half-life alone cannot compare two transitions, because most of the variation between them is phase space. Integrate the spectrum instead: f(Z, W₀) = ∫ F(Z, W) W p (W₀ − W)² dW from W = 1 to W₀, in units of m_ec², a dimensionless number growing roughly as W₀⁵ once W₀ ≫ 1. The product f t½ then isolates the matrix element, and it is quoted as log₁₀(f t½) with t½ in seconds because it spans ten orders of magnitude. Superallowed 0⁺ → 0⁺ Fermi decays cluster at log ft = 3.49, f t½ = 3072 s, and the tightness of that cluster across many nuclei is one of the cleanest tests of the weak vector coupling. Normal allowed decays sit at 4.5–6, first forbidden at 6–9. ³²P and ¹⁴C make the point: their half-lives differ by a factor of 1.5 × 10⁵, but their log ft values are 7.9 and 9.04, so phase space supplies about 10⁴ of that and nuclear structure barely more than ten.

Invert the shape. If dλ/dp ∝ F(Z, p) p² (Q − Tₑ)², then K = √[N(p)/(p²F(Z, p))] is proportional to Q − Tₑ: a straight line falling to zero at the endpoint. Two things make the transform worth doing. Counts near the endpoint are scarce and noisy, so fitting a line through the whole spectrum and extrapolating beats hunting for the last count; and any departure from straightness becomes diagnostic instead of hiding inside a bell curve. Curvature at low energy usually means a thick source, with electrons losing energy on the way out, or a second branch to an excited daughter level filling in underneath. Curvature across the whole plot means the decay is not allowed and a shape factor S(p, q) belongs in the denominator — ²¹⁰Bi is the classic first-forbidden case. A finite neutrino mass looks different again: K² ∝ (Q − Tₑ)√[(Q − Tₑ)² − (m_νc²)²], so the plot leaves the line only within the last few m_νc² and meets the axis vertically at Q − m_νc².