University Physics V · Radioactive Decay · 15.4
Beta Decay, Phase Space & the Kurie Plot
Three modes, one bookkeeping rule, and one spectrum shape. Learn to open or close each channel from a mass table, build the electron spectrum from the states available rather than from anything nuclear, then undo that shape to read the endpoint and rank the matrix element.
Build the model
Connect the measurement to the mechanism.
Beta decay is the weak interaction seen through a golden rule, and almost everything you can measure about it is counting rather than dynamics. The transition operator is a contact interaction, so once the lepton waves are frozen to their value across the nucleus — the allowed approximation, costing about (kR)² ≈ 8 × 10⁻⁴ in rate for a 1 MeV electron leaving a mass-32 nucleus — the nuclear matrix element becomes one energy-independent number and the whole spectrum shape comes from the density of final states: p² dp for the electron, (Q − Tₑ)² for the antineutrino it shares with. That is why the spectrum is continuous, why it vanishes at both ends, and why a fixed nuclear mass difference paired with a variable electron energy demanded a third particle.
Two corrections then sit on top: F(Z, p), the Coulomb distortion of the departing lepton by the daughter's charge, which is atomic physics rather than nuclear; and a shape factor that returns the moment the allowed approximation fails. The cost is that the model says nothing about the matrix element itself — you reach it only after dividing the phase-space integral out, as log ft — and the straightness of a Kurie plot certifies the allowed shape, not the neutrino mass.
- Simple definition
- A beta spectrum is continuous because the released energy is shared with a neutrino, and its shape is a count of how many ways that sharing can happen — phase space, scaled by the Coulomb pull of the daughter nucleus on the departing lepton.
- Example
- ³²P releases Q = 1711 keV, yet its electrons average only 695 keV; the rest leaves with the antineutrino. Divide the counts by p²F(Z, p), take the square root, and that lopsided spectrum becomes a straight line cutting zero at 1711 keV.
Capture opens the moment the atomic mass drops; β⁺ needs 1.022 MeV more, so ⁷Be at QEC = 862 keV captures and can never emit a positron.
M is the atomic mass, so the Z electron masses cancel; 2mec² = 1022.0 keV; Bₙ binds the captured electron
The shape is pure counting: p² counts electron states and (Q − Tₑ)² counts antineutrino states, while |Mfi|² is a single constant.
p is the electron momentum, Tₑ its kinetic energy, Q the endpoint; the daughter's recoil is neglected
At Tₑ = 200 keV with Zd = 16 it is 1.62 for an electron and 0.56 for a positron — a factor of 2.9 that is atomic, not nuclear.
+ for β⁻, − for β⁺; Zd is the daughter charge, β = v/c, α = 1/137.04; F is dimensionless
Straightens a bell-shaped spectrum into a line whose energy-axis intercept is the endpoint — the sharpest available measure of Q.
N(p) is counts per unit momentum; K carries arbitrary units, so only its intercept is read
The signal is a deficit inside the last few mνc², not a global bend: of order 10⁻¹³ of tritium decays fall in the last eV of its 18.59 keV spectrum.
the last electron energy is Q − mνc², where K meets the axis with infinite slope
Divides phase space out so matrix elements can be compared: ≈3.5 superallowed, 4.5–6 favoured allowed, 6–9 first forbidden.
f = ∫ F(Z, W) W p (W₀ − W)² dW from 1 to W₀ is dimensionless, W = E/mec²; t½ in seconds
Open the channels from a mass table, not a nuclear table
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 Qβ⁻ = [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: Qβ⁺ = QEC − 2mec², 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. ⁶⁴Cu shows all three at once: Qβ⁻ = 579.4 keV to ⁶⁴Zn, QEC = 1675.0 keV to ⁶⁴Ni, and therefore Qβ⁺ = 653.0 keV — three open channels taking 38.5%, 43.9% and 17.6% of decays. ⁷Be shows the toll biting: QEC = 861.9 keV is positive, Qβ⁺ = −160.1 keV is not, so it captures and never emits a positron.
The spectrum shape is a count of final states
Fermi's golden rule, λ = (2π/ħ)|Mfi|²ρ(Ef), 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 Eν = qc, so q = (Q − Tₑ)/c and dq/dEf = 1/c. Multiply the two densities and dλ/dp ∝ |Mfi|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.
F(Z, p): the lepton does not leave as a plane wave
The golden rule wants the lepton wavefunction evaluated at the nucleus, and a charged particle leaving a charge Zd 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 = Zd α/β, positive for β⁻ and negative for β⁺. Take a 200 keV lepton and a daughter with Zd = 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.
The allowed approximation and the two operators it leaves
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 = 0.028 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. Drop the leading term — through a parity change, or ΔJ = 2 — and the rate falls by roughly that (kR)². That is the whole meaning of forbidden.
log ft ranks matrix elements once phase space is removed
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 mec², 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. Favoured 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.
Read the endpoint off a Kurie plot
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 − (mνc²)²], so the plot leaves the line only within the last few mνc² and meets the axis vertically at Q − mνc².
Change one variable at a time
Make the relationship visible.
Set mνc² to 0 and the solid curve lies exactly on the faint line — that straightness is phase space, not evidence about the neutrino. Raise m to 150 keV and the bend appears only in the last 150 keV. Swing Zd from +30 to −30 to watch the dashed curve, the plot you get if F is never divided out.
ENDPOINT Q800 keV
CUT-OFF Q − mνc²800 keV
F(Zd) AT 100 keV2.06
PHASE-SPACE f (Z = 0)3.70
Live interpretationENDPOINT Q: 800 keV. CUT-OFF Q − mνc²: 800 keV. F(Zd) AT 100 keV: 2.06. PHASE-SPACE f (Z = 0): 3.70
Catch the common trap
Explain before calculating.
In a Kurie plot the measured N(p) is divided by p² F(Z, p) before the square root is taken. What does dividing by F remove?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAtomic masses are M(⁷Be) = 7.01692872 u and M(⁷Li) = 7.01600344 u, with 1 u = 931494.1 keV/c². Decide which beta channels are open for ⁷Be, and say what the emitted neutrinos look like.
- Electron capture first, since it pays no rest-mass toll: QEC = [M(⁷Be) − M(⁷Li)]c² = (7.01692872 − 7.01600344) u × 931494.1 keV/u = 0.00092528 × 931494.1 = 861.9 keV. Lithium's K-shell binding energy is about 0.06 keV, far below the precision here, so drop it.
- Now β⁺, which must create two electron rest masses: Qβ⁺ = QEC − 2mec² = 861.9 − 1022.0 = −160.1 keV. Negative, so that channel is shut.
- β⁻ would need ⁷Be to be heavier than ⁷B. It is not, so that channel is shut too.
- Capture is therefore the only route, and it has a two-body final state — daughter plus neutrino. So the neutrino energy is fixed rather than continuous: 861.9 keV to the ⁷Li ground state, and 861.9 − 477.6 = 384.3 keV on the 10.4% branch through ⁷Li's first excited state.
AnswerQEC = +861.9 keV and Qβ⁺ = −160.1 keV, so electron capture only. Its two-body final state gives two neutrino lines, at 861.9 and 384.3 keV, instead of the continuum a β⁺ branch would add.
Medium³²P decays to ³²S, so the daughter charge is Zd = 16. Evaluate the non-relativistic Fermi function F ≈ 2πη/(1 − e(−2πη)) with η = ±Zd α/β for a 200 keV electron, then for a 200 keV positron leaving the same daughter charge. Take mec² = 511.0 keV and α = 1/137.04.
- Momentum first, and relativistically: (pc)² = Tₑ² + 2mec²Tₑ = 200² + 2(511.0)(200) = 40000 + 204400 = 244400 keV², so pc = 494.4 keV.
- Total energy E = Tₑ + mec² = 711.0 keV, so β = pc/E = 494.4/711.0 = 0.6953.
- For the electron the daughter attracts, so η = +Zd α/β = 16 × 0.0072974/0.6953 = +0.1679 and 2πη = +1.0551.
- F₋ = 1.0551/(1 − e(−1.0551)) = 1.0551/(1 − 0.3482) = 1.0551/0.6518 = 1.62, so the attraction raises the count rate at this energy by 62%.
- The positron sees the same charge repelling it, so η only flips sign: F₊ = −1.0551/(1 − e(+1.0551)) = −1.0551/(1 − 2.8720) = −1.0551/(−1.8720) = 0.56.
- Ratio 1.62/0.56 = 2.9. Nothing nuclear changed between the two calculations — only the sign of the lepton's charge.
AnswerF₋ = 1.62 and F₊ = 0.56, a factor of 2.9 apart at the same 200 keV. This is precisely the distortion the Kurie transform divides out before any endpoint is read.
Hard³²P decays 1⁺ → 0⁺ with Q = 1711 keV and t½ = 14.27 d. Estimate log ft from f ≈ W₀⁵/30 with W₀ = 1 + Q/mec², classify the transition, then explain why ¹⁴C (log ft = 9.04, t½ = 5700 y) is 10⁵ times slower than ³²P (log ft = 7.9) although their matrix elements are within an order of magnitude.
- W₀ = 1 + 1711/511.0 = 4.348, so W₀⁵ = 1553 and f ≈ 1553/30 = 51.8. That form assumes Z = 0 and W₀ ≫ 1, so expect it to run low against the tabulated f ≈ 64 for Zd = 16.
- t½ = 14.27 × 86400 = 1.233 × 10⁶ s.
- f t½ = 51.8 × 1.233 × 10⁶ = 6.38 × 10⁷ s, so log ft = 7.81 — within 0.1 of the tabulated 7.9, which is all this approximation deserves.
- Classify it: ΔJ = 1 with no parity change, so the Fermi operator is excluded (it needs ΔJ = 0) and this is pure Gamow-Teller. log ft near 7.8 is allowed but hindered, well above the 4.5–6 of a favoured allowed decay and far above the 3.49 of a superallowed 0⁺ → 0⁺.
- Now the comparison. Half-lives: 5700 y = 1.799 × 10¹¹ s against 1.233 × 10⁶ s, a factor of 1.46 × 10⁵. Comparative half-lives: 10(9.04 − 7.9) = 13.8.
- So phase space supplies 1.46 × 10⁵ / 13.8 ≈ 1.1 × 10⁴ of that gap. ¹⁴C's Q is only 156 keV against ³²P's 1711 keV, and f climbs very steeply with W₀. It is log ft, not t½, that lets you say the two matrix elements are within a factor of fourteen.
Answerlog ft ≈ 7.81 against a tabulated 7.9: an allowed but hindered pure Gamow-Teller transition. Of the 1.5 × 10⁵ separating the two half-lives, about 10⁴ is phase space and only ≈14 is nuclear structure.