University Physics IV · Nuclear Physics · 13.5

Beta Decay & Fermi's Theory

Beta decay is where a mass table, a count of final states, and a particle that barely interacts all have to agree. Learn to pick the mode from atomic masses, get the spectrum's shape from state counting alone, and read the nuclear matrix element out of log ft.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Beta decay is the weak conversion of one nucleon into the other inside a bound nucleus, and Fermi's 1934 treatment is the golden rule applied to a contact interaction: a rate equal to a nuclear matrix element times a count of accessible final states. Two things follow. The energetics are pure bookkeeping in neutral-atom masses — Q(β⁻) is the atomic mass difference outright, β⁺ is charged an extra 2m_ec² = 1.022 MeV because the decay makes a positron and the daughter atom must shed an electron, and capture pays neither, opening a window where capture is the only mode a nuclide has.

The spectrum, meanwhile, is not dynamics at all. The leptons carry wavelengths near 140 fm against a 5 fm nucleus, so setting their wavefunctions to their values at the origin costs about 4% in amplitude, the matrix element goes constant, and everything left is phase space: N(p) ∝ p²(Q − T)², with Q its only parameter. The costs are explicit.

Coulomb distortion of the outgoing electron must be put back by hand as F(Z, E); the constant matrix element collapses whenever the allowed operators give zero, changing shape as well as rate; and G_F is fitted, not derived — the theory reproduces the spectrum while saying nothing about why the interaction picks out a handedness.

Simple definition: Beta decay is the weak conversion of a neutron into a proton or a proton into a neutron inside a nucleus, emitting an electron or a positron together with an antineutrino or neutrino that shares the released energy with it.
Example: A free neutron shows it bare: n → p + e⁻ + ν̄ₑ with Q = 939.565 − 938.272 − 0.511 = 0.782 MeV and a half-life near 610 s. Bound in a nucleus the same conversion can be blocked outright or run a thousand times faster.

Q values from neutral-atom massesQ(β⁻) = [M(X) − M(Y)]c² · Q(β⁺) = [M(X) − M(Y)]c² − 1.022 MeV

Picks the decay mode straight off a mass table, and shows why β⁺ needs 1.022 MeV of headroom that capture does not.

M are neutral-atom masses; 1 u = 931.494 MeV/c²; 2m_ec² = 1.022 MeV. Y is the Z ± 1 isobar.

Electron capture and its windowQ(EC) = [M(X) − M(Y)]c² − Bₙ

For 0 < ΔMc² < 1.022 MeV capture is the only route open — ⁷Be, at 0.862 MeV, has no β⁺ branch at all.

Bₙ is the binding energy of the vacated shell: 8.98 keV in copper, near 100 keV in the heaviest atoms.

Golden rule with a contact interactionλ = (2π/ħ)|M_fi|² dn/dE_f, M_fi = G_F ∫ ψ*_Y O ψ_X ψ*ₑ(0) ψ*_ν(0) dV

Splits the rate into a nuclear matrix element and a count of final states, so the spectrum's shape and the half-life come from different places.

G_F/(ħc)³ = 1.166 × 10⁻⁵ GeV⁻²; both lepton waves taken at the origin — the allowed approximation.

Allowed momentum spectrumN(p) dp ∝ F(Z, E) p² (Q − Tₑ)² dp

Gives the whole continuous shape from state counting alone, with Q the only free parameter and no dynamics needed.

p and Tₑ are the electron's momentum and kinetic energy in MeV/c and MeV; massless neutrino; |M_fi|² constant.

Kurie ordinateK(T) = √[ N(p) / (p² F(Z, E)) ] ∝ Q − Tₑ

Straightens the spectrum so Q is an intercept found by extrapolation, and any bend flags m_ν, a forbidden shape, or a second branch.

F = 2πη/(1 − e^(−2πη)) with η = ±Zα/β, plus for β⁻ and minus for β⁺; K in arbitrary units, Tₑ in MeV.

Comparative half-life ftf = ∫₁^(W₀) F(Z, W)(W₀ − W)² W √(W² − 1) dW, f ≈ W₀⁵/30

Divides out the Q⁵ so half-lives spread over 10²⁶ collapse into log ft from 3.5 to 22.5 — a direct read of the matrix element.

W = E/m_ec², W₀ = 1 + Q/m_ec²; f dimensionless, t½ in seconds; the approximation needs W₀ ≫ 1.

Three modes, one mass table, one electron ledger

Mass tables list neutral atoms, not nuclei, so do the electron bookkeeping once and reuse it. In β⁻ the parent has Z electrons and the daughter atom needs Z + 1; the emitted electron supplies exactly the missing one, the Z-dependence cancels, and Q(β⁻) = [M(X) − M(Y)]c² with atomic masses used straight. In β⁺ the daughter needs Z − 1, so the tabulated daughter carries one electron too many, and the decay must also create a positron — two debits of 0.511 MeV, 1.022 MeV in all. Electron capture consumes one of the parent's own electrons, so the counts already match on both sides and Q(EC) is the full mass difference less the binding energy of the vacated shell, 8.98 keV for copper. Take ⁶⁴Cu, which exceeds ⁶⁴Zn by 6.220 × 10⁻⁴ u and ⁶⁴Ni by 1.7982 × 10⁻³ u. Those two differences give Q(β⁻) = 0.579 MeV, Q(EC) = 1.675 MeV and Q(β⁺) = 0.653 MeV. All three channels are open, and ⁶⁴Cu duly uses all three.

A continuous spectrum needs a third particle

If β⁻ were a two-body decay, momentum conservation in the parent's rest frame would fix the electron's energy exactly and the spectrum would be a line. It is not: counts run continuously from zero to a sharp endpoint at Q, and the mean electron energy is only about Q/3. Spin makes the same point independently — n → p + e⁻ would put one spin-½ particle on the left and two on the right, which cannot balance. Pauli's 1930 letter proposed a light, neutral, spin-½ particle to carry off the balance of energy, momentum and angular momentum; Fermi named it the neutrino and built the rate around it. Lepton number then sorts the versions: β⁻ emits an antineutrino, while β⁺ and capture emit a neutrino. Reines and Cowan confirmed the particle in 1956 through the inverse reaction ν̄ + p → n + e⁺, tagged by a delayed coincidence between two 0.511 MeV annihilation photons and a neutron-capture gamma. The cross-section is of order 10⁻⁴³ cm², a mean free path in water of hundreds of light-years, which is why confirmation took twenty-six years.

Fermi's rate: a matrix element times a count of states

Fermi wrote the rate with the golden rule, λ = (2π/ħ)|M_fi|² dn/dE_f. The interaction he used is a contact one — all four fermions meeting at a single point with strength G_F — and we now know why that works: the exchanged W weighs 80.4 GeV, so its range is ħc/M_Wc² = 197.3/80400 fm = 2.5 × 10⁻³ fm, utterly unresolved by a nucleus. The second simplification is the allowed approximation. Expand each outgoing lepton wave as e^(ik⋅r) ≈ 1 + ik⋅r; for a 1 MeV electron pc = 1.42 MeV, so ħc/pc = 139 fm against a nuclear radius near 5 fm, and kR ≈ 0.04. Keep only the 1, evaluate both lepton wavefunctions at the origin, and |M_fi|² becomes a constant. Everything left is counting. The electron contributes 4πp² dp states, the neutrino 4πp_ν² dp_ν with p_νc = Q − Tₑ and dp_ν/dE = 1/c, so N(p) dp ∝ p²(Q − Tₑ)² dp. The shape of a beta spectrum is not dynamics at all: it is the number of ways two particles can share Q.

Coulomb distortion, and Kurie's straightening

The electron leaves through the daughter's Coulomb field, so its wavefunction at the nucleus is not a plane wave's. The repair is the Fermi function, F(Z, E) = 2πη/(1 − e^(−2πη)) in its non-relativistic form, with η = Zα/β for β⁻ and −Zα/β for β⁺. Attraction pulls slow electrons in and lifts the low-energy end: for the ³²P daughter, Z = 16, F runs from 1.82 at T = 0.1 MeV down to 1.44 at 1.2 MeV. Repulsion does the reverse, which is why positron spectra from the same Q look visibly harder. Kurie's move is to divide the measurement by everything the theory already knows and take a square root: K = √[N(p)/(p²F)] should be proportional to Q − T, a straight line meeting the axis at the endpoint. Two payoffs follow. Q comes from extrapolating a line fitted where counts are plentiful, rather than from chasing the last few counts. And departures are diagnostic: a low-energy bow means a missing F or self-absorption in the source, a high-energy break means an unresolved second branch or a forbidden shape factor.

ft values divide the phase space out

Beta half-lives run from milliseconds to 10¹⁶ years, but most of that range is phase space rather than nuclear physics: integrating p²(Q − T)² over the spectrum makes the rate climb roughly as Q⁵. To compare nuclei you divide it out. The Fermi integral f = ∫₁^(W₀) F(Z, W)(W₀ − W)² W √(W² − 1) dW, with W = E/m_ec² and W₀ = 1 + Q/m_ec², collects all of it; ft½ in seconds is the comparative half-life, and it is quoted as log₁₀(ft) because even after the division the spread is enormous. The bands are the useful part. Superallowed 0⁺ → 0⁺ decays cluster at log ft ≈ 3.5 — ¹⁴O gives 3.49; ordinary allowed transitions sit between 4.5 and 6; first-forbidden ones between 6 and 9; and ¹¹⁵In, fourth-forbidden with a half-life of 4.4 × 10¹⁴ years, reaches log ft ≈ 22.5. Each step of forbiddenness costs about (kR)² ≈ 10⁻³ in rate, three to four units of log ft, which is exactly what the ladder is measuring.

Fermi, Gamow-Teller, and where "allowed" stops

Because the leptons leave with ℓ = 0, an allowed transition cannot change the nuclear parity. Their spins couple two ways, and each way defines an operator. Antiparallel, S = 0, is Fermi: the operator is the identity, and it demands ΔJ = 0 with no parity change. Parallel, S = 1, is Gamow-Teller: the operator is σ, and it permits ΔJ = 0 or 1 but forbids 0 → 0, since a spin-1 lepton pair carries away angular momentum the nucleus must supply. The pure cases are the clean tests. ¹⁴O(0⁺) → ¹⁴N*(0⁺) can only be Fermi; ⁶He(0⁺) → ⁶Li(1⁺) can only be Gamow-Teller; the free neutron and the mirror decays mix the two. When both operators return zero — because ΔJ ≥ 2, or because the parity changes — the leading term is gone and the ik⋅r term has to be kept. That is not a small correction. The rate falls by roughly (kR)², and the shape factor is no longer flat, so the Kurie plot itself curves until that factor is divided out too.

Change one variable at a time

Make the relationship visible.

Interactive model

Endpoint energy Q1.2 MeV

Daughter proton number Z16

s = +1 for β⁻, −1 for β⁺1

Set s = +1 and push Z to 30: the curve bows above the line at low energy, where the daughter's Coulomb pull has piled up slow electrons. Flip s to −1 and the same Z pushes positrons away, so the curve sags below. Neither move shifts where the two meet the axis.

ENDPOINT Q1.20 MeV

MAX ELECTRON pc1.63 MeV/c

F(Z, E) AT T = 0.30 MeV1.55

√F AT T = 0.15 MeV1.300

Live interpretationENDPOINT Q: 1.20 MeV. MAX ELECTRON pc: 1.63 MeV/c. F(Z, E) AT T = 0.30 MeV: 1.55. √F AT T = 0.15 MeV: 1.300

Catch the common trap

Explain before calculating.

Misconception check

The 1.022 MeV charged to β⁺ is not the positron's rest energy counted twice for safety, and it is not a Coulomb barrier. In neutral-atom masses it is two separate debits. The decay creates a positron: 0.511 MeV. And the daughter atom, with charge Z − 1, needs one electron fewer than the parent, so the mass table has already charged you for an electron the daughter must shed: another 0.511 MeV. Electron capture pays neither, because the parent's own electron is consumed and the counts match on both sides, so its Q is the full atomic mass difference less only the keV-scale binding energy of the vacated shell. Hence the window: with 0 < ΔMc² < 1.022 MeV, capture proceeds and β⁺ is flatly forbidden. ⁷Be sits in that window at 0.862 MeV and decays by capture alone, with a half-life of 53.2 days.

An allowed β⁻ spectrum is turned into a Kurie plot, but the experimenter divides only by p² and forgets the Fermi function F(Z, E). Compared with the properly corrected plot, what does the result look like?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyNeutral-atom masses are M(⁶⁴Cu) = 63.929764 u, M(⁶⁴Ni) = 63.927966 u and M(⁶⁴Zn) = 63.929142 u, with 1 u = 931.494 MeV/c². Decide which of β⁻, β⁺ and electron capture are energetically open for ⁶⁴Cu, and give the Q value of each.

β⁻ goes to the Z + 1 isobar, ⁶⁴Zn, and for atomic masses the electron count cancels: Q(β⁻) = (63.929764 − 63.929142) u × 931.494 MeV/u = 6.220 × 10⁻⁴ × 931.494 = 0.579 MeV. Open.
Both routes to the Z − 1 isobar share one mass difference: 63.929764 − 63.927966 = 1.7982 × 10⁻³ u, worth 1.7982 × 10⁻³ × 931.494 = 1.675 MeV.
Electron capture consumes an atomic electron the parent already owned, so the electron counts match and Q(EC) is that whole 1.675 MeV, less only the 8.98 keV binding energy of the copper K shell. Open.
β⁺ pays 2m_ec² = 1.022 MeV on top — one 0.511 MeV to make the positron, one for the electron the daughter atom no longer needs: Q(β⁺) = 1.675 − 1.022 = 0.653 MeV. Open, but with an endpoint far below the capture Q drawn from the same mass difference.

AnswerQ(β⁻) = 0.579 MeV, Q(EC) = 1.675 MeV, Q(β⁺) = 0.653 MeV — all three open, and measurement bears it out: 38.5% β⁻, 43.9% capture, 17.6% β⁺.

MediumA magnetic spectrometer counts electrons from ³²P (daughter ³²S, Z = 16). Corrected for acceptance, two points survive: N = 1475 at T = 0.400 MeV where F(Z, E) = 1.508, and N = 998 at T = 1.200 MeV where F(Z, E) = 1.433. Build the Kurie ordinate at each point, extrapolate to the endpoint, and say what a 1 eV neutrino mass would do to that endpoint.

Convert kinetic energy to momentum with (pc)² = (T + m_ec²)² − (m_ec²)². At T = 0.400: 0.911² − 0.511² = 0.829921 − 0.261121 = 0.5688 MeV². At T = 1.200: 1.711² − 0.511² = 2.927521 − 0.261121 = 2.6664 MeV².
Form K = √[N/(p²F)], working in MeV² for p²c². At 0.400 MeV: 1475/(0.5688 × 1.508) = 1475/0.8578 = 1719.6, so K₁ = 41.47. At 1.200 MeV: 998/(2.6664 × 1.433) = 998/3.8210 = 261.2, so K₂ = 16.16.
The allowed shape makes K linear in T, so two points fix the line: slope = (16.16 − 41.47)/(1.200 − 0.400) = −25.31/0.800 = −31.63 MeV⁻¹.
Extrapolate to K = 0. From the first point, T = 0.400 + 41.47/31.63 = 0.400 + 1.311 = 1.711 MeV; from the second, 1.200 + 16.16/31.63 = 1.711 MeV. Consistent, so Q = 1.711 MeV.
With m_ν ≠ 0 the neutrino momentum becomes p_νc = √[(Q − T)² − (m_νc²)²], so K ∝ [(Q − T)√((Q − T)² − (m_νc²)²)]^½. The plot turns down and meets the axis at T = Q − m_νc². For m_νc² = 1 eV against Q = 1.711 MeV that is a shift of 5.8 × 10⁻⁷ of the endpoint.

AnswerK = 41.5 and 16.2 in the same arbitrary units; slope −31.6 MeV⁻¹, intercept Q = 1.711 MeV. A 1 eV neutrino mass moves that intercept by about 6 parts in ten million — invisible here, which is why the search uses tritium with Q = 18.6 keV.

Hard⁶He, ground state 0⁺, decays by β⁻ to the ⁶Li ground state, 1⁺, with Q = 3.508 MeV and t½ = 0.807 s. Use the large-W₀ approximation f ≈ W₀⁵/30, where W₀ = 1 + Q/m_ec², to find log ft. Then classify the transition against the allowed selection rules and say which operator drives it.

W₀ is the endpoint total electron energy in units of m_ec² = 0.511 MeV: W₀ = 1 + 3.508/0.511 = 1 + 6.865 = 7.865.
Raise it: 7.865² = 61.86, 7.865³ = 486.5, 7.865⁴ = 3826, 7.865⁵ = 3.009 × 10⁴. So f ≈ 3.009 × 10⁴/30 = 1.003 × 10³.
Multiply by the half-life: ft½ = 1.003 × 10³ × 0.807 s = 810 s, so log₁₀(ft) = 2.91. The tabulated value is 2.910, so the Z = 3 Coulomb correction dropped by the approximation is worth well under 1% here.
Apply the selection rules. J runs 0 → 1, so ΔJ = 1, and a Fermi transition is impossible because the identity operator cannot change J. Both states are even parity, consistent with the leptons leaving in ℓ = 0, so the decay is genuinely allowed rather than forbidden.
What is left is the σ operator with the lepton pair in a spin triplet: pure Gamow-Teller. A log ft near 2.9 sits in the superallowed band, which says the ⁶He and ⁶Li ground states have nearly the same spatial structure, so the matrix element is close to its maximum possible value.

Answerf ≈ 1.00 × 10³, ft½ ≈ 810 s, log ft = 2.91. With 0⁺ → 1⁺ and no parity change it cannot be Fermi: pure Gamow-Teller, and superallowed.