advanced34 min studyLesson 584 in Physics

Beta-decay spectra and Fermi theory

Derive the allowed spectral shape from lepton phase space and state the limits of selection rules and endpoint approximations.

On this page

What you’ll understand

  • Identify electron, neutrino, and recoil contributions to energy balance.
  • Explain the phase-space factors in an allowed beta spectrum.
  • Separate Fermi and Gamow–Teller selection rules from absolute transition strengths.

A beta electron has a continuous energy spectrum even when the parent and daughter nuclear states are discrete. The missing event-by-event energy is carried mainly by a neutrino, with a smaller daughter recoil. Fermi’s low-energy weak-interaction description combines nuclear matrix elements with this multi-particle phase space. Its simplest allowed approximation is powerful, but endpoint precision and forbidden transitions require more than a smooth curve with an adjustable normalization.

Three bodies share one available energy

In beta-minus decay a neutron changes into a proton while an electron and electron antineutrino are emitted. Beta-plus emission instead produces a positron and electron neutrino, while electron capture absorbs a bound or free electron. Energy, momentum, and angular momentum must all balance. Atomic-mass Q conventions differ among these cases, so the positron threshold cannot be inferred by copying the beta-minus formula. A daughter excitation also reduces the available energy of that particular branch.

For a heavy daughter and a nearly massless neutrino, the leading energy balance assigns Q between electron kinetic energy and neutrino energy, neglecting nuclear recoil. The electron endpoint approaches Q in this approximation, but finite recoil lowers it slightly and a nonzero neutrino mass changes the final endpoint region. Momentum conservation still matters when recoil energy is numerically small. It is inconsistent to omit the neutrino entirely and attribute a broad electron spectrum to random failure of energy conservation.

Q=Te+Eν+TR,pec=Te(Te+2mec2),TR≃∣pe+pν∣22MdQ=T_e+E_\nu+T_R,\qquad p_ec=\sqrt{T_e(T_e+2m_ec^2)},\qquad T_R\simeq\frac{|\mathbf p_e+\mathbf p_\nu|^2}{2M_d}
Q and the kinetic energies are in a common energy unit, pe and pν are momenta, and Md is the heavy daughter mass. The recoil formula is nonrelativistic for the daughter; the electron momentum relation retains its rest mass. Eν includes neutrino rest energy if it is retained.↔ Scroll wide equations horizontally.

Allowed decay factors the nuclear amplitude from lepton phase space

When lepton wavelengths are long compared with nuclear size, the leading nuclear transition amplitude can be treated as nearly independent of their momenta. Counting final electron and neutrino states then gives an electron spectral factor containing momentum, total electron energy, and the square of the remaining neutrino energy. A Coulomb correction, called the Fermi function, accounts for the charged lepton’s distorted wave near the daughter rather than replacing the nuclear weak matrix element.

For beta-minus emission the nuclear charge attracts the outgoing electron, while beta-plus emission repels the positron. Their Coulomb factors are therefore different. Recoil corrections, radiative effects, finite nuclear size, screening, neutrino mass, and forbidden shape factors can all matter depending on precision and energy. The endpoint and total rate are also distinct observables: integrating phase space gives a rate only after the relevant weak couplings and nuclear matrix elements are supplied.

dλdTe=C F(Zd,Te) (pec)(Te+mec2)(Q−Te)2(0≤Te≤Q)\frac{d\lambda}{dT_e}=C\,F(Z_d,T_e)\,(p_ec)(T_e+m_ec^2)(Q-T_e)^2\quad(0\le T_e\le Q)
C absorbs weak couplings, nuclear strength, and unit factors so dλ/dTe has units s⁻¹ per energy. The displayed leading allowed spectrum assumes negligible recoil and neutrino mass. F is dimensionless and Zd labels the daughter charge; the proper electron or positron sign must be used.↔ Scroll wide equations horizontally.

Selection rules classify leading operators, not guaranteed rates

The leading Fermi operator carries zero angular momentum and does not change parity, so it permits equal initial and final nuclear spin. The leading Gamow–Teller operator carries angular momentum one and also preserves parity: spin change can be zero or one, but a zero-to-zero transition is excluded for this operator. Nuclear isospin and structure constrain the matrix elements further. A spin-parity assignment satisfying a rule is therefore an allowed possibility, not proof of a large measured strength.

If the leading operators are excluded or suppressed, higher terms in the lepton-wave expansion can contribute. The name forbidden refers to suppression relative to the leading approximation, not an absolute impossibility. A useful spectrum diagnostic divides out the known momentum and Coulomb factors to form a Kurie plot, which is approximately linear only for the appropriate allowed, negligible-neutrino-mass model. Curvature can arise from several corrections and cannot alone be assigned uniquely to one new physical effect.

A transparent high-energy spectral limit

Use a toy allowed spectrum with Q = 4 MeV, negligible electron and neutrino masses, no recoil, and F = 1. Its shape is proportional to Te2(Q−Te)2T_e^2(Q-T_e)^2. Find the peak energy, the shape ratio at 1 MeV versus 2 MeV, and the mean electron energy.

  1. The shape is symmetric under Te↦Q−TeT_e\mapsto Q-T_e and vanishes at both endpoints. Its interior stationary point follows from differentiating: 2Te(Q−Te)(Q−2Te)=02T_e(Q-T_e)(Q-2T_e)=0, so the maximum is at 2 MeV.

  2. At 1 MeV the shape factor is 12(4−1)2=91^2(4-1)^2=9 in common arbitrary units; at 2 MeV it is 22(4−2)2=162^2(4-2)^2=16. Their ratio is 9/169/16, not the square of the electron-energy ratio alone.

  3. Symmetry about 2 MeV makes the mean equal to 2 MeV as well. Each event still shares energy differently; the equality of mean and peak is a feature of this massless symmetric limit, not a universal beta-spectrum property.

The result

The toy peak and mean are both 2 MeV, while the 1 MeV differential intensity is 9/16 of the 2 MeV value. Electron mass and Coulomb corrections break this simple symmetry.

Explore the model

Interactive lab

Interactive lab · function

Isolate the neutrino phase-space factor

Use x as a normalized available neutrino energy. Reset and set both controls to a = 1 and x = 1; predict y = 1. Reset and set both to a = 1 and x = 2; predict y = 4. This tests the squared available-energy factor only. The parabola omits electron momentum, Coulomb distortion, recoil, nuclear strength, and the full beta spectrum.

Isolate the neutrino phase-space factorThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forSquaring the neutrino energy is one phase-space ingredient, not the complete spectral shape.

Check your understanding

Which leading allowed nuclear beta operator can connect a 0⁺ state to another 0⁺ state?

Try it yourself

A toy beta branch has Q = 2.0 MeV. An emitted electron has kinetic energy 0.8 MeV. Neglect recoil only for the leading energy balance, and use mec2=0.5 MeVm_ec^2=0.5\,\mathrm{MeV}. Find the neutrino energy and electron momentum. Explain why these data do not fix the daughter recoil momentum uniquely.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use Q minus electron kinetic energy for the neutrino, but use the relativistic momentum relation for the electron.

Compare with the solution

The leading neutrino energy is 2.0−0.8=1.2 MeV2.0-0.8=1.2\,\mathrm{MeV}. The electron momentum satisfies pec=0.8(0.8+1.0)=1.2 MeVp_ec=\sqrt{0.8(0.8+1.0)}=1.2\,\mathrm{MeV}, hence pe=1.2 MeV/cp_e=1.2\,\mathrm{MeV}/c. For a massless neutrino its momentum magnitude is also 1.2 MeV/c, but their relative direction is unspecified. The daughter balances their vector sum, whose magnitude can range from zero to 2.4 MeV/c. Including its small recoil energy would slightly adjust the neutrino energy.

Take it with you

  • The continuous beta spectrum reflects multi-particle energy sharing.
  • Allowed phase space and nuclear transition strength play different roles.
  • Selection rules identify leading possibilities; forbidden does not mean impossible.

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