 Solution

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.