= 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:
$$
d\Gamma=Cp^2\,dp\int_0^\infty k^2\,dk\,\delta(E_0-E-k).
$$
The constant includes the allowed nuclear <matrix element>, weak-coupling and angular factors. Including the usual <Fermi function> for electron-nucleus Coulomb distortion gives
$$
\boxed{\frac{d\Gamma}{dp}=C F(Z,E)\,p^2(E_0-\sqrt{p^2+m_e^2})^2,\quad
0\leq p\leq\sqrt{Q(Q+2m_e)}.}
$$
For the kinetic-energy spectrum, $dp/dT=E/p$ instead yields
$$
\frac{d\Gamma}{dT}=C F(Z,E)\,pE(Q-T)^2.
$$
The 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 $E=p$, $E_0\simeq Q$, and
$$
\Gamma=C\int_0^Q p^2(Q-p)^2\,dp
=CQ^5\int_0^1x^2(1-x)^2\,dx
=\boxed{\frac{C}{30}Q^5.}
$$
Two <momentum> powers come from each final particle and one from the remaining integration. This is the origin of the <Sargent rule>. \b[Exact $Q^5$ dependence requires the relativistic-electron, constant-matrix-element limit.] For $Q\ll m_e$, the same approximation without Coulomb distortion has $pE\simeq m_e\sqrt{2m_eT}$ 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.
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