= Solution
At temperatures above a few MeV, <weak interactions> such as $n+\nu_e\leftrightarrow p+e^-$ and $n+e^+\leftrightarrow p+\bar\nu_e$ maintain <neutron-proton chemical equilibrium>. Negligible lepton chemical potentials give the <neutron-to-proton ratio> $r_{\rm eq}=n_n/n_p=\exp[-\Delta/(k_BT)]$, where $\Delta=(m_n-m_p)c^2\simeq1.293\,{\rm MeV}$. At high temperature this ratio is nearly one.
During <radiation domination>, the interaction time grows as $T^{-5}$, faster than the expansion time $T^{-2}$. Their crossing therefore produces <cosmological weak freeze-out>. If $\Theta=T/(10^{10}\,{\rm K})$, equating the displayed order-of-magnitude times gives $\Theta_f^{-5}\sim g_*^{-1/2}\Theta_f^{-2}$, so
$$
\boxed{T_f\sim10^{10}g_*^{1/6}\ {\rm K},\qquad
T_f\propto(Gg_*)^{1/6}G_{\rm weak}^{-2/3}}.
$$
Using $H^{-1}=2t$ instead of $t$ changes only the order-one coefficient. Before electron-positron annihilation, the usual effective relativistic count is $g_*=2+(7/8)(4+6)=10.75$. The factor $7/8$ weights fermionic relative to bosonic energy density.
After freeze-out, free neutrons decay while protons increase correspondingly. Define the neutron baryon fraction $X_n=n_n/(n_n+n_p)$. In the instantaneous-freeze-out estimate,
$$
X_{n,f}=\frac{r_f}{1+r_f},\qquad r_f=e^{-\Delta/(k_BT_f)},\qquad
X_n(t)=X_{n,f}e^{-(t-t_f)/\tau_n},
$$
where $\tau_n=t_{1/2}/\ln2$ is the neutron mean lifetime. Hence the ratio at the onset of nuclear synthesis is
$$
\boxed{r_N=\frac{r_f e^{-(t_N-t_f)/\tau_n}}
{1+r_f[1-e^{-(t_N-t_f)/\tau_n}]}}.
$$
This expression accounts for the protons produced by neutron decay.
Nuclear burning initially waits at the <deuterium bottleneck>: although deuterium is bound by about $2.2\,{\rm MeV}$, the many photons per baryon can still photodissociate it. Roughly, $\eta_b^{-1}e^{-B_D/(k_BT_N)}\sim1$ sets the bottleneck temperature, with refinements from reaction rates and phase space. Around $k_BT_N\sim0.1\,{\rm MeV}$, deuterium survives, reactions through deuterium, tritium and helium-3 rapidly build tightly bound helium-4, and nearly all surviving neutrons are captured.
Thus $n_{\rm He}\simeq n_n/2$ and the <primordial helium mass fraction> is
$$
\boxed{Y_{\rm He}\simeq2X_n(t_N)=\frac{2r_N}{1+r_N}}.
$$
For the standard estimate $r_N\simeq1/7$, neutron and proton constituent fractions are $1/8$ and $7/8$, respectively, and $Y_{\rm He}\simeq1/4$. Free neutrons are negligible after capture. The remaining hydrogen mass fraction is about $3/4$, and the helium-to-hydrogen number ratio is about $1/12$.
The printed numerical time coefficients are schematic, not a precision prediction of this quarter fraction: their literal crossing with $g_*=10.75$ gives $k_BT_f\simeq1.28\,{\rm MeV}$. Keeping physical weak-rate coefficients and the gradual freeze-out instead gives an effective freeze-out around $0.7$–$0.8\,{\rm MeV}$, followed by decay over a few minutes. The abundance formulas above explain the standard quarter fraction and its parameter dependence without treating the rounded time formulas as exact nuclear-rate data.
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