= Solution
The weak-conversion rate per expansion time falls as $T^3$. Below the stated threshold it becomes less than one, so neutron-proton interconversion cannot track the rapidly changing equilibrium ratio: <cosmological weak freeze-out> occurs. Free <neutrons> subsequently decay, reducing their abundance until <nuclei> can survive photodissociation. Once the <deuterium bottleneck> is passed, <nuclear reactions> rapidly assemble <neutrons> into the strongly bound <helium-4> <nucleus>. The short available time, low density and missing stable mass-five and mass-eight <nuclei> impede extensive production of heavier <nuclei>, leaving nearly all surviving <neutrons> in helium.
If $r=n_n/n_p$ just before helium formation, each helium <nucleus> uses two <neutrons> and two <protons>, so $n_{\rm He}=n_n/2$. Neglecting binding-energy corrections to nuclear <masses>,
$$
\boxed{Y_{4\rm He}=\frac{4(n_n/2)}{n_n+n_p}=\frac{2r}{1+r}=\frac29\simeq0.222\quad(r=1/8).}
$$
The remaining <protons> mainly become <hydrogen>. Counting all four helium nucleons, rather than only the captured <neutrons>, is essential to this <mass> fraction.
Back to article page
