 Solution

Let $n_n$ and $n_p$ count the available neutrons and protons immediately before their incorporation into nuclei. In the usual neutron-limited helium synthesis approximation, all surviving neutrons go into ${}^4\mathrm{He}$ and the remaining protons form ${}^1\mathrm{H}$. Each helium nucleus contains two neutrons and two protons, soUsing the equal-nucleon-mass approximation and neglecting trace nuclei and binding-energy corrections, the primordial helium mass fraction isThe ratio is the pre-assembly nucleon ratio, not the number ratio of free hydrogen nuclei to helium nuclei afterward. The derivation assumes enough protons to bind all the neutrons, as in standard Big Bang nucleosynthesis. The familiar $n_p/n_n\simeq 7$ gives $Y_p\simeq 1/4$ and hydrogen mass fraction $X_p\simeq 3/4$.