 Solution

In the hydrogen–helium approximation, take helium to be helium-4, and assume essentially all neutrons remaining at the onset of efficient nuclear burning enter it. Each helium-4 nucleus contains two neutrons and two protons, henceHere $n_p,n_n$ count constituent nucleons, not just free particles after burning. Neglecting the small nucleon mass difference and nuclear binding-energy correction, the helium mass is $4m_u{n}_{\mathrm{He}}$ and the total baryon mass is $m_u{n}_B$. Thus neutron-limited helium synthesis givesFor $n_n/n_p\simeq 1/7$, this yields $Y_p\simeq 1/4$ and the hydrogen mass fraction is $X_p\simeq 3/4$. The assumptions require $n_p\ge n_n$ and negligible free neutrons, deuterium and helium-3: the phrase hydrogen and helium must here mean predominantly hydrogen-1 and helium-4, not an arbitrary mixture of helium isotopes.