 Solution

In the nonrelativistic dilute regime, $E(p)=m{c}^2+p^2/(2m)+\cdots$ and $(m{c}^2-\mu )/kT\gg 1$, so the Bose or Fermi correction in the denominator is negligible. The common Maxwell-Boltzmann distribution approximation givesDifferentiating the quoted Gaussian integral with respect to its squared scale gives ${\int }_0^{\infty }{p}^2{e}^{-a{p}^2}dp=\sqrt{\pi }/(4a^{3/2})$. Thereforechemical equilibrium of $p+n\leftrightarrow D$ requires ${\mu }_D={\mu }_p+{\mu }_n$. Apply the density formula to each species to obtainUsing $g_n=g_p=2$, the specified $g_D=4$, and $m_n\simeq m_p$, $m_D\simeq 2m_p$, givesFor $X_a=n_a/n_B$ and $n_B=\eta n_{\gamma }$, the corresponding deuterium equilibrium abundance isThe very small baryon-to-photon ratio means that even the high-energy tail contains many photons per baryon capable of destroying deuterium when $kT$ is of order $B_D$. Substantial deuterium survives only after that tail has been exponentially suppressed, so $B_D/kT$ must be large enough to offset the small prefactor. This is the deuterium bottleneck, not a threshold set just by equality of thermal energy and binding energy. The degeneracy $g_D=4$ is used as stipulated for this calculation.