 Deuterium equilibrium abundance

Use natural units and a positive nuclear binding energy ${\epsilon }_D=m_n+m_p-m_D$. Chemical equilibrium of $n+p\leftrightarrow D+\gamma$ requires ${\mu }_D={\mu }_n+{\mu }_p$. The nonrelativistic Maxwell--Boltzmann number density givesHere $g_D=3$ and $g_n=g_p=2$ count the spin states. Taking $m_n\simeq m_p\simeq m_N$ and $m_D\simeq 2m_N$, then writing $X_i=n_i/n_B$, $n_B=\eta n_{\gamma }$ and using the photon number density yields the displayed abundance. The coefficient is approximately $8.14$. Omitting the spin-state ratio $3/4$ gives the rougher prefactor $16\zeta (3)/\sqrt{\pi }$. A convention $B=m_D-m_n-m_p<0$ instead writes the exponential as $e^{-B/T}$. The small baryon-to-photon ratio suppresses stable deuterium until the temperature is well below the binding scale, producing the deuterium bottleneck.