= Deuterium equilibrium abundance
{title2=$X_D/(X_nX_p)\simeq(12\zeta(3)/\sqrt\pi)\eta(T/m_N)^{3/2}e^{\varepsilon_D/T}$}
Use <natural units> and a positive <nuclear binding energy> $\varepsilon_D=m_n+m_p-m_D$. <Chemical equilibrium> of $n+p\leftrightarrow D+\gamma$ requires $\mu_D=\mu_n+\mu_p$. The <nonrelativistic Maxwell--Boltzmann number density> gives
$$
\frac{n_D}{n_nn_p}=\frac{g_D}{g_ng_p}\left(\frac{m_D}{m_nm_p}\right)^{3/2}\left(\frac{2\pi}{T}\right)^{3/2}e^{\varepsilon_D/T}.
$$
Here $g_D=3$ and $g_n=g_p=2$ count the spin states. Taking $m_n\simeq m_p\simeq m_N$ and $m_D\simeq2m_N$, 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>.
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