The number density of a species $\star$ of non-relativistic particles of mass $m$, in equilibrium at temperature $T$ and chemical potential $\mu$, is

$$n_{\star}=g_{\star}\left(\frac{2 \pi m k T}{h^{2}}\right)^{3 / 2} e^{\left(\mu-m c^{2}\right) / k T},$$

where $g_{\star}$ is the spin degeneracy. During primordial nucleosynthesis, deuterium, $D$, forms through the nuclear reaction

$$p+n \leftrightarrow D,$$

where $p$ and $n$ are non-relativistic protons and neutrons. Write down the relationship between the chemical potentials in equilibrium.

Using the fact that $g_{D}=4$, and explaining the approximations you make, show that

$$\frac{n_{D}}{n_{n} n_{p}} \approx\left(\frac{h^{2}}{\pi m_{p} k T}\right)^{3 / 2} \exp \left(\frac{B_{D}}{k T}\right)$$

where $B_{D}$ is the deuterium binding energy, i.e. $B_{D}=\left(m_{n}+m_{p}-m_{D}\right) c^{2}$.

Let $X_{\star}=n_{\star} / n_{B}$ where $n_{B}$ is the baryon number density of the universe. Using the fact that $n_{\gamma} \propto T^{3}$, show that

$$\frac{X_{D}}{X_{n} X_{p}} \propto T^{3 / 2} \eta \exp \left(\frac{B_{D}}{k T}\right)$$

where $\eta$ is the baryon asymmetry parameter

$$\eta=\frac{n_{B}}{n_{\gamma}}$$

Briefly explain why primordial deuterium does not form until temperatures well below $k T \sim B_{D}$.