The number density of particles of mass $m$ at equilibrium in the early universe is given by the integral

$$n=\frac{4 \pi g_{\mathrm{s}}}{h^{3}} \int_{0}^{\infty} \frac{p^{2} d p}{\exp [(E(p)-\mu) / k T] \mp 1}, \quad \begin{cases}- & \text { bosons } \\ + & \text { fermions }\end{cases}$$

where $E(p)=c \sqrt{p^{2}+m^{2} c^{2}}, \mu$ is the chemical potential, and $g_{\mathrm{s}}$ is the spin degeneracy. Assuming that the particles remain in equilibrium when they become non-relativistic $\left(k T, \mu \ll m c^{2}\right)$, show that the number density can be expressed as

$$n=g_{\mathrm{s}}\left(\frac{2 \pi m k T}{h^{2}}\right)^{3 / 2} e^{\left(\mu-m c^{2}\right) / k T} .$$

[Hint: Recall that $\left.\int_{0}^{\infty} d x e^{-\sigma^{2} x^{2}}=\sqrt{\pi} /(2 \sigma), \quad(\sigma>0) .\right]$

At around $t=100$ seconds, deuterium $D$ forms through the nuclear fusion of nonrelativistic protons $p$ and neutrons $n$ via the interaction $p+n \leftrightarrow D$. In equilibrium, what is the relationship between the chemical potentials of the three species? Show that the ratio of their number densities can be expressed as

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

where the deuterium binding energy is $B_{D}=\left(m_{n}+m_{p}-m_{D}\right) c^{2}$ and you may take $g_{D}=4$. Now consider the fractional densities $X_{a}=n_{a} / n_{B}$, where $n_{B}$ is the baryon density of the universe, to re-express the ratio above in the form $X_{D} /\left(X_{n} X_{p}\right)$, which incorporates the baryon-to-photon ratio $\eta$ of the universe.

[You may assume that the photon density is $n_{\gamma}=\left(16 \pi \zeta(3) /(h c)^{3}\right)(k T)^{3}$.]

Why does deuterium form only at temperatures much lower than that given by $k T \approx B_{D}$ ?