 Solution

Chemical equilibrium gives ${\mu }_{\mathrm{He}}=2{\mu }_p+2{\mu }_n$, since the photon chemical potential is zero. Set $A=(2\pi m_{p}kT/h^2{)}^{3/2}$. The degeneracies and approximate masses give prefactors $2A,2A,8A$ for protons, neutrons and helium. ThusHere the binding energy uses $2m_p+2m_n-m_{\mathrm{He}}$; the PDF prints $2n_n$ in that mass relation, a dimensional typo.

Baryon conservation gives $n_B\propto a^{-3}$. In adiabatic expansion without photon-producing entropy release, $T\propto a^{-1}$ and $n_{\gamma }\propto T^3\propto a^{-3}$, so $\eta =n_B/n_{\gamma }$ stays constant. This statement is for that epoch; entropy transfer such as electron-positron annihilation modifies the photon normalization.

Since the density ratio equals $n_B^3{X}_p^2{X}_n^2/X_{\mathrm{He}}$, substitution of the given photon density yieldsAt $m_p{c}^2\simeq 938$ MeV, $B_{\mathrm{He}}=30$ MeV and $kT=0.3$ MeV, the right side is about $1.22\times 10^{-29}{\eta }^{-3}$. Unity therefore corresponds to $\eta \simeq 2.3\times 10^{-10}$, illustrating the stated very rough temperature estimate for a cosmological baryon-to-photon ratio of order $10^{-10}$ to $10^{-9}$. A numerical value of $\eta$ is necessary; it is not specified in the question. The exponential makes the temperature estimate far less sensitive than the ratio itself to order-one prefactors.

Actual helium synthesis must proceed through light intermediates. The deuterium bottleneck persists because deuterium's much smaller binding energy allows the abundant high-energy photon tail to destroy it. Only at a lower temperature do enough deuterons survive to initiate efficient reactions leading to helium. Direct four-nucleon assembly at the equilibrium temperature is ineffective.