 Solution

Because $X_i=n_i/n_B$,Using $n_B=\eta n_{\gamma }$ and the given photon density in the deuterium equilibrium abundance formula givesAlthough the exponential favors deuterium once $k_{B}T<B_D$, the very small factor $\eta$ expresses the enormous excess of photons. The high-energy tail therefore photodissociates newly formed deuterium until the temperature is well below $B_D/k_B$; this is the deuterium bottleneck.