 Solution

There are about ${\eta }^{-1}\sim 10^9$ photons per baryon. Even when the typical photon energy is below the Hydrogen binding energy, the high-energy tail can provide abundant ionizing photons. The nonrelativistic electron's translational phase space also favours the ionized state. Quantitatively, the Saha equation contains both suppressing factors $\eta$ and $(T/m_e{)}^{3/2}$. Recombination needs their tiny product to be overcome by $e^{I/T}$, so $I/T$ must be several tens, not of order one.

For an illustrative ${\Omega }_B{h}^2=0.02$, the supplied baryon-to-photon relation gives $\eta =5.36\times 10^{-10}$. With $m_e=5.11\times 10^5\mathrm{eV}$, direct evaluation givesThus the equilibrium ionization fraction reaches a few per cent just below $0.3\mathrm{eV}$, while at $13.6\mathrm{eV}$ it is essentially unity. The exact temperature depends weakly on the baryon abundance and on the fraction used to define cosmological recombination. This is small baryon abundance delays hydrogen recombination. Recombination, photon decoupling and residual electron freeze-out are related but distinct: the last two involve rates and cannot be determined by chemical equilibrium alone.