Decoupled Temperature for photons: why is it 0.25 $\rm eV$ rather than 13.6 $\rm eV$? When calculating the decoupled temperature of photons using Saha' equation for the following process:
\begin{equation}
e^- p\longleftrightarrow H\gamma
\end{equation}
we find that $T_{dec}=3000$ K$=0.25$ eV. 
From my understanding, this phenomenon happens when it becomes thermodynamically favourable for protons and electrons to combine into neutral atoms. I was expecting it to be 13.6 eV (Rydberg energy) for this case, which is the Hydrogen's biding energy. Why is it less than that? 
 A: This is because there are hugely many more photons than charge-carriers per unit volume, roughly 10 billion photons for every electron in the universe.  
As an example, consider affairs when the universe cooled to a temperature of 1 eV, or around 10,000 K.  At this temperature, electrons are no longer relativistic and their density follows the Boltzmann distribution,
$$n_e = 2\left(\frac{m_e T}{2\pi}\right)^{3/2} \exp \left(\frac{\mu_e - m_e}{T}\right).$$
At $T = 10^4$ K, the electron density is $n_e \approx 10^4 \,{\rm cm}^{-3}$.  
Meanwhile, the number density of photons with an energy in excess of 13.6 eV can be found by integrating the Planck spectrum,
$$n_\gamma = \frac{1}{\pi^2}\int^\infty_{13.6}\frac{E^2}{\exp(E/T)-1}\, {\rm d}E,$$
giving $n_\gamma \approx 3 \times 10^9 \, {\rm cm}^{-3}$ at $T = 10^4$ K. In other words, there are around $3\times 10^5$ more photons than electrons per unit volume with energy greater than 13.6 eV! At these temperatures, there is no shortage of energetic photons available to re-ionize neutral hydrogen once it forms. 
The following illustration helps visualize this:

