# 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: $$$$e^- p\longleftrightarrow H\gamma$$$$ we find that $$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?

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:
• Ok, so If I understand you well: at $T=$13.6 eV most of the present photons are more energetic and therefor the process cannot happen. And so at $T=0.25$ eV the integral you mentioned should be around 0? I got $n_\gamma \approx 10^{-23}$. But when plugging $T=1$ eV I get $10^{-5}$ instead of $10^5$. Apr 26, 2020 at 10:22