# Calculation of mean energy value of photons in recombination

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eliot13

## Main Question or Discussion Point

I am interested in the calculation of the mean energy value of CMB (Cosmic Microwave Background) photons from which the recombination is performed.

The subject on French Wikipedia says :

"Intuitively, one might say that recombination occurs when energy average of photons is of the order of the ionization energy of hydrogen, 13.6 eV, or about 150 000 Kelvin. In practice, this estimate is incorrect, because the universe is a system that has a large number of photons by the atomic nucleus (of the order of one proton for one billion photons). Therefore, what matters is that the energy of a billionth of the most energetic photons is of the order of the ionization energy of the hydrogen atom. This occurs when the average energy of the "whole" of the photons is less than 13.6 eV.

Taking $$x_{e}$$ the ratio of the abundance of free electrons to the total abundance of hydrogen (both neutral and ionized), i.e :
$$x_{e}=\dfrac{n_{e}}{n_{p}+n_{H}}$$
with $$n_{e}$$ the number density of free electrons, $$n_{H}$$ that of atomic hydrogen and $$n_{p}$$ that of ionized hydrogen (i.e. protons). Saha equation yields :
$$\dfrac{x_{e}^{2}}{1-x_{e}}=\dfrac{5.8\,10^{21}}{\Omega_{b}h^{2}\,T^{3/2}}\,exp\big(\dfrac{-1.58\,10^{5}}{T}\big)$$
Then, recombination is done when average energy value of photons equals to 0.3 eV , so a temperature of 3000 K.
"

Why do the billionth of the most energetic photons has to be equal to 13.6 eV ?

I would like to find a demonstration which explains that the energy of a billionth of the most energetic photons must be of the order of the ionization energy of the hydrogen atom (13.6 eV).

To formulate a demonstration, I may start from a probability density function ( Maybe Bose-Einstein distribution for photons) and/or Saha equation.

It seems to be an issue to solve but I want to clarify with a mathematical point of view (with probabilities) this above affirmation.

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PeterDonis
Mentor
2019 Award
Why do the billionth of the most energetic photons has to be equal to 13.6 eV ?
Because at the time of recombination, only a billionth of the photons are photons that came from a proton and electron recombining into a hydrogen atom, because the ratio of the number of photons to the number of protons/electrons in the universe is a billion to one.

eliot13
Because at the time of recombination, only a billionth of the photons are photons that came from a proton and electron recombining into a hydrogen atom, because the ratio of the number of photons to the number of protons/electrons in the universe is a billion to one.
Do you mean that in the following reaction :

$$H^{+} + e^{-} \longrightarrow H + \gamma$$

the produced photon $$\gamma$$ has an energy equal or greater than 13.6 eV ?

Then, could we say photons that came from this reaction are part of the most energetic photons ? (that would imply that the rest (a large majority) of photons has an energy lower than 13.6 eV).

PeterDonis
Mentor
2019 Award
Do you mean that in the following reaction :
$$H^{+} + e^{-} \longrightarrow H + \gamma$$

the produced photon

$$\gamma$$

has an energy equal or greater than 13.6 eV ?
Yes.
Then, could we say photons that came from this reaction are part of the most energetic photons ? (that would imply that the rest (a large majority) of photons has an energy lower than 13.6 eV).
Yes.