Calculation of mean energy value of photons in recombination

In summary, recombination occurs when the average energy of the "whole" of photons is less than 13.6 eV, which happens when the energy of a billionth of the most energetic photons is also of this order. This is due to the large number of photons in the universe compared to the number of protons and electrons, with a ratio of one proton for one billion photons. This is further explained by the Saha equation, which takes into account the ratio of free electrons to total hydrogen abundance. This can be demonstrated using a probability density function, such as the Bose-Einstein distribution for photons.
  • #1

eliot13

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.

Thanks in advance
 
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  • #2
eliot13 said:
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.
 
  • #3
PeterDonis said:
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).
 
  • #4
eliot13 said:
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.
eliot13 said:
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.
 

What is the calculation of mean energy value of photons in recombination?

The calculation of mean energy value of photons in recombination is a method used to determine the average energy of photons released during the process of recombination. This is important in understanding the energy levels and transitions of electrons within a system.

Why is it important to calculate the mean energy value of photons in recombination?

Calculating the mean energy value of photons in recombination is important because it helps us understand the energy levels and transitions of electrons within a system. This can provide insights into the properties of materials and the behavior of particles in different environments.

What factors affect the mean energy value of photons in recombination?

The mean energy value of photons in recombination is affected by several factors, including the energy levels of the electrons involved, the type of material or medium in which the recombination occurs, and any external influences such as temperature or pressure.

How is the mean energy value of photons in recombination calculated?

The mean energy value of photons in recombination is calculated by multiplying the energy difference between the initial and final energy levels by the probability of recombination occurring at those levels. This calculation takes into account the different energy levels and their respective probabilities of recombination.

What are some real-world applications of calculating the mean energy value of photons in recombination?

The calculation of mean energy value of photons in recombination has many practical applications, including in the fields of optoelectronics, solar energy, and astrophysics. It also plays a crucial role in the development of new materials and technologies, such as quantum computing and photonics. Understanding the mean energy value of photons in recombination can also aid in the study of chemical reactions and the behavior of particles in different environments.

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