<p>The CMB is produced as the ionisation fraction of hydrogen falls from a high value to a very small value. Contrary to what is written in the Quora answer you may have been misled by, this happens at a temperature of about 3000 K.  The value given by 13.6 eV/<span class="math-container">$k_B$</span> (remember to multiply by the electric charge to put the energy in SI units) is not even &quot;roughly&quot; correct.</p>
<p>To understand why, firstly you need to consider that both photons and particles have a <em>distribution</em> of energies, with a significant fraction having energies several times the average energy (the average energy is also bigger than <span class="math-container">$k_B T$</span> - it is <span class="math-container">$2.82 k_BT$</span> for blackbody radiation photons). Secondly, to ionise a hydrogen atom does not necessarily need an energy of 13.6 eV if a significant fraction of the atoms are <em>already</em> in an excited state. Thirdly, and perhaps most importantly, there are about a billion photons for every proton, so even if a tiny fraction of them have energies high enough to ionise hydrogen, then the hydrogen will be ionised.</p>
<p>To get the temperature required such that equilibrium ionisation becomes small requires (numerical) solution of the <a href="https://en.wikipedia.org/wiki/Saha_ionization_equation" rel="nofollow noreferrer">Saha equation</a>. It is about 3000 K in the early universe.</p>

CC BY-SA 4.0
2022-10-16T09:48:18.723Z
I had a thermodynamic argument but it was wrong so I deleted my answer.
Kevin Kostlan
CC BY-SA 4.0
2022-10-16T10:25:21.727Z
@KevinKostlan I think it was nearly right, but you should have compared the number of energy states per unit volume rather than just the volume. I may add something to this answer.
ProfRob
CC BY-SA 4.0
2021-12-26T12:13:42.717Z
2023-07-16T12:02:21.910Z
2023-07-16T12:02:21.910Z
ProfRob
<p>If the ionization (or reionization, or Recombination) energy of atomic hydrogen is 13.6 EV, which corresponds to a black body temperature of 30,000K, why did the CMB not begin to appear then?</p>
<p>Why did the universe wait until it had cooled down to 3,000K to release the Relic Radiation (the CMB)?</p>
<p>I may have found a clue on Quora:  How is Hubble Constant calculated from Cosmic Microwave Background Radiation measurements?</p>
<p>The redshift is known quite precisely from the theory, since a hot big bang model (the standard model of cosmology) predicts the point at which the plasma de-ionized to form neutral hydrogen. Roughly, this is when the universe cooled to a temperature below the binding energy of hydrogen,  13.6eV/kb .</p>
<p>From <a href="https://www.quora.com/How-is-Hubble-constant-calculated-from-Cosmic-Microwave-Background-Radiation-measurements/answer/Brent-Follin?ch=10&amp;oid=6880453&amp;share=bfe08ff2&amp;target_type=answer" rel="nofollow noreferrer">Brent Follin's Quora answer to <em>How is Hubble constant calculated from Cosmic Microwave Background Radiation measurements?</em></a></p>
<p><a href="https://i.sstatic.net/Zejgy.png" rel="nofollow noreferrer">(Screenshot)</a></p>
<blockquote>
<p>The redshift is known quite precisely from the theory, since a hot big
bang model (the standard model of cosmology) predicts the point at
which the plasma de-ionized to form neutral hydrogen. Roughly, this is
when the universe cooled to a temperature below the binding energy of
hydrogen, <span class="math-container">$13.6\frac{eV}{k_b}$</span>.</p>
<p>To answer the titular question, the hubble constant is determined
through obtaining the angular diameter distance to the last scattering
surface. That's not a direct observable; instead you infer it through
trigonometry. We can directly measure the angular scale of the Baryon
Acoustic oscillations in the CMB--it's the distance between troughs in
the power spectrum shown in Leo C. Stein's answer. In the standard
<span class="math-container">$\lambda$</span>CDM</p>
<p>cosmological model, we also know* the physical scale of the BAO
feature, known as the sound horizon length. The angular diameter
distance is then defined as</p>
<p><span class="math-container">$D_A=\frac{r_s}{\theta_s}$</span></p>
<p>where the numerator is the known physical scale and the bottom is the
measured angular scale (this is just basic trig). The angular diameter
distance is a well-known function of the hubble rate, and you can
infer the hubble rate from getting the angular diameter distance
(assuming the only pertinent species of particles in the universe are
dark matter, baryons, photons, neutrinos, and the cosmological
constant).</p>
</blockquote>
<p>Doesn't dividing 13.6 eV by a number much smaller than 1 give an enormously large number, not a lower one?</p>
<p>I am confused...</p>

CC BY-SA 4.0
2021-12-27T21:05:43.607Z
Rule #1: Never rely on Quora (aka Yahoo Answers 2: Electric Boogaloo).
Nick S
CC BY-SA 4.0
2023-03-27T22:31:12.410Z
Because it decided too
Starship
<observational-astronomy><cosmology><big-bang-theory><cosmic-microwave-background>
Why was the Cosmic Microwave Background (CMB) released at a blackbody temperature of 3000K rather than 30,000K?
Kurt Hikes
# Why was the Cosmic Microwave Background (CMB) released at a blackbody temperature of 3000K rather than 30,000K?

(Asked by: username_0 on 2021-12-25)

If the ionization (or reionization, or Recombination) energy of atomic hydrogen is 13.6 EV, which corresponds to a black body temperature of 30,000K, why did the CMB not begin to appear then?

Why did the universe wait until it had cooled down to 3,000K to release the Relic Radiation (the CMB)?

I may have found a clue on Quora: How is Hubble Constant calculated from Cosmic Microwave Background Radiation measurements?

The redshift is known quite precisely from the theory, since a hot big bang model (the standard model of cosmology) predicts the point at which the plasma de-ionized to form neutral hydrogen. Roughly, this is when the universe cooled to a temperature below the binding energy of hydrogen, 13.6eV/kb .

From [Brent Follin's Quora answer to *How is Hubble constant calculated from Cosmic Microwave Background Radiation measurements?*](https://www.quora.com/How-is-Hubble-constant-calculated-from-Cosmic-Microwave-Background-Radiation-measurements/answer/Brent-Follin?ch=10&oid=6880453&share=bfe08ff2&target_type=answer)

[(Screenshot)](https://i.sstatic.net/Zejgy.png)

> The redshift is known quite precisely from the theory, since a hot big
> bang model (the standard model of cosmology) predicts the point at
> which the plasma de-ionized to form neutral hydrogen. Roughly, this is
> when the universe cooled to a temperature below the binding energy of
> hydrogen, $13.6\frac{eV}{k\_b}$.
>
> To answer the titular question, the hubble constant is determined
> through obtaining the angular diameter distance to the last scattering
> surface. That's not a direct observable; instead you infer it through
> trigonometry. We can directly measure the angular scale of the Baryon
> Acoustic oscillations in the CMB--it's the distance between troughs in
> the power spectrum shown in Leo C. Stein's answer. In the standard
> $\lambda$CDM
>
> cosmological model, we also know\* the physical scale of the BAO
> feature, known as the sound horizon length. The angular diameter
> distance is then defined as
>
> $D\_A=\frac{r\_s}{\theta\_s}$
>
> where the numerator is the known physical scale and the bottom is the
> measured angular scale (this is just basic trig). The angular diameter
> distance is a well-known function of the hubble rate, and you can
> infer the hubble rate from getting the angular diameter distance
> (assuming the only pertinent species of particles in the universe are
> dark matter, baryons, photons, neutrinos, and the cosmological
> constant).

Doesn't dividing 13.6 eV by a number much smaller than 1 give an enormously large number, not a lower one?

I am confused...

**Tags:** observational-astronomy, cosmology, big-bang-theory, cosmic-microwave-background

> **Comments:**
> **username_1**: Rule #1: Never rely on Quora (aka Yahoo Answers 2: Electric Boogaloo).  
> **username_2**: Because it decided too  

---

## 1 answers

### Answer by username_3 on 2021-12-26. Score: 28 | Accepted answer

The CMB is produced as the ionisation fraction of hydrogen falls from a high value to a very small value. Contrary to what is written in the Quora answer you may have been misled by, this happens at a temperature of about 3000 K. The value given by 13.6 eV/$k\_B$ (remember to multiply by the electric charge to put the energy in SI units) is not even "roughly" correct.

To understand why, firstly you need to consider that both photons and particles have a *distribution* of energies, with a significant fraction having energies several times the average energy (the average energy is also bigger than $k\_B T$ - it is $2.82 k\_BT$ for blackbody radiation photons). Secondly, to ionise a hydrogen atom does not necessarily need an energy of 13.6 eV if a significant fraction of the atoms are *already* in an excited state. Thirdly, and perhaps most importantly, there are about a billion photons for every proton, so even if a tiny fraction of them have energies high enough to ionise hydrogen, then the hydrogen will be ionised.

To get the temperature required such that equilibrium ionisation becomes small requires (numerical) solution of the [Saha equation](https://en.wikipedia.org/wiki/Saha_ionization_equation). It is about 3000 K in the early universe.

> **Comments:**
> **username_4**: I had a thermodynamic argument but it was wrong so I deleted my answer.  
> **username_3**: @KevinKostlan I think it was nearly right, but you should have compared the number of energy states per unit volume rather than just the volume. I may add something to this answer.  

---

