# Why was the Cosmic Microwave Background (CMB) released at a blackbody temperature of 3000K rather than 30,000K?

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 .

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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...

• Rule #1: Never rely on Quora (aka Yahoo Answers 2: Electric Boogaloo). Dec 27, 2021 at 21:05
• Because it decided too Mar 27 at 22:31

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.