Wolfram Alpha:

```Cosmic Background Radiation and Decoupling
------------------------------------------

Photons get scattered in a plasma (the reason we can't see through
the sun). When looking into space the further the distance the further
back in time we see and the higher the temperature.  Eventually, we
get to point where temperature is high enough for ionization (~3000K)
and the universe becomes opaque.  This is the SURFACE OF LEAST
SCATTERING.  The reason why we don't see this glow in the sky is that
the boundary is moving very fast and the waves are red shifted to the
point where their energy and temperature are very low (~3K).  What
from the SLS.

Image courtesy of Wikipedia.

I = EV= (8πh/c3)(ν3/ehν/kT - 1)  ... 1.

where I is the Intensity per frequency and EV is the energy per
unit volume per frequency (I = Power/A = E/AΔt = E/V).

The radiation has a blackbody spectrum.  The peak is at λ ~ hc/kBT

The temperature and density variation of the CMB (δT/T and δρ/ρ)
is approximately equal to 10-5.

Decoupling Temperature and Time
-------------------------------

Prior to the time of decouping, highly energetic photons prevented
electrons and protons from forming neutral atoms of Hydrogen (and
subsequently Helium).  This can be represented by:

H + γ <--> p + e

As the universe cooled, the photons no longer had enough energy
to cause ionization, and neutral atoms started to form.  This
caused a rapid reduction of free electrons and photons no longer
were scattered.  The result is that the very first elements are
formed and, in the process, the universe becomes transparent.
This process is referred to as DECOUPLING.  The question is at
what temperature and time did this occur?

From Plancks law, the photon number density is found by dividing
equation 1. by the energy, hν,  and integrating from 0 to ∞.
Thus,

nγ = (8π/c3)∫(ν2/ehν/kT - 1)dν

This is a tricky integral to solve so we shall just state the
result:

nγ = 0.243(kBT/hc)3  ... 2.

To get the total energy density we integrate equation 1. from 0
to ∞.

I = (8πh/c3)∫(ν3/ehν/kT - 1)dν

This is another tricky integral that yields:

ETotal = (π2/15)(kBT)4/(hc)3

= 0.658(kBT)4/(hc)3  ... 3.

This can also be written as:

ETotal = αT4 = 4σ/c where α is the radiation constant and σ is
Stephan's constant.

Dividing equation 3. by equation 2. gives the energy of a photon from
the CMB:

Eγ ~ 2.7kBT

We next consider the SAHA equation which describes the degree of
ionization of a plasma as a function of the temperature, density,
and ionization energies of the atoms.  The Saha equation follows
directly from the Boltzmann distribution.

(1 - X)/X = np(kBTme/2πh2)-3/2exp(Q/kBT)

Where X = np/(np + nH) = ne/(np + nH) due to charge neutrality.

When X = 1 the gas is fully ionized.  When X = 0 the gas is neutral.

We can also write:

nH = (X - 1)np/X

Now define the baryon to photon ratio, η:

η = nB/nγ = np/Xnγ  (from np/[nγnp/(np + nH)] = nB/nγ)

∴ np = ηXnγ

Substituting into equation 2. gives:

np = 0.243ηX(kBT/hc)3

According to present observational measurements of the baryonic
density, Ωm,0 from the CMB, nγ is approximately equal to 1.7 x 109.

The Saha equation becomes:

(1 - X)/X = 0.243ηX(kBT/hc)3(kBTme/2πh2)-3/2exp(Q/kBT)

Which simplifies to:

(1 - X)/X2 ~ 3.84η(kBT/mec2)3/2exp(Q/kBT)

If we assume X = 0.5 and Q for Hydrogen to be 13.6 eV (2.2 x 10-18 J)
and solve for T we get that the temperature at the time of decoupling

Now consider the equation for redshift:

z + 1 = TE/TT = aT/aE = λT/λE = ρE/ρT = tT2/3/tE2/3

where T = Today, E = Emitted and D = Detected

TT = 2.7 K ... the temperature of the CMB.
TE ~ 3000 K

For the mass dominated period when decoupling occurred, we can write:

z + 1 = TE/TT = 1100 = tT2/3tE2/3

∴  z + 1 = tT/tE = 11003/2 ~ 37,000

∴ tE ~ 365,000 years if the universe is about 13.5 x 109 years old.

Therefore, decoupling occurred about 350,000 years after the Big Bang.```