122 The hot universe
After most of the helium is converted into He+ions, the singly charged ions
capture a second electron and become neutral. The second electron also ends up in
the first orbit. The electron–electron interaction substantially reduces the binding
energy: it is only 24.62 eV for the second electron. Therefore, the second stage of
helium recombination occurs at a lower temperature than the first. For example,
atT12 000 K,the density of neutral helium atoms is still negligible; only after
the temperature decreases belowT∼5000 K does helium become neutral and
decouple from radiation. At this time hydrogen is still fully ionized and the universe
remains opaque to radiation.
Problem 3.23Assuming chemical equilibrium, derive the expression for the ratio
of the number densities of He+and neutral He. Verify that forη 10 5 this ratio is
equal to unity atT6800 K and is about 10−^4 atT5600 K.
Problem 3.24Explain why the recombination temperature is significantly smaller
than the corresponding ionization potential energies.
3.6.2 Hydrogen recombination: equilibrium consideration
The main reaction responsible for maintaining hydrogen and radiation in equilib-
rium is
p+e−H+γ, (3.183)where H is a neutral hydrogen atom. For the ground( 1 S)state, the binding energy
of neutral hydrogen,
BH=mp+me−mH= 13 .6 eV, (3.184)corresponds to a temperature of 158 000 K.In this case, the Saha formula can be
derived in the same manner as (3.178) and takes the form
npne
nH=
(
Tme
2 π) 3 / 2
exp(
−
BH
T
)
, (3.185)
wherenHis the number density of the hydrogen atoms in the ground state and we
have taken into account that the corresponding ratio of the statistical weightsgi
is equal to unity. At equilibrium, neutral hydrogen atoms are also present in the
excited states: 2S, 2 P.... However, atT<5000 K,their relative concentrations
are negligible: for example,
n 2 P
nH=
g 2 P
g 1 Sexp