Big Bang Nucleosynthesis 135At energies of the order of푚n−푚p= 1 .293MeV or less, this ratio is dominated by the
exponential. Thus, atkT= 0 .8MeV, the ratio has dropped from 1 to^15. As the Universe
cools and the energy approaches 0.8MeV, the endothermic neutron-producing reac-
tions stop, one by one. Then no more neutrons are produced but some of those that
already exist get converted into protons in the exothermic reactions.
Nuclear Fusion. Already, at a few MeV, nuclearfusion reactionsstart to build up
light elements. These reactions are exothermic: when a neutron and a proton fuse into
a bound state some of the nucleonic matter is converted into pure energy according
to Einstein’s formula [Equation (3.8)]. This binding energy of thedeuterond,
푚p+푚n−푚d= 2 .22 MeV, (6.74)is liberated in the form of radiation:
n+p→d+훾. (6.75)The deuteron is also written^2 H+in general nuclear physics notation, where the super-
script퐴=2 indicates the number of nucleons and the electric charge is given by the
superscript ‘+’. The bound state formed by a deuteron and an electron is thedeuterium
atom^2 H, which of course is electrically neutral. Although the deuterons are formed
in very small quantities, they are of crucial importance to the final composition of
matter.
As long as photons of 2.22MeV or more are available, the reaction in Equa-
tion (6.75) can go the other way: the deuteronsphotodisintegrateinto free protons and
neutrons. Even when the mean temperature of radiation drops considerably below
2.22MeV, there is still a high-energy tail of the Planck distribution containing hard훾
rays which destroy the deuterons as fast as they are produced.
All evidence suggests that the number density of baryons, or equivalently nucleons,
is today very small. In particular, we are able to calculate it to within a factor훺Bℎ^2 ,
푁B=
휌B
푚B
=
훺B휌c
푚B≃ 11. 3 훺Bℎ^2 m−^3. (6.76)At the end of this section we shall discuss the value of the baryon density parameter
훺B, which is a small percentage.
The photon number density today is푁훾= 4. 11 × 108 per m^3 from Equation (6.12).
It is clear then that푁B∕푁훾is such a small figure that only an extremely tiny fraction
of the high-energy tail of the photon distribution may contain sufficiently many hard
훾rays to photodisintegrate the deuterons. However, the 2.22MeV photons created
in photodisintegration do not thermalize, so they will continue to photodisintegrate
deuterium until they have been redshifted below this threshold. Another obstacle to
permanent deuteron production is the high entropy per nucleon in the Universe. Each
time a deuteron is produced, the degrees of freedom decrease, and so the entropy must
be shared among the remaining nucleons. This raises their temperature, counteract-
ing the formation of deuterons. Detailed calculations show that deuteron production
becomes thermodynamically favored only at 0.07MeV. Thus, although deuterons are