104 The hot universe
is the sole remaining cause for a change in the number of neutrons. As a result, the
neutron concentration decreases fort>t∗as
Xn(t)=X∗nexp(−t/τn). (3.117)Note that after freeze-out one can neglect the degeneracy of the leptons, which
would increase the neutron lifetime, and useτn,as quoted above.We will see that
nucleosynthesis, in which nearly all free neutrons are captured in the nuclei (where
they become stable), occurs att∼250 s.This is a rather substantial fraction of
the neutron lifetime and hence the neutron decay significantly influences the final
abundances of the light elements.
3.5.2 “Deuterium bottleneck”
Complex nuclei are formed as a result of nuclear interactions. Helium-4 could,
in principle, be built directly in the four-body collision:p+p+n+n→^4 He.
However, the low number densities during the period in question strongly suppress
these processes. Therefore, the light complex nuclei can be produced only through a
sequence of two-body reactions. The first step is deuterium(D)production through
the reaction
p+nD+γ. (3.118)There is no problem with this step because fort< 103 s the corresponding reaction
rate is much larger than the expansion rate.
Let us calculate the deuterium equilibrium abundance. We define the abundance
by weight:
XD≡ 2 nD/nN,wherenNis the total number of nucleons (baryons) including those in complex
nuclei. The relation betweenXDand the abundances of the free neutrons,Xn≡
nn/nN,and protons,Xp≡np/nN, can be found using (3.61) for each component.
Because the deuterium nucleus with spin zero is metastable, its total statistical
weight isgD= 3 .Taking into account thatgp=gn=2 and the chemical potentials
satisfy the conditionμD=μp+μn,we find
XD= 5. 67 × 10 −^14 η 10 TMeV^3 /^2 exp(
BD
TMeV)
XpXn, (3.119)where
BD≡mp+mn−mD 2 .23 MeV