Physical Foundations of Cosmology

3.5 Nucleosynthesis 117

3.5.5 The other light elements

Now we can calculate the final abundances of the other light elements by simply
using the quasi-equilibrium conditions.

Helium-3 The expression for the quasi-equilibrium concentration of^3 He follows
from (3.148):

X (^3) He≈

3

2

(

λDD1

λ (^3) HeD

XD+ 2

λDp

λ (^3) HeD
Xp

)(

1 + 2

λ^3 Hen

λ (^3) HeD
Xn
XD

)− 1

. (3.166)

If the baryon density is not too large, the rate of the dominant reaction in which^3 He is
destroyed is larger than the rate of the deuterium destruction. Therefore, the freeze-
out of^3 He occurs a little bit later than the freeze-out of deuterium. After deuterium
freeze-out, a small leakage from the D reservoir to the^3 He reservoir still maintains
a stationary flow through the^3 He reservoir and the quasi-equilibrium condition for

(^3) He is roughly satisfied at the time of its freeze-out. SubstitutingXnX 2
D/R^1 into
(3.166) and the experimental values for the ratios of the corresponding reaction
rates, taken for definiteness atT 0 .06 MeV, we obtain
X 3 fHe
0. 2 XDf+ 10 −^5
1 + 4 × 103 XDf

, (3.167)

whereXDfis given in (3.162). This result is in good agreement with the numerical
calculations shown in Figure 3.8. For example, forη 10 = 1 ,we haveXDf 4 ×
10 −^4 andX 3 fHe 3 × 10 −^5 ,that is, the final^3 He abundance is 10 times smaller
than the deuterium abundance.
The difference betweenXDfandX 3 fHedecreases for largerη 10 .Forη 10  10 ,the
freeze-out concentrations of the deuterium and helium-3 are about the same and
equal to 10−^5 .In a universe withη 10 > 10 ,the reaction Dp→^3 Heγdominates
in producing^3 He around the freeze-out time and nearly all deuterium is destroyed
in favor of^3 He, which thus becomes more abundant than deuterium. In this case,
the freeze-out of^3 He is determined by two competing reactions, Dp→^3 Heγand

(^3) HeD→ (^4) Hen,and, irrespective of how largeXDis, they give rise to the final
(^3) He abundance,X 3 f
HeλDp/λ^3 HeD^10
− (^5). The weak dependence ofX 3 f
Heon the
baryon density forη 10 >10 is due to the temperature dependence of the reaction
rates, which we have ignored.
TritiumThe quasi-equilibrium condition (3.147) gives
XT=

(

3

2

λDD2

λDT

+ 2

λ^3 Hen

λDT

Xn

X^2 D

X (^3) He

)

XD. (3.168)