118 The hot universe
Assuming that tritium freeze-out occurs at about the same time as for deuterium,
and substitutingXnX^2 D/R 1 into (3.168), we find
XTf(
0. 015 + 3 × 102 X 3 fHe)
XDf, (3.169)where the experimental valuesλDD2/λDT 0 .01 andλ (^3) Hen/λDT1 have been
used. Forη 10 1 ,we haveXTf 10 −^5 .Note that, for anyη 10 ,the tritium final
abundance is several times smaller than the deuterium abundance.
Problem 3.21When does tritium freeze-out take place? For whichη 10 can we use
XDfin (3.168) to estimateXTf? Which value ofXDshould be used otherwise?
Problem 3.22Explain why the^3 He concentration increases monotonically in time
(see Figure 3.7) but the tritium concentration first rises to a maximum and then
decreases until it freezes out.
Lithium-7 and beryllium-7 The quasi-equilibrium conditions for^7 Li and^7 Be result
from the dominant reactions in which^7 Li and^7 Be are produced and destroyed (see
Figure 3.6):
7
12
λ (^4) HeTX (^4) HeXT+λ (^7) BenX (^7) BeXn=λ (^7) LipX (^7) LiXp, (3.170)
7
12
λ (^4) He (^3) HeX (^4) HeX (^3) He=λ (^7) BenX (^7) BeXn. (3.171)
One can check that other reactions, such as^7 Li+D→ 24 He+nand^7 Be+D→
24 He+p,can be ignored forη 10 > 1 .It follows from these equations that
X^7 Li=
7
12
X (^4) He
Xp
(
λ (^4) HeT
λ (^7) Lip
)(
XT+
λ (^4) He (^3) He
λ (^4) HeT
X^3 He
)
. (3.172)
The ratioλ (^4) HeT/λ (^7) Lipis nearly constant over a broad temperature interval, increasing
only from 2. 2 × 10 −^3 to 3× 10 −^3 as the temperature drops from 0.09 MeV to
0 .03 MeV,while
r(T)≡
λ (^4) He (^3) He
λ^4 HeT
changes significantly over the same temperature interval, namely,r 5 × 10 −^2
forT 0 .09 MeV andr 6 × 10 −^3 forT 0 .03 MeV. With these values of the
reaction rates we obtain
X (^7) Li∼ 10 −^4 (XT+r(T)X (^3) He). (3.173)
To estimate the freeze-out concentration for^7 Li we must know the values ofXT,
r(T)andX^3 Heat^7 Li freeze-out. For 5>η 10 > 1 ,freeze-out occurs after the