Physical Foundations of Cosmology

3.5 Nucleosynthesis 109

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10 0

10 −^4

10 −^9

10 −^14

time [minutes]

temperature

mass fraction

p

n^4 He

D

T

(^3) He
1010 K (0.86 MeV) 109 K (0.86 keV) 108 K(8.6 keV)
Fig. 3.7.
alone cannot preventXDfrom further growth.) Although the deuterium concentra-
tion ceases to grow, the concentration of free neutrons strongly decreases because
they go first to the deuterium reservoir and then, without further delay, proceed
down the pipes towards the heavier elements. For most of the neutrons, the final
destination is the^4 He reservoir. In fact, the binding energy of^4 He (28.3 MeV)
is four times larger than the binding energies of the intermediate elements,^3 He
( 7 .72 MeV)and T( 6 .92 MeV)and, therefore, if^4 He were in equilibrium with these
elements, it would dominate at low temperatures. A system always tends to equi-
librium in the quickest possible way. Therefore, most of the free neutrons will form
(^4) He to fulfil its largest equilibrium demand.
Problem 3.19Verify that atT∼ 0 .1 MeV the equilibrium concentrations of D,
(^3) He and T are many orders of magnitude smaller than the (^4) He concentration.
The reactions in which^4 He is formed proceed as follows. First, deuterium is
converted into tritium and^3 He according to (3.123). Next, tritium combines with