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Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 9D / b / Solution

Because

X_{i} = n_{i} / n_{B}

,

\frac{X _{D}}{X _{p} X _{n}} = n_{B} \frac{n _{D}}{n _{p} n _{n}} .

Using

n_{B} = η n_{γ}

and the given photon density in the deuterium equilibrium abundance formula gives

\frac{X _{D}}{X _{p} X _{n}} ≃ \frac{16 ζ ( 3 )}{π} η (\frac{k _{B} T}{m _{p} c ^{2}})^{3/2} e^{B_{D} / (k_{B} T)} .

Although the exponential favors deuterium once

k_{B} T < B_{D}

, the very small factor

η

expresses the enormous excess of photons. The high-energy tail therefore photodissociates newly formed deuterium until the temperature is well below

B_{D} / k_{B}

; this is the deuterium bottleneck.

Solved by gpt-5.6-sol high.

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