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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-2/9b/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 2 / 9B / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

Write

n_{b} = n_{e} + n_{H} ≃ η n_{γ}

. Charge neutrality gives

n_{p} = n_{e}

, and therefore

n_{e} = X_{e} n_{b}, n_{H} = (1 - X_{e}) n_{b} .

Substitution into the Saha ionization equation yields

\frac{1 - X _{e}}{X _{e}^{2}} = n_{b} (\frac{2 π ℏ ^{2}}{m _{e} k _{B} T})^{3/2} e^{I / (k_{B} T)} .

Using the given photon density,

\frac{1 - X _{e}}{X _{e}^{2}} = η \frac{2 ζ ( 3 ) ( 2 π ) ^{3/2}}{π ^{2}} (\frac{k _{B} T}{m _{e} c ^{2}})^{3/2} e^{I / (k_{B} T)} .

The baryon-to-photon ratio is extremely small. Even when

k_{B} T ≪ I

, the high-energy tail of the enormous photon population contains enough ionizing photons per baryon to delay cosmological recombination. The exponential factor must become large enough to overcome both

η \sim 1 0^{- 9}

and the phase-space factor, which occurs near

k_{B} T ≃ 0.3 eV

rather than near

I

.

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