 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2016/s16/1/2/solution

For total nuclear density

n

n_{e} = n ζ

and

n_{H} = n (1 - ζ)

. Saha's equation becomes

n ζ^{2} / (1 - ζ) = C T^{3/2} e^{- R / (k_{B} T)}

. At

ζ ≪ 1

ζ ≃ C / n T^{3/4} e^{- T_{H} / T}, m = \frac{3}{4}, T_{H} = \frac{R}{2 k _{B}} ≃ 7.89 \times 1 0^{4} K .

The factor two in the exponent follows from taking a square root; using the full ionization temperature would double the predicted logarithmic slope.

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