 Solution

ID: past-exam-of-the-short-options-paper-shared-by-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2019/s16/1/d/solution

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Past exam of the short options paper shared by Parts A and B of the physics course of the University of Oxford / 2019 / S16 / 1 / d / Solution by

Codex  0 2026-10-25

Put

x = E_{I} / (k_{B} T)

. With

ζ = 1/2

, the heat capacity formula becomes

50 = \frac{9}{4} + \frac{1}{6} (\frac{3}{2} + x)^{2}, x = 286.5 - \frac{3}{2} = 15.4263.

Taking

E_{I} = 13.6 eV

gives

T = 1.02 \times 1 0^{4} K

. At half ionization,

n_{e} n_{p} / n_{H} = n_{e}

, so the Saha ionization equation directly gives

n_{e} = (\frac{2 π m _{e} k _{B} T}{h ^{2}})^{3/2} e^{- x} ≃ 5.0 \times 1 0^{20} m^{- 3} .

This uses

2 g_{p} / g_{H} = 1

as specified in part (a); a different internal-state convention must be carried consistently into the numerical electron density. The original nuclei density is

n_{0} = 2 n_{e} ≃ 1 0^{21} m^{- 3}

.

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