 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/2018/s26/2/c/ii/solution

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Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2018 / S26 / 2 / c / ii / Solution by

Codex  0 2026-10-25

Fully ionized helium supplies three particles per four hydrogen masses, so

μ_{He} ≃ 4/3

. The same small

Z

, but negligible hydrogen fraction, gives

κ_{0, He} \propto Z

, half the hydrogen-star coefficient. The relative luminosity is

\frac{L _{He}}{L _{H}} = \frac{κ _{0, H}}{κ _{0, He}} (\frac{4/3}{1/2})^{7} (\frac{0.5}{1})^{5} (\frac{1 0 ^{8}}{1.5 \times 1 0 ^{7}})^{1/2} .

(1)

Consequently

L_{He} / L_{H} ≃ 2 (\frac{8}{3})^{7} \frac{1}{32} \frac{20}{3} ≃ 155.

(2)

The very strong molecular-weight dependence outweighs the reduction in mass. Treating the two opacities as identical would miss a factor two, and using neutral-gas molecular weights would be inconsistent with the stated hot ionized stars.

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