 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ii/paper-3/9c/a/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 3 / 9C / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-05

Chemical equilibrium imposes

μ_{e} + μ_{p} = μ_{H} + μ_{γ}

. For blackbody photons,

μ_{γ} = 0

. Define the positive Hydrogen binding energy by

I = (m_{e} + m_{p} - m_{H}) c^{2}

, approximately

13.6 eV

. Multiplying the two Maxwell-Boltzmann distribution number densities and dividing by the neutral-hydrogen density gives

\frac{n _{e} n _{p}}{n _{H}} = \frac{g _{e} g _{p}}{g _{H}} (\frac{2 πk T}{h ^{2}} \frac{m _{e} m _{p}}{m _{H}})^{3/2} e^{- I / (k T)} .

Use

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

and

m_{p} / m_{H} ≃ 1

to obtain the Saha ionization equation

\frac{n _{e} n _{p}}{n _{H}} = (\frac{2 π m _{e} k T}{h ^{2}})^{3/2} e^{- I / (k T)} .

The assumptions are nonrelativistic, dilute, nondegenerate species at a common temperature, ground-state hydrogen with the stated degeneracy factors, negligible mass correction in the prefactor, and sufficiently rapid reactions to maintain chemical equilibrium.

 Read the full article

 New to topics? Read the docs here!