Chemical equilibrium for $e^-+p\leftrightarrow H+\gamma$ gives $\mu_e+\mu_p=\mu_H$, since $\mu_\gamma=0$. Insert the supplied nonrelativistic equilibrium densities. Neglecting $m_e/m_p$ in the translational reduced mass and using the stated degeneracy convention gives
$$
\frac{n_H}{n_en_p}
=\left(\frac{2\pi}{m_eT}\right)^{3/2}
e^{(m_e+m_p-m_H)/T}.
$$
Charge neutrality gives $n_p=n_e$, so
$$
\boxed{\frac{n_H}{n_e^2}=\left(\frac{2\pi}{m_eT}\right)^{3/2}e^{B_H/T}}.
$$
This is the hydrogen <Saha ionization equation>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 310 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-310/2/a/solution