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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2025

Paper 3

Define
$$
K(T)=\frac1{n_B}\left(\frac{2\pi m_ek_BT}{h^2}\right)^{3/2}
=\frac{\sqrt{2\pi}}{8\eta\zeta(3)}
\left(\frac{m_ec^2}{k_BT}\right)^{3/2}.
$$
Using $n_e=n_B\mathcal F$, $n_{j^+}=X_{j^+}n_j$, and $n_{j^0}=(1-X_{j^+})n_j$, the two Saha relations become the closed system
$$
\boxed{
\mathcal F\frac{X_{H^+}}{1-X_{H^+}}
=K(T)e^{-I_H/(k_BT)}},
$$
$$
\boxed{
\mathcal F\frac{X_{He^+}}{1-X_{He^+}}
=2K(T)e^{-I_{He}/(k_BT)}},
$$
where
$$
\mathcal F=(1-Y_p)X_{H^+}+\frac{Y_p}{4}X_{He^+}.
$$
These are two equations for the two ionization fractions.

When $Y_p\to0$, $\mathcal F=X_{H^+}$, and the hydrogen equation reduces to
$$
\boxed{
\frac{X_{H^+}^2}{1-X_{H^+}}
=\frac1{n_B}\left(\frac{2\pi m_ek_BT}{h^2}\right)^{3/2}
e^{-I_H/(k_BT)}},
$$
the usual <hydrogen-only Saha equation>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 14E / b / iii / Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 14E / b / iii / Solution

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