# Harvard Astronomy 201b

## CHAPTER: The Saha Equation

In Book Chapter on March 5, 2013 at 3:21 am

(updated for 2013)

How do we deal with the distribution over different states of ionization r? In thermodynamic equilibrium, the Saha equation gives: $\frac{ n^\star(X^{(r+1)}) n_e } { n^\star (X^{(r)}) } = \frac{ f_{r+1} f_e}{f_r}$,

where $f_r$ and $f_{r+1}$ are the partition functions as discussed in the previous section. The partition function for electrons is given by $f_e = 2\big( \frac{2 \pi m_e k T} {h^2} \big) ^{3/2} = 4.829 \times 10^{15} (\frac{T}{K})^{3/2}$

For a derivation of this, see pages 103-104 of this handout from Bowers and Deeming.

If $f_r$ and $f_{r+1}$ are approximated by the first terms in their sums (i.e. if the ground state dominates their level populations), then $\frac{ n^\star ( X^{ (r+1) } ) n_e } {n^\star ( X^{ (r) } ) } = 2 \big(\frac{ g_{r+1,1} }{g_{ r,1}}\big) \big( \frac{ 2 \pi m_e k T} {h^2} \big)^{3/2} e^{-\Phi_r / kT}$,

where $\Phi_r=E_{r+1,1}-E_{r,1}$ is the energy required to ionize $X^{(r)}$ from the ground (j = 1)  level. Ultimately, this is just a function of $n_e$ and $T$. This assumes that the only relevant ionization process is via thermal collision (i.e. shocks, strong ionizing sources, etc. are ignored).