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Saha Equation

The Saha equation for ionization equilibrium may be written

\begin{displaymath}\frac{N_+}{N_0} = A (kT)^{3/2} N_e^{-1} \exp(-\chi_0/kT)\end{displaymath}

where


\begin{displaymath}A = \frac{Q_+}{Q_0} \;\; \frac{2\,(2\pi m_e )^{3/2}}{h^3} \end{displaymath}

where $Q$ is a partition function which normalizes a distribution among all the states of the neutral or ion. For hydrogen

\begin{displaymath}\frac{Q_+}{Q_0} = \frac{1}{2}\end{displaymath}

Substituting for $m_e$ and $h$ in SI units we have

\begin{displaymath}A = \frac{(2\pi \; 9.1096\times10^{-31})^{3/2}}{(6.6262\times10^{-34})^3}\end{displaymath}


\begin{displaymath}A = 4.70679 \times 10^{+55}\end{displaymath}


\begin{displaymath}\log_{10} A = +55.6727\end{displaymath}



John Kielkopf
2005-10-12