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Why do I take the logarithm of the Saha formula when calculating the ratio between ionized and neutral atoms of an element? What is the physical meaning?

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closed as unclear what you're asking by Kyle Oman, Jon Custer, Rob Jeffries, Mo_, Cosmas Zachos Jan 21 '18 at 0:56

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The Saha equation relates one ionization state to another ionization state (often the ground state): $$ \frac{n_{i+1}}{n_i}\sim T^{3/2}\exp\left[-\varepsilon/k_BT\right] $$ where $n_i$ are the $i$th ionization state populations and the other variables take their normal meaning.

To see the reason why one would take the logarithm, note that $\log[x]\geq0$ for $x\geq1$ and $\log[x]<0$ for $x<1$. Since the Saha equation gives relative populations of two ionization states, then taking the logarithm can show more quickly which state is more populous: $$ \log\left[\frac{n_{i+1}}{n_i}\right]\to \begin{cases} >0 & \text{if }\, n_{i+1}>n_i \\ \sim0 & \text{if }\, n_{i+1}\sim n_i \\ <0 & \text{if }\, n_{i+1}<n_i \end{cases} $$ So any positive value means the higher state is more populated than the lower state, while the negative values indicate the lower state is more populated.

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  • $\begingroup$ Don't see how comparing with zero is a simpler process compared to comparing with one. $\endgroup$ – Weijun Zhou Jan 30 '18 at 18:19
  • $\begingroup$ @WeijunZhou whether it is a simpler process or not, which is probably subjective, is irrelevant to the fact that it does happen and the answer I gave is the why. $\endgroup$ – Kyle Kanos Jan 30 '18 at 18:24

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