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$\begingroup$ So, what we have to do is find the recombination coefficient for all levels except for the ground state and sum them up. But can't an electron have an infinite number of exited states? And can we approximate the cross section to just depend on the principal quantum number (Just use the atomic radius)? $\endgroup$– ChandrahasCommented Feb 16, 2017 at 5:54
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$\begingroup$ Can I just consider the velocity distribution to be constant if the electrons didn't have sufficient time to attain thermal equilibrium. Approximately how will it take to attain thermal equilibrium? Thanks $\endgroup$– ChandrahasCommented Feb 16, 2017 at 5:57
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$\begingroup$ Good questions. 1) There are indeed an infinite number of excited states, but the recombination coefficients decrease rapidly enough at higher energy states that the sum converges to a finite value. For ionized hydrogen at 5000 K, this sum is $6.8 \times 10^{-13}$, and subtracting out the ground state leaves you with $4.5 \times 10^{-13}$. 2) The electron-electron collision cross section is generally much larger than the electron-ion recombination cross section, so the electrons thermalize rapidly and it is usually appropriate to assume a thermal distribution for their velocities. $\endgroup$– kleingordonCommented Feb 16, 2017 at 17:18
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$\begingroup$ Regarding quantum numbers: it depends on what level of detail you are looking for. If you care about tracking everything exactly, it turns out there is a difference in the rate of recombination to different $l$ states at a given $n$ value. $\endgroup$– kleingordonCommented Feb 16, 2017 at 17:21
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$\begingroup$ I also forgot to mention that, at sufficiently high densities, collisional ionization (electron collides with an atom/ion, knocking off a bound electron) will also be playing an important role in keeping the plasma ionized. Depending on your application, either photoionization or collisional ionization can be the dominant effect that keeps the plasma ionized. To find the equilibrium situation, you need to balance the appropriate ionization rate with the recombination rate, which you calculate as I've discussed in my answer $\endgroup$– kleingordonCommented Feb 16, 2017 at 23:06
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