The cutoff frequency $\nu_{\rm
max}$ is much higher than any
radio frequency; it corresponds to infrared radiation. As we will soon
calculate, the typical electron speed in a $T \sim 10^4$ K HII
region is $v \approx 7 \times 10^7$ cm s$^{-1}$ and the typical impact
parameter is $b \approx 10^{-7}$ cm, so $\nu_{\rm max} \approx 10^{14}$
Hz, corresponding to the near-infrared
wavelength $\lambda_{\rm min} \approx 3\,\mu$m.

Next we need to find the distributions
of $v$ and $b$ to evaluate
the radio emission from an HII region. The distribution of $v$ depends
on the electron temperature $T$. The distribution of $b$ depends
on
the electron number density $N_{\rm e}$ (cm$^{-3}$) and on the ion
number
density $N_{\rm I}$ (cm$^{-3}$).

In LTE, the average kinetic energies
of electrons and ions are
equal. Since the electrons are much less massive, their speeds are much
higher, and we can make the approximation that the ions are nearly
stationary during an interaction.