 Solution

For $h\nu \ll k_{B}T$, expand the exponential in the thermal free-free emissivity:The energy emitted per unit frequency is therefore approximately flat, with the weak Gaunt factor dependence superposed. At these low photon energies, most of the thermal electrons can participate in emission; there is no exponentially small population threshold.

For $h\nu >k_{B}T$, the spectrum turns down approximately as $e^{-h\nu /(k_{B}T)}$, apart from the Gaunt factor. Conservation of energy requires an emitting electron to supply at least $h\nu$. The high-energy tail of its Maxwell-Boltzmann distribution is exponentially rare, which produces this cutoff.

“Flat” refers to $j_{\nu }$, an energy spectrum per unit frequency. The emitted photon number spectrum instead contains $1/(h\nu )$, while energy per logarithmic frequency interval is $\nu j_{\nu }$. At sufficiently low frequency the plasma can also become optically thick to free-free absorption; the given optically thin emissivity is then not the escaping spectrum.