 Solution

Take the gas pressure $P$ to vanish at the exterior boundary and the atmosphere to be thin enough that surface gravity of a star is constant. With constant electron-scattering opacity, negligible radiative acceleration, and hydrostatic equilibrium in optical depth,If the radiative force is retained, $dP/dr=-\rho (g-\kappa F/c)$ for gas pressure, and the same result holds with the constant positive effective gravity $g_{\mathrm{eff}}=g-\kappa F/c$. A nonzero exterior gas pressure instead gives $d\log P/d\tau =1/(\tau +\kappa P(0)/g_{\mathrm{eff}})$. Total gas-plus-radiation pressure has the nonzero radiative surface term from part (ii), so the advertised exact $1/\tau$ formula presumes gas pressure with zero exterior gas pressure.