I think you can get a estimate like this.
For a semiconductor with no split in the quasi-Fermi levels, the electrons and holes take their intrinsic values (carrier density) $n_0$ and $p_0$ ($cm^{-3}$). The charge carriers are in equilibrium with the thermal photons being absorbed and emitted inside the material. So if we calculate the emission rate of thermal photons then we know the time constant for how long the thermally generated carriers will last before recombining (because at equilibrium upwards rates and downward rates must balance).
Let's assume perfect bimolecular recombination, then the rate of thermal emission is,
$\frac{\partial n}{\partial t} = Bn_0p_0$,
where $B$ is the bimolecular recombination coefficient, for GaAs, $B=7\times10^{-10}$$cm^{6}s^{-1}$, and the intrinsic carrier density is, $n_i=n_0=p_0=2\times10^{6}$$cm^{-3}$. This gives a transition rate of 2800 $s^{-1}$.
This seems a bit slow. But it's correct for the assumptions, namely because we assumed an un-doped intrinsic semicondutor (the carrier density is very low).
For more information I recommend 'Light-Emitting Diodes by E. Fred Schubert’, search for the vanRoobroeck-Shockley equation.