Lasers/lifetime of metastable states on brightness of beam

[CAF123]

If we consider two laser beams with identical 3 level energy level schemes except that the metastable state for one has a lifetime of 1μs and the other, 100μs. Which will give the brighter (more intense) beam?

I gather that, since all things are identical up to the metastable state, the number of excited atoms in the energy level before the metastable state will be the same for both lasers(Is this correct, or is it a matter of probability?). From this excited state, they will spontaneously decay (stimulated emission unlikely here because of short lived state - please correct me if i am wrong) to the metastable state. Which will then give a brighter beam given the lifetimes above?

Many thanks.

 

[Cthugha]

I gather that, since all things are identical up to the metastable state, the number of excited atoms in the energy level before the metastable state will be the same for both lasers(Is this correct, or is it a matter of probability?). From this excited state, they will spontaneously decay (stimulated emission unlikely here because of short lived state - please correct me if i am wrong) to the metastable state. Which will then give a brighter beam given the lifetimes above?

This is a bit hard to say because "all things are identical" is a bit hard to interpret. You are intending to use cw pump, right? One important thing is the $\beta$-factor of the laser which basically gives the spontaneous emission from the metastable state to the lasing mode of interest divided by all spontaneous decay channels (radiative decay to other modes and non-radiatice decay). Basically, it is an indicator of how large losses are. Now the question is "all things are identical" whether you want these two lasers to have the same $\beta$ factor or the same radiative and non-radiative loss rates. Obviously both cannot be the same at the same time as $\beta$ will depend on the ratio of the timescale of the loss channel decay rates to the decay rate to the cavity mode.