Stimulated Emission: Understanding Its Changing Rate

[BareFootKing]

I am having trouble understand why

is true I would think that the rate in which N2 is changing is the rate of stimulated emission and spontanous emission together. Why is it just the rate of stimulated emission


It is in the mathematical model section
http://en.wikipedia.org/wiki/Stimulated_emission

 

[Jano L.]

That's because the article is about stimulated emission only. The spontaneous emission is neglected.

 

[BareFootKing]

Why can we neglect spontaneous emission?

 

[mfb]

There are some systems where spontaneous emission can be neglected, and some where this is not possible. The article considers the first type only.

 

[lightarrow]

I am having trouble understand why

is true I would think that the rate in which N2 is changing is the rate of stimulated emission and spontanous emission together. Why is it just the rate of stimulated emission

Actually, it's not that we neglect spontaneous emission in this case, it's that *we don't consider it* by definition, because we are computing the rate of decay by stimulated emission only.

If we consider *all 3 processes happening*, as it happens in every real system, then dN₂/dt is:

dN₂/dt = -N₂ A₂₁ - N₂B₂₁ ρ(\nu) + N₁B₁₂ ρ(\nu).

The 3 coefficients A₂₁, B₂₁ and B₁₂ are defined by the relations:

1. Spontaneous emission dN₂/dt = -A₂₁N₂
2. Stimulated emission dN₂/dt = -B₂₁N₂ ρ(\nu)
3. Absorption: dN₁/dt = -B₁₂N₁ ρ(\nu)

To find the relationship between the 3 coefficients we can consider a system at thermodynamic aequilibrium, in which ρ(\nu) is that of the blackbody:

ρ(\nu) = 8πh\nu³/c³ * 1/{exp(h\nu/kT)-1}

and in which N₂/N₁ is given by Boltzmann' statistic:

N₂/N₁ = exp{-(E₂-E₁)/kT} = exp{-h\nu/kT}.

At the aequilibrium, dN₂/dt = 0 (as well as dN₁/dt) so:

0 = -N₂ A₂₁ - N₂B₂₁ ρ(\nu) + N₁B₁₂ ρ(\nu)

Substituting the relations for ρ(\nu) and N₂/N₁ and understanding that the coefficients don't have to depend on T, it's easy to find:

i) A₂₁ = 8πh\nu³/c³
ii) B₁₂ = B₂₁