# Stimulated emission.

1. Mar 14, 2013

### 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

2. Mar 14, 2013

### Jano L.

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

3. Mar 14, 2013

### BareFootKing

Why can we neglect spontaneous emission?

4. Mar 14, 2013

### Staff: Mentor

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

5. Apr 3, 2013

### lightarrow

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$\nu$3/c3 * 1/{exp(h$\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$\nu$3/c3
ii) B12 = B21

Last edited: Apr 3, 2013