Deriving Expression for Population Fraction (Statistical Mechanics)

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SUMMARY

The discussion focuses on deriving the population fraction for a quasi-3 level solid-state laser gain medium, specifically under non-optical pumping conditions. The population fraction formula is established as \(\frac{n_{i}}{N} = \frac{g_{0}e^{0}}{g_{0}e^{0}+g_{1}e^{-\frac{10meV}{K_{B}T}}}\). The user seeks clarification on the degeneracies of the energy levels, with consensus that \(g_{0}=g_{1}=1\) is a valid assumption for this calculation. The temperature of 25°C is noted as relevant for determining the ground state population fraction.

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Collisionman
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Homework Statement



A quasi-3 level solid-state laser gain medium consists of a ground state manifold containing two energy levels within which a single electron can be promoted, with the second energy 10meV above that of the lowest level.

A. Where the gain medium is not optically pumped, derive the relationship for the population fraction held in the ground energy level.

B. Given the value for the ground state population fraction where the coolant holds the gain medium to a temperature of 25°C.

Homework Equations



Population Fraction: \frac{n_{i}}{N} =\frac{g_{i}e^{-\frac{1}{K_{B}T}E_{i}}}{\sum_{i}g_{i}e^{-\frac{1}{K_{B}T}E_{i}}}



The Attempt at a Solution



For the ground energy level, E_{0}=0 and, for the second energy level, E_{1}=10meV. So, for the ground energy level, the population fraction is;

\frac{n_{i}}{N} = \frac{g_{0}e^{0}}{g_{0}e^{0}+g_{1}e^{-\frac{10meV}{K_{B}T}}}

I'm unsure of what the degeneracies are for E_{0} and E_{1}. Would I be right in assuming they're just g_{0}=g_{1}=1?

If I knew this I'd have no problem finishing the rest of the question which is just a matter of substitution.

Any help appreciated and thanks in advance.

Regards,

Collisionman
 
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Collisionman said:
Would I be right in assuming they're just g_{0}=g_{1}=1?

Yes.

Note that you don't really care about all possible degeneracies. Like for example, you don't care at all that electrons have spin 1/2, as those factors cancel out. All you care about is the relative number of states corresponding to each energy.
 

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