In calculating the ratio of first state excited to unexcited chemical species using the Boltzmann distribution, are we assuming that we only have species in the unexcited and first state excited states? It seems to me that this must be the case, because there's no part of the equation that accounts for the population of e.g. the second excited state.

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    $\begingroup$ we don't need to know the population of any other states to determine the ratio between the population of two states. $\endgroup$ – Tyberius Nov 10 '18 at 20:46

The expression for the Boltzmann equation is:

$\eta_{i} = \dfrac{g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}{\sum_{i} g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}$,

wherein $\eta_{i}$ the probability is to encounter the ensemble in state $i$ is in the ensemble, $g_{i}$ the degeneracy of state $i$, and $\epsilon_{i}$ its energy (I reckon the other variables are familiar to you).

To calculate the ratio between two states, you can simply divide the above expression for two different states by which you obtain:

$\dfrac{\eta_{i}}{\eta_{j}} = \dfrac{g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}{\sum_{i} g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)} \cdot \dfrac{\sum_{i} g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}{g_{j} \exp \left( \dfrac{-\epsilon_{j}}{k_{b}T} \right)} = \dfrac{g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}{g_{j} \exp \left( \dfrac{-\epsilon_{j}}{k_{b}T} \right)}$

The expression above shows that because states $i$ and $j$ are of the same ensemble, you do not have to use (or even know) the energy levels of the other states in that ensemble if you are only interested in the probability ratio between the two states.


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