Thermal occupation of energy levels

When radiation is absorbed, inducing a transition from $E_i$ to $E_f$, the degree of attenuation depends on the population of the energy level $E_i$. If many molecules in the system populate this energy level, then there will be more attenuation due to more molecules that can undergo the transition $E_i$ to $E_f$. Since most spectra are recorded at a finite temperature $T$, it is the temperature that determines the population. In general, given a large number of molecules at temperature $T$, we can only determine the probability that a given molecule is in its energy level $E_i$. This probability $P(E_i)$ is given by

$$P(E_i) = {g(E_i) e^{-E_i/k_{\rm B}T} \over \sum_j g(E_j) e^{-E_j/k_{\rm B}T}}$$

Here $g(E_i)$ is the degeneracy of the energy level $E_i$ and $k_{\rm B}$ is Boltzmann's constant. The denominator is needed to normalize the probabilities so that

$$\sum_i P(E_i) = 1$$

The relative probability of the population of the energy level $E_i$ to the population of $E_f$ is then

$${P(E_i) \over P(E_f)} = {g(E_i) e^{-E_i/k_{\rm B}T} \over \sum_j g(E_j) e^{-E_j/k_{\rm B}T}} {\sum_j g(E_j)e^{-E_j/k_{\rm B}T} \over g(E_f) e^{-E_j/k_{\rm B}T}}$$

$$= {g(E_i) \over g(E_f)} e^{-(E_i-E_f)/k_{\rm B}T}$$

At $T=300$ K, $k_{\rm B}T \approx$ 0.6 kcal/mol or 2.5 kJ/mol. The spacing between typical rotational and vibrational levels is much less than this, so at room temperature $T=300$ K, there are many rotational and vibrational levels that are populated. By contrast, the spacing between electronic energy levels is considerably larger than 2.5 kJ/mol, so at room temperature, usually only the ground electronic state is occupied.