A laser passes though a cell containing vapor of an alkaline element which causes atomic transitions. I was told that the formula for the atomic density in a Rubidium cell, for example, is given by:
$$N=-\frac{1}{\sigma L}\ln \frac{I}{I_0}$$ where:
$\sigma$ is the cross section and is inversely proportional to the temperature T
$L$ is the cell length
$I/I_0$ is the signal transmission on the peaks
But I have some questions:
Does "atomic density" correspond to "atoms in a volume unit that participate in the stimulated emission process"?
Is $\sigma$ the cross section for a particular transition? (for example the Rubidium transition $D_2$)
$N$ depends on temperature. Is this due to the fact that if the temperature is higher, the atomic electrons can have more energy and occupy more external levels?
If the cell is enlightened with blue light, the peaks are deeper (Light-induced atomic desorption (LIAD) effect) and, with $T$ constant, $N$ assumes an higher value. Generally some rubidium atoms attach on cell walls and blue light give them energy for being "free". $N$ is calculated in relation to the atoms that effectively make the atomic transition (thanks to peaks depth and sigma, is it right?) and the atoms that are attached to the walls aren't be counted. Is it correct? So blue light permits to (a part of) unattached atoms to give the transition and be counted in $N$. Probably this reasoning is confusing, I'm trying to respond to the question "Why the peaks are deeper and N is higher if I use blue light?"