Atomic transitions and density: clarification 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?"
 A: 
1) does "atomic density" correspond to "atoms in a volume unit that partecipate to the stimolated emission process"?

"Atomic density" means number of total rubidium atoms per volume of gas, not just the ones that happened to participate in the absorption/emission process.

2) is $\sigma$ the cross section for a particular transition? (for example the Rubidium transition $D_2$)

Yes, it's the cross-section at that particular laser wavelength and gas temperature/pressure/concentration.

3) $N$ depends on temperature. Is this due to the fact that if the tempertaure is higher, the atomic electrons can have more energy and occupy more external levels?

Actually, $N$ does not depend on temperature, since there are always the same number of Rb atoms per volume in the cell, irregardless of how hot it is. While $\sigma$ depends on temperature, this is counterbalanced by the temperature-dependence of $I$. 
This may seem confusing, but it has a simple explanation: an increase in temperature decreases $\sigma$ because the linewidth of the transition increases (due to a mixture of Doppler or pressure broadening). But that also means that $I$ will decrease, since the absorptive cross-section decreased.
Alternately, you can see this from the fact that the equation
$$N=-\frac{1}{\sigma L}ln \frac{I}{I_0}$$
is just a rearrangement of Beer's law (from high school chemistry):
$$I=I_0e^{-LN\sigma}$$
where $N$ is the number of atoms per volume, a constant. 
EDIT: Based on a link sunrise gave in the comments, the setup is actually a cell with excess solid Rb, which is heated to give a vapor pressure. Thus in this setup, $N$ is not constant, but instead depends on the vapor pressure of the solid/liquid Rb as a function of temperature. So it has nothing to do with the atomic electrons and levels, but rather is from high school chemistry.

4) If the cell is enlighened with blue light, the peaks are deeper (LIAD effect) and -mantaining T constant - N assumes an highter value. Generally some rubidium atoms attak 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 attacked to the walls aren't be counted. Is it correct? So blue light permits to (a part of) ex-attacked 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 highter if I use blue light?"

If you shine blue light on the cell, solid Rb on the cell walls will get knocked off and go into the vapor phase, and thus $N$ increases. Thus $I$ will decrease (because there are more Rb atoms absorbing the laser), making the absorption peaks deeper.
