# Deriving the absorption coefficient from state transitions

This source states that the absorption coefficient $$\gamma$$ is proportional to the difference in particle number density $$\Delta n$$ per unit volume between the initial and the final rotational state and also proportional to the dipole moment $$\mu_T$$ $$\gamma \propto \Delta n\cdot \mu_T$$ According to the Beer-Lambert Law the absoprtion coefficient is also equal to: $$\gamma=\log_{10}\bigg(\frac{I_0}{I}\bigg)\cdot \frac{1}{c\cdot l}$$ Where $$I_0$$ and $$I$$ are the incidental and transmitted intensity, $$c$$ is the concentration of the solution and $$l$$ the traveled length through the solution by the incidental beam.

From this Beert-Lambert law I woud deduce that the absorption coefficient is in terms of per 1 unit length and per 1 unit concentration.

I would assume that the difference in number density $$\Delta n$$ is analogous to the concentration of a solution.

But how is $$\gamma$$ then proportional to the $$\Delta n$$ if it should be per 1 unit of $$\Delta n$$ just as it is per 1 unit of sample concentration?

• $\Delta n$ is the population difference between the two rotational states of the same molecule per unit volume and is mostly a function of temperature as dictated by the Boltzmann distribution. – Paul Jun 19 '20 at 19:31
• @Paul In that case, I would assume that the total number density of particles $N$ is the concentration $c$ instead and that the absorption coefficent $\gamma$ should be per 1 unit of $N$, correct? Such that the absorptivity $\log_{10}\bigg(\frac{I_0}{I}\bigg)$ is proportional to: $$\log_{10}\bigg(\frac{I_0}{I}\bigg)\propto \Delta n \cdot \mu_T \cdot N \cdot l$$ Is this correct? – Phy Jun 19 '20 at 19:51
• When the population difference is zero there is no net absorption because of stimulated emission, i.e 'absorption' from the upper state to the lower state plus a photon. This is common in rotational spectroscopy where levels have very similar populations at room temperature, unlike in the uv -vis and near ir. The $\gamma$ in the second equation is the extinction coefficient, a property of the molecule and proportional to the transition dipole, and not the same as in your first eqn, it seems that $\Delta n \mu\equiv \gamma c l$ but you should check. – porphyrin Jun 19 '20 at 22:07
• @porphyrin The source mentions right beneath Example 1 that it can also be the absorption coefficient instead of the extinction coefficient. How is that possible? – Phy Jun 20 '20 at 14:02
• These are different words for the same thing, it's best to look at how it is defined and then decide what it is. Not all authors use the same definition. Extinction coefficient is still very widely used in the scientific literature and textbooks. – porphyrin Jun 21 '20 at 7:24