I am having difficulties understanding the relation between absorption spectrum and gas concentration. The online resources I found, including many questions here, do not clarify my doubts.

Take for example this picture from wikipedia. It lacks any information about what concentration/partial pressure of each species results in the given absorption spectrum:


I don't understand how the absorption spectrum in the picture above relates to the concentration of the species in the atmosphere. For example, assuming that the concentration used for that spectrum is the average earth atmospheric concentration (i.e. 340ppm of CO2), I wonder what happens when the concentration halves. Do we get only 50% absorbance in the 4.3 microns band for CO2 or do we get a bit more than that because CO2 will still be fairly opaque at this concentration/wavelength?


1 Answer 1


The Beer-Lambert Law says that the transmittance of a sample is given by $T = e^{-n \sigma \ell}$, where $\ell$ is the path length, $\sigma$ is the absorption (or scattering) cross-section of the scattering species, and $n$ is the concentration of the species (molecules per volume.) If you halve the concentration ($n' = n/2$), then we can see from the above formula that the new transmittance is $T' = e^{-n' \sigma \ell} = (e^{-n \sigma \ell})^{1/2} =\sqrt{T}$.

In the case of the graphs above, it's important to note that what you're calling "100%" absorption is really something like 99.999999% absorption and 0.000001% transmission. There's never going to be perfect absorption in this sort of situation, because the scattering/absorption process is probabilistic. In general, the above argument says that if a fraction of $1 - T$ of the photons are absorbed before you halve the concentration, then $1 - \sqrt{T}$ will be absorbed afterwards. Exactly how big of a difference this makes depends on exactly how close you are to 100%. If you were scattering 99% of the light before, then you'll be scattering only 90% after. If it was 99.999999% $(1 - 10^{-8})$ before, then it will be 99.99% after ($1 - 10^{-4}$).

The graph below shows the new absorbance $A'$ (vertical axis) as a function of the old absorbance $A$ (horizontal axis). The fact that the slope of the graph becomes infinite as $A \to 1$ reflects the idea that it really matters how close to 100% absorbance you are. On the other hand, at very low absorbance (near $A = 0$), you can see that halving the concentration simply halves the absorbance.

enter image description here

  • $\begingroup$ That is extremely clear, thank you! From wikipedia it seemed that the beer-lambert law was not exponential, maybe because it is assuming a logarithmic scale for A? $\endgroup$
    – Redirectk
    Dec 21, 2022 at 21:09

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