Beer's law: UV-Vis absorbance values exceeding 2

Question

In classical analytical chemistry, absorbance values in Beer's law can theoretically range from 0 to infinity.

As general rule of thumb neither high, nor very low absorbance are used for constructing calibration curves. The reason for avoiding high absorbance is that when absorbance is 2, only 1% light reaches the detector and with 3, only 0.1% light reaches the detector. Analytical chemists were taught to avoid absorbance > 1.5. Low absorbance values are avoided because it hard to distinguish the light beam with and without the cuvet in place. The question then arises how much is the real absorbance and how much is contributed by noise.

Nowadays (= within the last decade), I am seeing UV-Vis spectrophotometers have absorbance ranges from 0 to 3, and some have even more. One day a PhD student showed an absorbance > 5 for a spectra, apparently without even thinking twice on the meaning of absorbance. Similarly, researchers showed an absorbance spectrum with the maximum at absorbance of 3.

My question is this a fundamental flaw in Beer's law that high absorbance values are erroneous or it is limitation in detector technology and the intensity of light sources? Please ignore the chemical reasons for deviations in Beer's law.

Imagine, if we had very high intensity light source, such as xenon arc lamp, instead of a typical deuterium or a tungsten lamp, would we get less error in the absorbance measurements of concentrated solutions? Absorbance is a ratio technique, but the number of photons reaching the detector will be high, even with conc. solutions, with intense light sources.

Can any spectroscopist shed some light on this issue (no pun intended) whether high intensity sources would beneficial in UV-Vis spectrophotometry of concentrated solutions?

Answer 1

Beer's law limitation can be more easily understood if the coefficient $\epsilon$ is considered as a cross-section effect. Let's start with a broad beam or a bundle of rays having a section $A$ (in $m^2$). Its measured intensity is $I_o$. Now suppose there exist plenty of tiny obstacles suspended in the bundle of rays, and that each such particle behave like a dark disk perpendicular to the bean, with a surface or a cross-section $\sigma$. $\sigma$/A is the fraction of $I_o$ that is blocked off by one such particule. If there are $n$ particules in the whole cylindrical bundle, the fraction of $I_o$ that is captured is $\Delta$$I$/$I_o$ = $n·\sigma$/$A$.

Now let l be the length of the bundle where the particules are present. The volume of this bundle filled with dark particles is $V=A·l$. So the fraction of $I_o$ absorbed by the particules is $\Delta$$I$/$I_o$ = $n·\sigma ·l/A·l$ = $(n/V)·\sigma ·l$ = $c·\sigma ·l$, if $c = n/V$ is the concentration of the particules in the volume crossed by the beam.

Going to the limit, this expression becomes : $dI/I_o$ = - $\sigma ·l·dc$. Integration gives : $ln(I/I_o)$ = - $\sigma·c·l$

This last expression is similar to Beer's law : $log(I/I_o) = - \epsilon ·c'·l$. I have written $c'$ because in the beginning of my text $c$ is the number of particules per volume unit. And the concentration in Beer's law is in mol/L. $c' = c/N_o$, where $N_o$ is Avogadro's number.

In the end we obtain the final formula : $\epsilon = N_o·\sigma /2.3$

Now if a particule is a disk of $0.2$ $nm$ diameter (dimension of a small molecule), the corresponding $\epsilon$ is about $10^5 L mol^{-1} cm^{-1}$. This is the order of magnitude of $\epsilon$ for the usual dye.

But if $\epsilon$ = $N_o·\sigma /2.3 = 10'400 m^2/mol$. And this is the point where I would like to go today. It means that if one mole of the dye molecules were all to be touched by the beam, they would have to cover a surface of $10'400 m^2$. This is as large as a football ground. It also means that when the concentration becomes too high, Beer's law does not apply any more. The majority of the molecules of the dye are hidden behind others.

You may even calculate that when $c > 1/ (\epsilon · cm), $ then $ : I/I_o < 0.01$, the absorbance is $> 2$, and this is the limit of application of Beer's law, because most of the molecules are hidden behind some other ones at higher values of $c$.

Sorry for the length of the demonstration.