6
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ High intensity light sources, per se, do not solve the problem. Some expensive instruments can get to 5 or 6. But Tom O'Haver has a paper (I will send it if I can find it again!) where he discusses curve fitting transmission profiles and this effectively gets you to much higher absorbances: 100 or higher! Of course, there are no photons at the center of the profile, but the width carries information. But this is NOT a panacea. If I cannot find the paper, just ask Tom: it is very nice and I have computer sims demonstrating it. $\endgroup$ – Ed V Feb 8 at 19:25
  • $\begingroup$ Okay, great to know. Do you recall the title of the paper. Absorbance of 100! This is amazing. He spent a significant portion of his life on xenon light source. $\endgroup$ – M. Farooq Feb 8 at 19:27
  • $\begingroup$ Of course, this is really for atomic profiles! No way UV-vis with solutions could use this idea. I will send along the paper's title ASAP! $\endgroup$ – Ed V Feb 8 at 19:35
  • $\begingroup$ High intensity UV/Vis light sources are terrible for your analyte, because it gets heated and/or destroyed. $\endgroup$ – Karl Feb 8 at 20:49
  • $\begingroup$ @Karl, of course lasers are to be avoided, but xenon arc lamps are common in spectrofluorometers and at least commercial UV-Vis. The trick is not to irradiate the sample but have brief pulses. $\endgroup$ – M. Farooq Feb 8 at 20:56
-2
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Maurce, in a real spectrophotometer a very small portion of the solution is irradiated. The solid angle so formed encloses a relatively small number of molecules. $\endgroup$ – M. Farooq Feb 9 at 4:29
  • $\begingroup$ Maurice, Interesting perspective on cross-sections. Do you have a reference which shows this limit in terms of cross sections. $\endgroup$ – M. Farooq Feb 9 at 13:25
  • 3
    $\begingroup$ I think that explaining absorption as if molecules are discs shading others behind them is a misconception. The important property is the transition dipole, (think of a vector) and this takes various orientations in space, depending on molecular orientation, which determines the probability of absorption along with its size. This can vary greatly, i.e. the extinction coefficient varies by many orders of magnitude between different types of transitions. The experimental limit on the Beer-lambert law is only signal to noise and not for any fundamental reason. $\endgroup$ – porphyrin Feb 10 at 19:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.