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Does the wavelength setting affect the absorbance reading of a spectrophotometer? For example, if I am using a spectrophotometer at 550 nm to determine the protein concentration of my samples, would my results be different if I analyze them at 500 nm?

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  • $\begingroup$ I don't understand your question, can you clarify it? $\endgroup$
    – Chris
    Oct 11, 2015 at 18:46
  • $\begingroup$ The change in absorption due to concentration (Beer's Law) is the most at the wavelength of maximum absorption. But using another wavelength should give you similar results, but they would be so accurate. $\endgroup$
    – TanMath
    Oct 11, 2015 at 20:03
  • $\begingroup$ correct me if I'm wrong. The change of the wavelength would result in a lower absorption, but it will be similar to the one with the bigger wavelength. $\endgroup$
    – micheal
    Oct 11, 2015 at 20:13
  • $\begingroup$ Molecules will have a certain range of absorbance with a peak at a certain point. A wavelength longer than the peak absorbance and shorter than the peak absorbance will result in more light being recorded by the detector. You can determine peak absorbance by taking several readings of the same sample and varying the wavelength of the spectrophotometer. you will need to make very small changes around the peak to find the exact peak wavelength, but you should be able to find it this way. On the opposite sides of the peak, absorbance will fall off to near 0 if you have calibrated for solvent. $\endgroup$
    – AMR
    Oct 11, 2015 at 20:32
  • $\begingroup$ @AMR perfectly described what I mean by wavelength of maximum absorption. You should do your analysis at this wavelength. But if you do it at another wavelength, then it will give you similar results but the result would be less accurate. $\endgroup$
    – TanMath
    Oct 11, 2015 at 20:54

2 Answers 2

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Light absorption happens because of interaction of light with electrons (in a way) such that the energy of the light is transferred to some electrons that moves them to a higher energy state.

As per quantum chemistry (Bohr's model of hydrogen can be referred for a basic understanding), there are distinct energy states (solutions of the Schrödinger equation) and there are no intermediate levels. Therefore, to excite an electron to a higher energy level the energy of the incident light should match the energy gap between the ground state and the excited state.

In a molecule, there are different electrons, that reside in different orbitals and environments; only a few electrons can be exited by the energy that UV-Visible range of the electromagnetic spectrum can provide. These electrons are typically referred to as chromophores. You should refer to a book on spectroscopy for more details. A molecule can have multiple chromophores and each may absorb a different wavelength.

Having said that, there are energy bands (multiple closely spaced energy levels) in the molecule instead of fixed energy state which allows absorption in a range of wavelengths around a value at which the absorption is maximum. The wavelength vs absorption graph will give a peak at the maximum (which usually looks symmetrical but not always necessarily be so).

enter image description here

From ChemWiki

The width of this absorption band would differ between different chromophores; so in your hypothetical case, some chromophore which has an absorption maximum at 580nm may absorb 530nm while another chromophore with same maximum, may not.

The Beer-Lambert's law will hold for all wavelengths that can be absorbed i.e. the absorbance would be proportional to the concentration of the molecule (unless there are complex phenomena like energy transfers that may be dependent on concentration; this is unlikely in dilute solutions).

As a practical suggestion, I would say that if your spectrophotometer has fixed options then choose the wavelength that is closest to the absorption maximum.

If you see the above example graph and let's say your aim is to quantify nitrophenolate in a mixture of nitrophenol and nitophenolate. Then a measurement at 350nm instead of 400nm would give you erroneous result. In other cases a 50nm difference may not matter (for e.g. measuring at 450 for the nitrophenolate). It all depends on what you are analysing. You should know the molecule that you are analysing.

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  • $\begingroup$ The Bohr model is only valid for hydrogen. However the Schrödinger Equation takes into account all the rest... so I would probably edit, to just mention Schrödinger. $\endgroup$
    – AMR
    Oct 12, 2015 at 7:59
  • $\begingroup$ From Chemistry 8th Ed - Zumdahl & Zumdahl "At first Bohr's model appeared very promising. The energy levels calculated by Bohr closely agreed with the values obtained from the hydrogen emission spectrum. However, when Bohr's model was applied to atoms other than hydrogen, it did not work at all. Although some attempts were made to adapt the model using elliptical orbits, it was concluded that Bohr's model was fundamentally incorrect." $\endgroup$
    – AMR
    Oct 12, 2015 at 8:08
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    $\begingroup$ @AMR We are not even arguing which is the best atomic model. I had mentioned Bohr's model because it is easy to understand for a beginner in spectroscopy. Though it is obsolete, the fundamentals that describe energy levels are not incorrect even though they are applicable only for hydrogen atom. I cannot take the OP through a course on quantum mechanics. Nothing that I had said is misleading I guess; however, for your satisfaction I made the edit :) $\endgroup$
    – WYSIWYG
    Oct 12, 2015 at 8:15
  • $\begingroup$ I would probably have mentioned something about conjugated systems reducing the energy of the photon needed to get the electrons to change energies, resulting in "absorbing" longer wavelengths and probably not worried about mentioning Bohr and the Man with the Cat, as the OPs lab exam is more likely to have a question about why PNP can absorb light in the visible spectrum that about quantum theory,... but then that is just me. $\endgroup$
    – AMR
    Oct 12, 2015 at 9:04
  • $\begingroup$ @AMR Conjugation is just one case but there can be several ways including splitting of orbitals, interaction with solvent or other dipoles in the molecule, etc. That will go in a different direction altogether. Best to avoid that IMO. $\endgroup$
    – WYSIWYG
    Oct 12, 2015 at 9:18
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In addendum, keeping the wavelength around the absorption maximum for the light-attenuating species avoids introducing error into Beer's Law. There are limitations to the law, outlined here where we can see that around the lambda max, there's little change in absorption per unit wavelength:

enter image description here

In case B, this would end up returning a bad extinction coefficient! You can experimentally determine this point either by testing the solution at different wavelengths or more accurately, running a program where the machine goes through it's entire wavelength range and measures absorption for you. You'll get a parabola where the maximum is your lambda max, so it's solvable by calculating the maxima if necessary (take the first derivative, set y'= 0, solve for y when x = (y'=0) for the associated ordered pair). You also want to be careful when you're using something like the Bradford assay where the bound form of the dye shifts the lambda max!

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  • $\begingroup$ Nice information +1. However, in an experiment I would not expect the wavelength of the incident light to change (if the light is monochromatic). Also, I do not understand Figure B. Why would absorbance vs concentration become non-linear for non-λmax wavelengths? $\endgroup$
    – WYSIWYG
    Oct 13, 2015 at 5:55
  • $\begingroup$ I think this is the explanation here. As far as the math goes I'll admit I'm not actually 100%! $\endgroup$
    – CKM
    Oct 13, 2015 at 19:08

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