Beer-Lambert Law: Calculating Light Absorption in Solutions

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In summary: I'm not an expert in this area. So, in summary, I don't think the energy of the excess (x-y)% light went anywhere.
  • #1
sid_galt
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Let's say a compound having a molar absorptivity of 20000 and a concentration of 3e-5 M is kept in a container 5 cm long.
This gives that each centimetre of solution will absorb 50% of the light passing through it since

[tex]ln\frac{I}{I_0}=-ebc[/tex]
where
I is the final intensity of light, [tex]I_0[/tex] is the initial intensity of light, e is the maximum molar absorptivity, b is the path length and c is the concentration.

The molecule is such that 20 J/cm2 of intensity of light is enough to excite all the molecules in a 1 cm range.

The frequency of the light is such that the energy of one photon is just enough to excite the first fundamental frequency of vibration in the compound molecule.

100J/cm2 is the intensity of light actually passed through the solution.

Since according to the beer lambert law, 50% i.e. 50J/cm2 of light has been absorbed by the 1rst cm of solution, where did the rest 30J/cm2 go when we know that its photons do not have enough energy to excite further fundamental frequencies of vibration in the molecule and that only 20J/cm2 is required to excite the first fundamental frequency?
 
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  • #2
Is this the entire question reproduced verbatim, or is some part of it the question and some of it your doubts ? Also, about the 50%/cm number - is that your calculation or is it given ? You realize that this can be calculated from the given data ? I get 54.9%/cm for this number, so I'm guessing that 50%/cm just a rough approximation. And to get this number, I'm assuming the molar absorbtivity is 20,000 M-1 cm-1 - because the units are not provided.

Please confirm/correct all of the above.
 
  • #3
Thank you for the reply.

Gokul43201 said:
Is this the entire question reproduced verbatim, or is some part of it the question and some of it your doubts ?

I was reading about the beer-lambert law and this doubt popped up in my mind. I thought the best way to phrase the doubt would be by asking such a question.

Since QM allows only certain vibration modes, I can't understand where the rest of the energy aside from the 20J/cm2 necessary to put the molecules in the first fundamental frequency of vibration is going?

The rest of the photons cannot excite the other vibrational modes as their frequency is resonant for only the first fundamental frequency.


Calculations
The concentration is in Moles, the length in cm and the molar absorptivity in M-1 cm-1

Also, about the 50%/cm number - is that your calculation or is it given ?

My calculation.
I assumed a concentration of 3e-5 and a molar absorptivity of 20000.

I get 54.9%/cm for this number, so I'm guessing that 50%/cm just a rough approximation.

I checked my calculations again. You are correct, it should be 54.9%/cm.
 
  • #4
Please can anyone help?
 
  • #5
I think this might be a better way to express myself.

If the intensity of a particular frequency of infrared light is so high that although according to the beer lambert law, x% of light should be absorbed, but the actually only y% of the light is necessary to put all the molecules in excited state where y < x, where did the energy of the excess (x-y)% light go?
 
  • #6
Your question is interesting, and I believe it probes an unstated assumption of the B-L Law. I'll have to do a quick calculation (just to make sure I'm not talking nonsense) before I go ahead and answer...perhaps later tonight.
 
  • #7
Okay, I think I may have slightly misunderstood your original question. I thought the numbers you presented were real values for a certain system, not numbers you made up.
sid_galt said:
I think this might be a better way to express myself.

If the intensity of a particular frequency of infrared light is so high that although according to the beer lambert law, x% of light should be absorbed, but the actually only y% of the light is necessary to put all the molecules in excited state where y < x, where did the energy of the excess (x-y)% light go?
The Beer-Lambert Law has regimes where it applies, and conditions under which it applies. If you look at the derivation of the law, you will see no restriction on the number of photons that can be absorbed "at a time" by a single molecule. This assumes either (i) a photon density that's lower or comparable to the molecular density (concentration), or (ii) sufficient number of available modes per molecule.

In your case you have a photon density of about 10^25 /m^2 and a molecular density lower than 10^16 /m^2, with only 1 available mode per molecule (actually you have also provided a number which may prove that the number of modes per molecule is much lower than 1, but I didn't do the calculation because of a problem with your units). It looks to me like B-L will not work in this regime.

Note : If you're throwing out numbers off the top of your head, that may easily be a reason why there's an apparent contradiction. For instance the real extinction coefficient may in fact be low enough that you never get anywhere near 20 J/cm^2 (actually should be W/cm^2), forget about going above it. The extinction rate depends on the micro-energetics of the system and you can not conjure up numbers for the two independently of each other.
 

1. What is the Beer-Lambert Law?

The Beer-Lambert Law, also known as the Beer-Lambert-Bouguer Law, is a mathematical equation that describes the relationship between the concentration of a solute in a solution and the amount of light it absorbs. It is commonly used in chemistry and spectroscopy to calculate light absorption in solutions.

2. How is the Beer-Lambert Law used in scientific research?

The Beer-Lambert Law is used to determine the concentration of a solute in a solution by measuring the absorbance of light at a specific wavelength. This information can then be used to study the properties of the solute, such as its concentration, molar absorptivity, and extinction coefficient.

3. What are the key assumptions of the Beer-Lambert Law?

The Beer-Lambert Law assumes that the solution is dilute, meaning that the concentration of the solute is low enough that it does not significantly affect the refractive index of the solution. It also assumes that the solute is uniformly distributed throughout the solution and that the light source is monochromatic and collimated.

4. How is the Beer-Lambert Law derived?

The Beer-Lambert Law is derived from the principles of Beer's Law and Lambert's Law. Beer's Law states that the amount of light absorbed by a solution is directly proportional to the concentration of the solute, while Lambert's Law states that the amount of light absorbed is also directly proportional to the path length of the light through the solution. Combining these two laws results in the Beer-Lambert Law.

5. Can the Beer-Lambert Law be applied to all types of solutions?

The Beer-Lambert Law is most accurate when applied to solutions that are colorless and transparent, and when the solute does not interact with the solvent. It can also be used for solutions that absorb visible light, but it becomes less accurate for solutions that absorb UV or IR light. Additionally, the law is not applicable to solutions with large concentrations of suspended particles or colloids.

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