# Beer-Lambert Law

1. Sep 22, 2005

### sid_galt

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

$$ln\frac{I}{I_0}=-ebc$$
where
I is the final intensity of light, $$I_0$$ 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?

2. Sep 22, 2005

### Gokul43201

Staff Emeritus
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. Sep 22, 2005

### sid_galt

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

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

I checked my calculations again. You are correct, it should be 54.9%/cm.

4. Sep 22, 2005

### sid_galt

5. Sep 23, 2005

### sid_galt

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. Sep 23, 2005

### Gokul43201

Staff Emeritus
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. Sep 25, 2005

### Gokul43201

Staff Emeritus
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.
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.