5. In the fluorescence experiment, the optics are set up such that the fluoresce

Question

5. In the fluorescence experiment, the optics are set up such that the fluorescence is collected from roughly the middle of the cuvette. The excitation light has to penetrate 5 mm into the cuvette and the emitted fluorescence has to pass through 5 mm of liquid to get out of the cuvette. Therefore, a sample that is ten times more concentrated will not produce a ten-fold larger signal because the more concentrated sample absorbs more of the incoming light and more of the outgoing light. Assuming excimer formation is negligible at the lower concentrations, how much stronger would you expect the fluorescence to be from the 10* pyrene sample compared to the 10 sample? How does this compare to what you measure? Be as quantitative as possible.

Explanation / Answer

According to the Beer-Lambert's law of absorption of light through liquids, we have:

A = e * l * c

where, A = Absorbance of light, e = extinction coefficient, l = lenght of path travelled by light in the medium, c = concentration of the medium/liquid

Thus, here we have two liquids of different concentrations and assuming their excimer formations are negligible, the absorbances would be given by:

A1 / A2 = e * 5 * 10-4 / e * 5 * 10-5

This gives the ratio:

A1/A2 = 10

or A2*10 = A1

Thus, the diluted sample will have 10 times lower absorbace as compared to the concentrated sample of pyrene.

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