Question

Fluorescein is a highly molecule that is often used for analytical chemistry.

A. If fluourescein is excited with light at 494 nm do you expect the emitted fluorescene to be at a wavelength that is longer or shorter than 494 nm? Briefly explain why.

B. Write the equation that shows the relqationship between fluorescene and concentration.

C. If you generated a calibration curve for fluorescein using excitation at 485 nm and then changed the excitation wavelength to 494 nm to measure a fluorescein sample, do you expect the concentation you calculate for the sample using the calibaration curve will be accurate? why?

Answer 1

A) The emitted fluorescence will be at a wavelength longer than 494 nm. This happens because the fluorescence wavelength is independent of excitation wavelength and always occurs from the lowest excited state to the ground state. This means that it does not matter whether we send the molecule to State 1, state 2, state 3 etc. ( energy in increasing order ). The fluorescence will always occur from State 1 to the ground state ( State 0 ). Since the energy gap between state 1 and 0 is less than the excitation energy we used, the emitted fluorescence occurs at a longer wavelength ( lesser energy ).

B)    This is the relation between I_f ( fluorescence intensity ) and concentration.

Is=K1. ebc

• k is a proportionality constant attributed to the instrument
• I_o is the incident light intensity
• ϵ is the molar absorptivity,
• b is the path length, and
• c is the concentration of the substrate.

C) The concentration measured must be accurate since the excitation wavelength would not change the fluorescence wavelength.