Fluorescence spectroscopy has better detection limits than UV-Vis spectroscopy.
ID: 524276 • Letter: F
Question
Fluorescence spectroscopy has better detection limits than UV-Vis spectroscopy. Why? (An answer without an explained reason would carry zero points) (b) Two fluorescent compounds X and Y are present in a mixture. How would you quantify the components X and Y in mixture, given steps clearly; (i) If excitation spectrum peaks of X and Y with peaks at lambda_ex, X = 300nm and lambda_ex, Y = 350nm at lambda_em, X = 400nm and lambda_em, Y = 500nm. (ii) If excitation spectrum peaks of X and Y with peaks at lambda_ex, X = 300nm and lambda_ex, Y = 305nm at lambda_em, X = 400nm and lambda_em, Y = 500nm. (c) The Excitation and emission spectra of quinine and anthracene are given on the next page. Use only those labels shown on the pots, appropriate to your answer to refer to excitation, emission maxima or any other spectral parameters that you would use to accomplish the task at hand.Explanation / Answer
Fluorescence spectroscopy is better than UV-Visible Spectroscopy because
1) Higher Detection limits even at very low concentrations (i.e upto 10-7).
2) The radiant power of the source is directly proportional to the intensity of the fluoroscent species.
3) In this technique, the measurement takes place at zero background levels.
b) The energy difference between the absorption and emission level of X is higher (since wavelenth difference is greater here and wavelength is inversly proportional to the Energy)
(i) X will emit with more extent compared to Y, therefore can be quantified.
(ii) X will emit with more extent compared to Y, therefore can be quantified.