Q.

A specific dye when dissolved in water was found to absorb maximum light at 4500Ao. Its maximum fluorescence emission is at 5000Ao. On the average, the number of fluorescence quota is 52% of the number of quanta absorbed. Calculate the percentage of absorbed light that is emitted as fluorescence. Use wavelengths of maximum absorption and emission.

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answer is 46.9.

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Detailed Solution

(i) Energy (E) absorbed by each photon=hv=hcλ. Hence:E=6.626×10−34Js×3.0×108ms−4500×10−10m=4.4×10−19J(ii) Energy (E) emitted by each photon=hcλ=6.626×10−34Js×3.0×108ms−5000×10−10m=3.97×10−19J∴ Energy fraction emitted=3.97×10−19J4.4×10−19J×52100×100=46.9%
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