A certain dye absorbs light of $λ = 4000 Å$ and then fluorescences light of $λ = 5000 Å$ . Assuming that under given conditions 50% of the absorbed energy is re-emitted out as fluorescence, calculate the ratio of the number of quanta emitted out to the number of quanta absorbed

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$\frac{3}{8}$

$\frac{5}{8}$

$\frac{8}{3}$

$\frac{8}{5}$

answer is A.

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Detailed Solution

Number of quanta emitted out to the number of quanta absorbed can be calculated as;

$N_{1} \times \frac{h c}{4000} \times 0.5 = N_{2} \times \frac{h c}{5000} \Rightarrow \frac{N_{2}}{N_{1}} = \frac{5}{8}$

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