A certain dye absorbs light of $λ = 4530 ˚ A$ and then fluorescence light of $5080 ˚ A$ . Assuming that under given conditions $47 %$ of the absorbed energy is re-emitted out as fluorescence, calculate the ratio of quanta emitted out to the no. of quanta absorbed. ( $X \times 10$ in this form to nearest integer)

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Solution

Let $n$ , photon be absorbed by dye and $n_{2}$ photons are re-emitted out as fluorescence. Then,

Energy re-emitted out $= \frac{47}{100} \times E n e r g y a b s o r b e d$

$n_{2} \times \frac{h c}{λ_{e m i t t e d}} = \frac{47}{100} \times \frac{n_{1} \times h c}{λ_{a b s o r b e d}}$

$\frac{n_{2}}{n_{1}} = \frac{47}{100} \times \frac{λ_{e m i t t e d}}{λ_{a b s o r b e d}}$

$= \frac{47}{100} \times \frac{5080}{4530} = 0.527$

Hence, the ratio of quanta emitted out to the no.of quanta absorbed is $0.527 \times 10 = 5$

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