The dye acriflavine when dissolved in water has its maximum light absorption at $4530 A^{0}$ and has maximum florescence emission at $5080 A^{0}$ . The number of fluorescence quanta is about 53% of the number of quanta absorbed. What percentage of absorbed light energy is emitted as fluorescence?(round off to the nearest integer)

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answer is 47.

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Detailed Solution

$E_{a b s} \times \frac{x}{100} = E_{e m i t} \Rightarrow n_{1} . \frac{h c}{λ_{1}} . \frac{x}{100} = n_{2} . \frac{h c}{λ_{2}} ∴ x = \frac{n_{2}}{n_{1}} \times \frac{λ_{1}}{λ_{2}} = \frac{53}{100} \times \frac{4530}{5080} = 47.3$

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