The dye acriflavine when dissolved in water has its maximum light absorption at 4530 Å and has maxim

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The dye acriflavine when dissolved in water has its maximum light absorption at 4530 Å and has maximum florescence emission at 5080 Å. The number of fluorescence quanta is about 53% of the number of quanta absorbed. What percentage of absorbed light energy is emitted as fluorescence?

Accepted Answer

$E_{abs} 	imes \frac{x}{100} = E_{emit} \Rightarrow n_{1}.\frac{hc}{\lambda_{1}}.\frac{x}{100} = n_{2}.\frac{hc}{\lambda_{2}}$$	herefore x = \frac{n_{2}}{n_{1}} 	imes \frac{\lambda_{1}}{\lambda_{2}} = \frac{53}{100} 	imes \frac{4530}{5080} = 47.3$

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