Atomic StructureHard
Question
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?
Options
A.41%
B.47%
C.74%
D.63%
Solution
$E_{abs} \times \frac{x}{100} = E_{emit} \Rightarrow n_{1}.\frac{hc}{\lambda_{1}}.\frac{x}{100} = n_{2}.\frac{hc}{\lambda_{2}}$
$\therefore x = \frac{n_{2}}{n_{1}} \times \frac{\lambda_{1}}{\lambda_{2}} = \frac{53}{100} \times \frac{4530}{5080} = 47.3$
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