# A certain dye absorbs light of $\lambda = 4530 A^{\large\circ}$ and then fluorescence light of $5080 \;A^{\large\circ}$ Assuming that under given conditions $47\%$ of the absorbed energy is re-emmited out as fluorescence, Calculate the ratio of Quantum emitted out to the no. of quanta absorbed

$(a)\;0.527\qquad(b)\;1.327\qquad(c)\;0.132\qquad(d)\;0.232$

Energy of light absorbed in an photon = $\large\frac{hc}{\lambda_{absorbed}}$
Let $n_1$ photons are absorbed
$\therefore$ Total energy absorbed = $\large\frac{n_1hc}{\lambda_{absorbed}}$
Now E of light re-emitted out in one photon = $\large\frac{hc}{\lambda_{emitted}}$
Let $n_2$ photons are re-emitted then
$\therefore$ Total energy re-emitted out = $\large\frac{n_2hc}{\lambda_{emitted}}$
$E_{absorbed}\times \large\frac{47}{100} = E_{re-emitted out}$
$\large\frac{hc}{\lambda_{absorbed}}\times n_1\times \large\frac{47}{100} = n_2\times \large\frac{hc}{\lambda_{emitted}}$
$\large\frac{n_2}{n_1} = \large\frac{47}{100}\times \large\frac{\lambda_{emitted}}{\lambda_{absorbed}}$
$= \large\frac{47}{100}\times \large\frac{5080}{4530}$
$\large\frac{n_2}{n_1}$ = 0.527
answered Feb 1, 2014 1 flag