A certain dye absorbs light of $\lambda =4000{\text{\hspace{0.17em}\hspace{0.17em}A}}^o$ and then fluoresces light of $5000{\text{\hspace{0.17em}\hspace{0.17em}A}}^o.$ Assuming that under given conditions 50% of the absorbed energy is re-emitted out as fluorescence, calculate the ratio of number of quanta emitted out to the number of quanta absorbed.

$\begin{array}{l}{\mathrm{E}}_{\mathrm{emitted}}=\frac{50}{100}\times {\mathrm{E}}_{\mathrm{absorbed}}\\ \mathrm{No}.\mathrm{of}\mathrm{emitted}\mathrm{photons}\times \mathrm{Energy}\mathrm{of}\mathrm{emitted}\mathrm{photon}=\\ \frac{50}{100}\times \mathrm{No}.\mathrm{of}\mathrm{absorbed}\mathrm{photon}\times \mathrm{Energy}\mathrm{of}\mathrm{absorbed}\mathrm{photon}.\\ \therefore {\mathrm{n}}_{\mathrm{e}}\times \frac{12400}{5000}=\frac{50}{100}\times {\mathrm{n}}_{\mathrm{a}}\times \frac{12400}{4000}\\ (\mathrm{since} \mathrm{E}=\frac{\mathrm{hC}}{\mathrm{\lambda }}; \mathrm{hC} =12400\mathrm{eV}.{\mathrm{A}}^{\mathrm{o}})\\ \therefore \frac{{\mathrm{n}}_{\mathrm{e}}}{{\mathrm{n}}_{\mathrm{a}}}=\frac{5}{8}.\end{array}$