Question

A certain dye absorbs light of $\lambda = 4000 \,\mathring{A} $ and then fluoresces light of $5000 \,\mathring{A} $. 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.

• A. $\frac{5}{8}$
• B. $\frac{8}{5}$
• C. $\frac{3}{8}$
• D. $\frac{8}{3}$

Answer 1

Answer: (A)

$E_{emitted}=\frac{50}{100}\times E_{absorbed}$
No. of emitted photons $\times$ Energy of emitted photon $=\frac{50}{100}$
$\times$ No. of absorbed photon $\times$ Energy of absorbed photon.
$\therefore n_{e} \times\frac{12400}{5000}=\frac{50}{100}\times n_{a}\times\frac{12400}{4000} $
$\therefore \frac{n_{e}}{n_{a}} =\frac{5}{8}.$