Question

A certain dye absorbs 4700 $\stackrel{o}{A}$ and fluoresces at 5080 $\stackrel{o}{A}$, these being wavelengths of maximum absorption that under given conditions 47% of the absorbed energy is emitted. The ratio of the no. of quanta emitted to the number absorbed is:

• A. 0.508
• B. 0.896
• C. 0.453
• D. 0.324

Answer 1

The correct option is A 0.508

According to given condition,

$E_{emit}=0.47E_{absorb}$ and,

$\frac{n_{e}hc}{5080}=\frac{0.47n_{a}hc}{4700}$

$\frac{n_e}{n_a}=\frac{0.47\times 5080}{4700}$

$\frac{n_e}{n_a}=0.508$

Hence, the correct option is $A$