Question

A certain dye absorbs $4700\stackrel{\circ }{A}$ and fluoresces at $5080\stackrel{\circ }{A}$, these are being wavelengths of maximum absorption that under given conditions 47% of the absorbed energy is emitted. Calculate the ratio of the number of quanta emitted to the number of quanta absorbed.

• A. $0.108$
• B. $0.508$
• C. $0.72$
• D. $1.72$

Answer 1

The correct option is B $0.508$

According to the question,
$E_{emitted}=\frac{47}{100}\times E_{absorbed}$
Again, we know, $E=nh\nu =nh\frac{c}{\lambda }$
$\therefore$$\frac{n_{emitted}hc}{{\lambda }_{emitted}}=\frac{47}{100}\times \frac{n_{absorbed}hc}{{\lambda }_{absorbed}}\Rightarrow \frac{n_{emitted}}{n_{absorbed}}=\frac{47}{100}\times \frac{{\lambda }_{emitted}}{{\lambda }_{absorbed}}=\frac{47}{100}\times \frac{5080}{4700}=0.508$