Question

A certain dye absorbs light of $\lambda =4530A^0$ and wavelength of fluorescence light is $5080A^0$. Assuming that under given condition 47% of the absorbed energy is remitted out as fluorescence the ratio of quanta emitted out to the number of quanta absorbed is:

• A. 0.527
• B. 1.5
• C. 52.7
• D. 3

Answer 1

The correct option is A 0.527

Energy of light absorbed in an photon $=hc/{\lambda }_{absorbed}$

Let $n_1$ photons are absorbed

Total energy absorbed $=n_{1}hc/{\lambda }_{absorbed}$

Now E of light re-emitted out in one photon $=hc/{\lambda }_{emitted}$

Let $n_2$ photons are re-emitted then

Total energy re-emitted out $=n_{2}hc/{\lambda }_{emitted}$

$E_{absorbed}\times 47/100=E_{re-emittedout}$

$hc/{\lambda }_{absorbed}\times n_1\times 47/100=n2\times hc{\lambda }_{emitted}$

$n_2/n_1=47/100\times {\lambda }_{emitted}{\lambda }_{absorbed}$

$=47/100\times 5080/4530$

$n_2/n_1=0.527$