A certain dye absorbs light \[\lambda = 4530A^\circ \] of and wavelength of fluorescence light is \[5080A^\circ \]. Assuming that under given condition \[47\% \] of 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. \[3\]

As we know the formula for the energy of a photon is \[\dfrac{{hc}}{\lambda }\].And if we are having \[n\] photons then energy of these photons will be \[\dfrac{{nhc}}{\lambda }\]

And we are given the percentage of absorbed energy is \[47\% \].We know the value of wavelength absorbed as, \[4530\]. And also given with wavelength of remitted as, \[5080\].

Let energy be absorbed as \[{E_1} = \dfrac{{{n_1}hc}}{\lambda_1 }\].

And energy re-emitted as \[{E_2} = \dfrac{{{n_2}hc}}{{{\lambda _2}}}\].

Therefore from the given data

\[{E_1} \times 0.47 = {E_2}\]

\[\Rightarrow \dfrac{{{n_1}hc}}{\lambda_1 } \times 0.47 = \dfrac{{{n_2}hc}}{{{\lambda _2}}}\]

\[\Rightarrow \dfrac{{{n_1}}}{\lambda_1 } \times 0.47 = \dfrac{{{n_2}}}{{{\lambda _2}}}\]

\[\Rightarrow \dfrac{{{n_2}}}{{{n_1}}} = \dfrac{\lambda_2 }{{{\lambda _1}}} \times 0.47\]

\[\Rightarrow \dfrac{{{n_2}}}{{{n_1}}} = \dfrac{{5080}}{{4530}} \times 0.47\]

\[\therefore \dfrac{{{n_2}}}{{{n_1}}} = 0.527\]

Photon vitality is the vitality conveyed by a solitary photon. The measure of vitality is straightforwardly relative to the photon's electromagnetic recurrence and accordingly, comparably, is conversely corresponding to the frequency. The higher the photon's recurrence, the higher its energy. Photon vitality is more valuable than heat vitality, as the vast majority of the sunlight based pillar will in general change into warmth of ambients. The most effective work of sun based bars will be finished by using the quantum parts of it as in the photosynthetic elements of plants. It is imperative to examine the utilization of sunlight based photon energies to part water accepting plants as a model.