$(a)\;0.527\qquad(b)\;1.327\qquad(c)\;0.132\qquad(d)\;0.232$

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Energy of light absorbed in an photon = $\large\frac{hc}{\lambda_{absorbed}}$

Let $n_1$ photons are absorbed

$\therefore$ Total energy absorbed = $\large\frac{n_1hc}{\lambda_{absorbed}}$

Now E of light re-emitted out in one photon = $\large\frac{hc}{\lambda_{emitted}}$

Let $n_2$ photons are re-emitted then

$\therefore$ Total energy re-emitted out = $\large\frac{n_2hc}{\lambda_{emitted}}$

$E_{absorbed}\times \large\frac{47}{100} = E_{re-emitted out}$

$\large\frac{hc}{\lambda_{absorbed}}\times n_1\times \large\frac{47}{100} = n_2\times \large\frac{hc}{\lambda_{emitted}}$

$\large\frac{n_2}{n_1} = \large\frac{47}{100}\times \large\frac{\lambda_{emitted}}{\lambda_{absorbed}}$

$= \large\frac{47}{100}\times \large\frac{5080}{4530}$

$\large\frac{n_2}{n_1}$ = 0.527

Hence the answer is (a)

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