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A monochromatic beam of light of wavelength $400\,nm$ incident normally upon a photosensitive surface ($25\%$ reflects and rest absorbs), produces a pressure of $5 \times 10^{-7}\,Nm^{-2}$ on it. If $0.1\%$ of the incident photons produce electrons then corresponding saturation current will be..............$\mu A$ (consider area of photosensitive surface $= 5\,cm^2$)
$0.48$
$9.6$
$19.2$
$12$
Solution
Radiation pressure $\mathrm{P}=\frac{25}{100}\left(\frac{2 \mathrm{I}}{\mathrm{c}}\right)+\frac{75}{100}\left(\frac{\mathrm{I}}{\mathrm{c}}\right)=\frac{125 \mathrm{I}}{\mathrm{c}}$
$\Rightarrow \mathrm{I}=\frac{\mathrm{Pc}}{1.25}=\frac{5 \times 10^{-7} \times 3 \times 10^{8}}{1.25}=120 \mathrm{Wm}^{-2}$
Now number of photons incident $=\frac{\mathrm{IA} \lambda}{\mathrm{hc}}$
$=5 \times 10^{24} \times 120 \times 5 \times 10^{-4} \times 400 \times 10^{-9}$
$=12 \times 10^{16} \mathrm{s}^{-1}$
Given $n_{e}=\left(\frac{0.1}{100}\right) n_{p}=10^{-3} \times 12 \times 10^{16}$
$ \therefore $ Current $\mathrm{i}_{s} =\left(\mathrm{n}_{\mathrm{e}}\right) \mathrm{e}=12 \times 10^{13} \times 1.6 \times 10^{-19}$
$=19.2 \times 10^{-6} \mathrm{A}=19.2\, \mu \mathrm{A} $
