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Ultraviolet light of wavelength $300 \ nm$ and intensity $1.0 \ watt/m^2$ falls on the surface of a photosensitive material. If $1\%$ of the incident photons produce photoelectrons, then the number of photoelectrons emitted from an area of $1.0\ cm^2$ of the surface is nearly
$9.61 \times {10^{14}}per\;\sec $
$4.12 \times {10^{13}}per\;\sec $
$1.51 \times {10^{12}}per\;\sec $
$2.13 \times {10^{11}}per\;\sec $
Solution
(c) Intensity of light
$I = \frac{{Watt}}{{Area}} = \frac{{nhc}}{{A\lambda }}$
==> Number of photon $n = \frac{{IA\lambda }}{{hc}}$
Number of photo electron = $\frac{1}{{100}} \times \frac{{IA\lambda }}{{hc}}$
$ = \frac{1}{{100}}\frac{{1 \times {{10}^{ – 4}} \times 300 \times {{10}^{ – 9}}}}{{6.6 \times {{10}^{ – 34}} \times 3 \times {{10}^8}}} = 1.5 \times {10^{12}}$
