Ultraviolet light of wavelength 300 nm and intensity 1.0 watt/m² 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 cm² of the surface is nearly

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$9.61 \times {10^{14}}per\;\sec$

$4.12 \times {10^{13}}per\;\sec$

$2.13 \times {10^{11}}per\;\sec$

$1.51 \times {10^{12}}per\;\sec$

answer is C.

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Detailed Solution

Intensity of light
$I = \frac{{Watt}}{{Area}} = \frac{{nhc}}{{A\lambda }}\,$ ⇒ Number of photon $n = \frac{{IA\lambda }}{{hc}}$
$\therefore$ Number of photo electron = $\frac{1}{{100}} \times \frac{{IA\lambda }}{{hc}}$
$= \frac{1}{{100}}\frac{{1 \times {{10}^{ - 4}} \times 300 \times {{10}^{ - 9}}}}{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}} = 1.5 \times {10^{12}}$