Question

Ultraviolet light of wavelength $300nm$ and intensity $1.0$ $W/m^2$ falls on the surface of a photoelectric material. If one percent of the incident photons produce photoelectrons, then the number of photoelectrons emitted from an area of $1.0$ $c{m}^2$ of the surface is nearly (in per second)

• A. $9.61\times {10}^{14}$
• B. $4.12\times {10}^{13}$
• C. $1.51\times {10}^{12}$
• D. $2.13\times {10}^{11}$

Answer 1

Answer: (C)

$1.51\times {10}^{12}$

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