Question

UV light of wavelength 300 nm and intensity 1 $W{m}^{-2}$ falls on the surface of photo electric metal. If one percent of incident photons, produce photo electrons, then the number of photo electrons emitted from an area of 1.0 $c{m}^2$ of the surface is nearly

• A. $2.13\times {10}^{11}{s}^{-1}$
• B. $1.5\times {10}^{12}{s}^{-1}$
• C. $3.02\times {10}^{12}{s}^{-1}$
• D. $3\times {10}^{12}{s}^{-1}$

Answer 1

The correct option is B $1.5\times {10}^{12}{s}^{-1}$

$E=\frac{hc}{\lambda }=\frac{12400}{3000}=4.133eV=6.6\times {10}^{-19}J$
Power of source $P=I\times A=1\times {10}^{-4}={10}^{-4}W$
$\therefore$ No of photons per second $=\frac{P}{E}=\frac{{10}^{-4}}{6.6\times {10}^{-19}}$
$\therefore$ Number of electrons $=\frac{1}{100}\times \frac{{10}^{-4}}{6.6\times {10}^{-19}}=1.5\times {10}^{12}{s}^{-1}$