When a beam of 10.6 eV photons of intensity $2.0 \mathrm{~W} / \mathrm{m}^2$ falls on a platinum sur

Question

When a beam of 10.6 eV photons of intensity $2.0 \mathrm{~W} / \mathrm{m}^2$ falls on a platinum surface of area $1.0 	imes 10^{-4} \mathrm{~m}^2$ and work function $5.6 \mathrm{eV}, 0.53 \%$ of the incident photons eject photoelectrons. The number of photoelectrons emitted per second is $x 	imes 10^{11}$, where the value of $x$ is $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

(6.25) : Incident energy
$
E=10.6 \mathrm{eV}=10.6 	imes\left(1.6 	imes 10^{-19}ight) \mathrm{J}
$
$=16.96 	imes 10^{-19} \mathrm{~J}$
$\begin{aligned} & 	ext { Given : } \frac{	ext { Energy incident }}{	ext { area x time }}=2 \mathrm{~W} / \mathrm{m}^2 \ & 	herefore \quad \frac{	ext { Number of incident photons }}{	ext { area x time }}\end{aligned}$
$=\frac{2}{16.96 	imes 10^{-19}}$
$\begin{aligned} & =1.18 	imes 10^{18} \ & 	herefore \quad \frac{	ext { Incident photons }}{	ext { time }}\end{aligned}$
$=\left(1.18 	imes 10^8ight) 	imes$ area
$=1.18 	imes 10^{18} 	imes\left(1.0 	imes 10^{-4}ight)=1.18 	imes 10^{14}$
$	herefore \quad \frac{	ext { Number of photoelectrons }}{	ext { time }}$
$=\left(\frac{0.53}{100}ight) 	imes\left(1.18 	imes 10^{14}ight)$
or $n=6.25 	imes 10^{11}$