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 workfunction $5.6 \mathrm{eV}, 0.53 \%$ of the incident photons eject photoelectrons. Find the number of photoelectrons emitted per second. [Take $1 \mathrm{eV}=1.6 	imes 10^{-19} \mathrm{~J}$.]

Professor PC Thomas Classes · Answer

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