Question

A beam of electromagnetic radiation of intensity 6.4$\times {10}^{-5}W/c{m}^2$ is comprised of wavelength, $\lambda =310nm$. It falls normally on a metal (work function $\varphi =2eV$) of surface area $1c{m}^2$. If one in ${10}^3$ photons ejects an electron, total number of electrons ejected in 1$s$ is ${10}^x$ ($hc=1240eV-nm,1eV=1.6\times {10}^{-19}J$), then x is

Answer 1

$$\begin{array}{cc}P& =\text{Intensity}\times \text{Area}\\ & =6.4\times {10}^{-5}W-c{m}^{-2}\times 1{cm}^2\\ & =6.4\times {10}^{-5}W\end{array}$$
For photoelectric effect to take place, energy should be greater than work function
Now,
$$E=\frac{1240}{310}=4eV>2eV$$
Therefore, photoelectric effect takes place
Here n is the number of photons emitted.

$n\times E=I\times A⟹n=\frac{IA}{E}=\frac{6.4\times {10}^{-5}}{6.4\times {10}^{-19}}={10}^{14}$
Where, n is number of incident photon
Since, 1 out of every 1000 photons are sucessfull in ejecting 1 photoelectron
Therefore, the number of photoelectrons emitted is
$$=\frac{{10}^{14}}{{10}^3}$$
$$\therefore x=11$$