# A beam of electromagnetic radiation of intensity

Question:

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

Solution:

(11.00) Energy of proton

$E=\frac{h c}{\lambda}=\frac{1240}{310}=4 \mathrm{eV}>2 \mathrm{eV}[=\phi]$

(so emission of photoelectron will take place)

$=4 \times 1.6 \times 10^{-19}=6.4 \times 10^{-19}$ joule

$N=\frac{6.4 \times 10^{-5} \times 1}{4 \times 6.4 \times 10^{-19}}=10^{14}$

No. of photoelectrons emitted per second

$=\frac{10^{14}}{10^{3}}=10^{11}\left(\because 1\right.$ in $10^{3}$ photons ejects an electron)

$\therefore$ Value of $X=11.00$