Question

A metal surface having a work function $\varphi =2.2\times 10^{-19}J$, is illuminated by the light of wavelength $1320\mathring{A}$. What is the maximum kinetic energy (in $eV$ ) of the emitted photoelectron? [Take $h=6.6\times 10^{-34}Js$ ]

Answer 1

Answer: 8

K.E. $=\frac{hc}{\lambda }-{\varphi }_0$
$=\frac{6.6\times 10^{-34}\times 3\times 10^8}{1320\times 10^{-10}}-2.2\times 10^{19}$
$=15\times 10^{-19}-2.2\times 10^{-19}$
$=12.8\times 10^{-19}J$
K.E. $=\frac{12.8\times 10^{-19}}{1.6\times 10^{-19}}eV=8eV$