Light of wavelength 4000 $\stackrel{\mathrm{o}}{\mathrm{A}}$ is incident on a metal plate whose work function is 2 eV. What is maximum kinetic energy of emitted photoelectron?

If the maximum kinetic energy of photoelectrons emitted from metal surface is E_o and W is the work function of metal then
${\mathrm{E}}_{\mathrm{k}}=\mathrm{hv}-\mathrm{W}$

where Iv is the energy of photon absorbed by the electron in metal.

$\begin{array}{l}\therefore {\mathrm{E}}_{\mathrm{k}}=\frac{\mathrm{hc}}{\mathrm{\lambda }}-\mathrm{W}\\ \mathrm{wherev}=\frac{\mathrm{c}}{\mathrm{\lambda }}\\ \mathrm{Putting}\mathrm{the}\mathrm{numerical}\mathrm{values},\mathrm{we}\mathrm{have}\\ {\mathrm{E}}_{\mathrm{k}}=\left[\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{4000\times {10}^{-10}\times 1.6\times {10}^{-19}}-2\right]\mathrm{eV}\\ {\mathrm{E}}_{\mathrm{k}}=3.1-2=1.1\mathrm{eV}\end{array}$