Light described at a place by the equation $E=\left(100V{m}^{-1}\right)[\sin (5\times {10}^{15}{s}^{-1})t+\sin (8\times {10}^{15}{s}^{-1})t]$ falls on a metal surface having work function $2.0 \mathrm{eV}$. Calculate the maximum kinetic energy ( in eV ) of the photoelectrons.

The light contains two different frequencies. The one with larger frequency will cause photoelectrons with largest kinetic energy. This larger frequency is $v=\frac{\omega }{2\pi }=\frac{8\times {10}^{15}{s}^{-1}}{2\pi }$

The maximum kinetic energy of the photoelectrons is 

$K_{\max }=hv-\phi =(4.14\times {10}^{-15}eVs)\times \left(\frac{8\times {10}^{18}}{2\pi }{s}^{-1}\right)-2.0eV$

$=5.27eV-2.0eV=3.27eV.$