The surface of a metal of work function ${\varphi }_0$is illumi nated by light whose electric field component varies with time as $E=E_0[1+\cos (\omega t\left)\right]\sin \left({\omega }_{0}t\right)$. Calculate the maximum kinetic energy of photoelectrons emitted from the surface.

The given electric field component is

$E=E_0\sin \left({\omega }_{0}t\right)+E_0\sin \left({\omega }_{0}t\right)\cos \left(\omega t\right)$

$\Rightarrow E=E_0\sin \left({\omega }_{0}t\right)+\frac{E_0}{2}\left[\sin \left({\omega }_0+\omega \right)t+\sin \left({\omega }_0-\omega \right)t\right.$

So, the given light sample that comprises three different frequencies i.e. 

$\omega ,\left({\omega }_0+\omega \right),\left({\omega }_0-\omega \right)$

The maximum kinetic energy will be due to most energetic photon i.e. of frequency 

$\left(\frac{\omega +{\omega }_0}{2\pi }\right)$. Since $K_{\max }=hv-{\varphi }_0$

$\Rightarrow K_{\max }=\frac{h\left(\omega +{\omega }_0\right)}{2\pi }-{\varphi }_0$