The electric field at a point associated with a light wave is given by $E=200[\sin (6\times {10}^{15})t+\sin (9\times {10}^{15})t]{\mathrm{Vm}}^{-1}$$(Given:h=4.14\times {10}^{-15 }\mathrm{eVs})$. If this light falls on a metal surface having a work function of 2.50eV, the stopping potential is 

Given, $E=200[\sin (6\times {10}^{15})t+\sin (9\times {10}^{15})t]$ 
Work function,  $\varphi =2.5eV$
The frequency of e.m. waves $=\frac{6\times {10}^{15}}{2\pi }Hz$  and  $\frac{9\times {10}^{15}}{2\pi }Hz$
So, energy of one photon will be h$\upsilon$  and another will be  $h\upsilon \text{'}$
  $$\begin{gathered} h\upsilon =4.14\times {10}^{15}\times \frac{6}{2\pi }\times {10}^{15}eV =3.95eV \\ h\upsilon \text{'}=4.14\times {10}^{-15}\times \frac{9}{2\pi }\times {10}^{15}eV=5.93eV \end{gathered}$$
  
So, K.Emax  $h{\upsilon }_{\max }-$  work function
$K.E_{\max }=5.93-2.50=3.43eV \Rightarrow e{V}_0=3.43eV \Rightarrow V_0=3.43V$