The electric field at the point associated with a light wave is given by
E = 200 $\left[\sin \left(6 \times 10^{15}\right) t+\sin \left(9 \times 10^{15}\right) t\right] V m^{-1}$
Given : h = 4.14 × $10^{-15} \mathrm{eVs}$
If this light falls on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be :