The electric field at a point associated with a light wave is given by $E=200\left[\sin \left(6\times {10}^{15}\right)t+\sin \left(9\times {10}^{15}\right)t\right]{\mathrm{Vm}}^{-1}$

Given :$h=4.14\times {10}^{-15}\mathrm{eVs}$

If this light falls on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be