The electric field at a point associated with a light wave is given...

Given: $E = 200\left[\sin \left(6 \times {10}^{15}\right)t + \sin \left(9 \times {10}^{15}\right)t\right]V{m}^{- 1}$

Here, angular velocity, ${\omega}_{1} = 6 \times {10}^{15}$and ${\omega}_{2} = 9 \times {10}^{15}$

Maximum kinetic energy of the photoelectron is ${KE}_{max .} = E - ϕ$

$= hf - ϕ = \dfrac{h\omega}{2\pi} - ϕ$

$= \dfrac{4 . 14 \times {10}^{- 15} \times 9 \times {10}^{15}}{2 \times 3 . 14} - 2 . 5 = 5 . 92 - 2 . 50 = 3 . 42eV$