The electric field at a point associated with a light wave is `E=100Vm^-1 sin [(3*0xx10^15 s^-1)t] sin [(6*0xx10^15s^-1)t].` If this light falls on a metal surface having a work function of `2*0 eV`, what will be the maximum kinetic energy of the photoelectrons?

Explanation

Given ` E = 100 sin [(3 xx 10(-15) s ^(-1) ) t] sin `
` [ (6 xx 10^(-15) s^(-1)) t] `
` = (100) 1 / 2 [cos [(9 xx 10^(15) s^(-1))t]`
` - cos [(3 xx 10^(15)s^(-1) ) t]`
The w are ` 9 xx 10^(15)` and `3 xx 10^(15) `
For larget K.E.
` I_max = W_max / 2pi = 9 xx 10^(15) / 2pi `
` E = phi_0 = K.E. `
`hf - phi_0 = K.E. `
`rArr 6.63 xx 10^(-34) xx 9 xx 10^(15) / 2 pi xx 1.6 xx 10^(-19) = -2 = KE`
`rArr KE = 6.63 xx 9 / 2 xx 1.6 xx 3.14 - 2`
` = 3.938 eV = 3.93 eV`