Find the maximum kinetic enegry of photoelectrons liberated form the surface of lithium by electromagnetic radiation whose electric component varies with time as `E = a(1+cos omegat) cos omega_(0)t`, where `a` is a constant, `omega = 6.0.10^(14)s^(-1)` and `omega_(0) = 360.10^(15)s^(-1)`.

Explanation

We write
`E =a (1+cos omegat) cos omega_(0)t`
`= acos omega_(0)t +(a)/(2)[cos(omega_(0)-omega)t+cos(omega_(0)+omega)t]`
It is abvious that light has there frequencies and the maximum `K.E.` of photon electrons ejected is
`h(omega +omega_(0))-A_(Li)`
where `A_(Li) = 2.39eV`. Substituting we get `0.37ev`.