The electric field associated with a monochromataic beam of light becomes zero, `2.4xx10^(15)` times per second. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is 2.0eV, `h=6.63xx10^(-34)Js`.

Explanation

The elctric field become ` 1.2 xx 10^(15) / 5 ` times per second .
` :. Frequency = 1.2 xx 10^(15) / 2 = 0.6 xx 10^(15) `
` hv = phi_0 + KE `
` rArr hv = phi_0 = KE `
` rArr KE = 6.63 xx 10^(-34) xx 0.6 xx 10(16) / 1.6 xx 10^(-19)`
` - 2 = 0.482 ev = 0.48 eV`.