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`.

Answer

`2.97 eV`

Explanation

In one complete vibration twice the electric field becomes zero, so frequency of the incident light,
`v=1/2xx2.4xx10^(15)=1.2xx10^(15)cps`.
Max K.E. `=hv-phi_(0)`
`=(6.63xx10^(-34)xx1.2xx10^(15))/(1.6xx10^(-19))-2.0`
`=2.97 eV`