Work function of metal is 2 eV.Light of intensity `10^(-5)Wm^(-2)` is incident on 2`cm^(2)` area of it.If `10^(17)` electrons of these metals absorb the light,in how much time does the photo electric effect start?Consider the waveform of incident light.

Explanation

Intesity of incident light is `10^(-5)Wm^(-2)`
`therefore` Energy incident on `1m^(2)` area in 1 s in `10^(-5)` J.
`therefore` Energy incident on area of `2cm^(2)`
`=2xx10^(-4)m^(2)`
`=2xx10^(-4)xx10^(-5)`
`=2xx10^(-9)J`
This energy is absorbed by `10^(17)` electrons.
`therefore` Average energy absorbed by each electron
`(2xx10^(-9)J)/(10^(17))`
`-2xx10^(-26)J`
Now ,electron may get emitted when it absorbs energy equal to work function of its metal. In this given problem work function is `2eV=2xx1.6xx10^(-19)J`.Thus,electron requires `(2xx1.6)xx10^(-19)J` of energy to get emitted .To absorb `2xx10^(-26)J` of energy,time required is 1s, therefore to absorb energy
`2xx1.6xx10^(-19)`J,time required is ,
`t_(e)=(2xx1.6xx10^(-19))/(2xx10^(-26))=1.6x10^(7)`S