The electric field of light wave is given as `vec(E)=10^(-3)cos((2 pi x)/(5xx10^(-7))-2pi xx 6xx10^(14)t) hat(x) (N)/(C)`. This light falls on a metal plate of work function 2 eV. The stopping potential of the photo-electrons is:
Given, E (in eV) `= (12375)/(lambda ( "in" Å ))`

Explanation

Here `omega=2 pi xx6xx10^(-14)`
`impliesf=6xx10^(14)Hz`
Wavelength
`lamda=c/f=(3xx10^(8))/(6xx10^(14))=0.5xx10^(-6)m=5000Å`
Give `E=12375/5000=2.48eV`
Using `E=W+eV_(s)`
`implies2.48=2+eV_(s)`
or `V_(s)=0.48V`