The electric field associated with a light wave is given by `E=E_(0)sin[(1.57xx10^(7)m^(-1))(ct-x)]`. Find the stopping potential when this light is used in an experiment on a photoelectric effect with the emitter having work function 2.1eV. `h=6.62xx10^(-34)Js`.

Answer

`1.0V`

Explanation

Here, `E=-E_(0)sin[(1.57xx10^(7)m^(-1))(ct-x)]`
Comparing it with the equation of harmonic wave in electric field , `E=E_(0)sin(2pi)/lambda(vt-x)`
we have , `(2pi)/lambda=1.57xx10^(7) or lambda=(2pi)/(1.57xx10^(7))m`,
`V_(0) =(hc)/(lambdae)-(phi_(0))/e`
`=(6.63xx10^(-34)xx3xx10^(8)xx1.57xx10^(7))/(2xx3.142xx1.6xx10^(-19))-2.1`
`=1.0V`