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

Explanation

Here , `E =E_(0) " sin " [ (1.57 xx10^(7) m^(-1))(ct-x)]`
Comparing it with the equation harmonic wave in electric field ,
We have , `(2pi)/lambda = 1.57 xx 10^(7) or lambda = (2pi)/(1.57 xx 10^(7))m `
`V_(0)=(hc)/(lambda_(e ))-(phi_(0))/e`
`= (6.63 xx10^(-34)xx3xx10^(8) xx 1.57 xx10^(7))/(2xx3.142 xx1.6 xx10^(-19))-2.1 =1V`