The electric field associated with a light wave is given by `E= E_0 sin [(1.57 xx 10^7 m^(-1)(x-ct)].` Find the stopping potential when this light is used in an experiment on photoelectric affect with a metal having work - function 1.9 eV.

Options

A. `0.6 V`
B. `1.2 V`
C. `1.8 V`
D. `2.4 V`

Answer

B

Explanation

`E =E_(0) sin (1.57xx10^(7))(x - ct)`
`E =E_(0) sin .(omega)/(c)(x - ct)`
`(omega)/(c) =1.57xx10^(7)`
`(2pi upsilon)/(upsilon lambda) =1.57xx10^(7) =(pi)/(2)xx10^(7)`
`lambda =4xx10^(-7) m=400 nm`
`(1242)/(lambda) =(1242)/(400) =3.1 eV`