The electric field associated with a light wave is given by `E= E_0 sin [(1.57x 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.

Answer

A, B

Explanation

` f = (omega)/(2pi) = (1.57xx10^7)_c)/(2pi)`
`E = (hf)/(1.6xx10^(-19) eV`
`=((6.63xx10^(-34)(1.57xx10^7)(3xx10^8)/(1.6xx10^(-19)xx2xxpi)`
=3.1 eV
`K_(max) = E -W = 1.2 eV.`