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.

Explanation

Given, E = E_0 sin [(1.57 xx 10(7) m ^(-1) ) ( x - ct )]`
` w = 1.57 xx 10^(7) xx c `
` rArr f = 1.57 xx 10^(7) xx 3 xx 10^(8) / 2 pi Hz `
` W_0 = 1.9eV `
Now eV_0 = hv - W_0 `
` =4.14 xx 10^(-15) xx (1.57 xx 3 xx 10^(15) / 2pi ) - 1.9eV`
`= 3.105 - 1.9 = 1.205 V `
V_0 = 1.205 xx 1.6 xx 10^(-19) / 1.6 xx 10^(-19) `
` 1.205 V `