Find the frequency of light which ejects electrons from a metal surface. Fully stopped by a retarding potential of `3 V`, the photoelectric effect begins in this metal at a frequency of `6xx10^(14)Hz`. Find the work function for this metal. (Given `h=6.63xx10^(-34)Js`).

Explanation

The threshold frequency for the given metal surface is
`v_(th) = 6 xx 10^(14) Hz`
Thus the work function for metal surface is
`phi= hv_(th) = 6.63 xx 10^(-34) xx 6 xx 10^(14) =3.978 xx 10^(-19) J`
As stopping potential for the ejected electrons is 3V, the maximum kinetic energy of ejected electrons will be
`KE_("max") = 3eV`
`= 3 xx 1.6 xx 10^(-19) J = 4.8 xx 10^(19) J`
According to photo electric effect equation, we have
`hv= hv_(th) + KE_("max")`
or frequency of incident light is
`v= (phi + KE_("max"))/( h ) `
`= ( 3.978 xx 10^(-19) + 4.8 xx 10^(-19))/(6.63 xx 10^(-34)) = 1.32 xx 10^(15) Hz`