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`).

Answer

`1.327xzx10^(15)s^(-1), 2.48 eV`

Explanation

Here, `V_(0)=3V, v_(0)=6xx10^(14)s^(-1), v=? phi_(0)=?`
`eV_(0)=hv-hv_(0) or hv=hv_(0)+eV_(0)`
or `v=v_(0)+(eV_(0))/h=6xx10^(14)+((1.6xx10^(-19))3)/(6.6xx10^(-34))`
`=6xx10^(14)+7.27xx10^(14)=13.27xx10^(14)s^(-1)`
`phi_(0)=hv_(0)=(6.6xx10^(-34))xx(6xx10^(14))J`
`=(6.6xx10^(-34)xx6xx10^(14))/(1.6xx10^(-19)) eV=2.48 eV`