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

According to Einstein 's photoelectric equation,
`E_(k) = h V - W`
or If `V_(s)` is retarding or stopping potential and `V_(0)` the theshold frequency, then above equation becomes
`eV_(s) = h v - h v_(0)`
or `h v = eV_(s) + h v_(0)`
or `v = (eV_(s))/(h) + v_(0)`
Hence , `e = 1.6 xx 10^(-10) coulomb, V_(s) = 3 V, and v_(0) = 6 xx 10^(-14) HZ`
Therefore, required frequency
`v = (1.6 xx 10^(-10) xx 3)/(6.63 xx 10^(-34)) + 6 xx 10^(14)`
`= 7.24 xx 10^(14) + 6 xx 10^(14) = 13.24 xx 10^(14) HZ`
`= 1.324 xx 10^(15) HZ`