Calculate the retarding potential required to stop the photoelectrons emitted from a metal surface of work function `1.2 eV`, when light of frequency `6.3xx 10^(14)` Hz is incident on it

Explanation

Given `w= 1.2 eV , v = 6.3xx10^(14)`Hz
we know that `E = (hv)/e = (6.625xx10^(-34)xx6.3xx10^(14))/(1.6xx10^(-19))=2.608`eV
From Einstein.s equation, K.E. of emitted electrons
is given by, `E_(k) =E-w = 2.608-1.2 eV = 1.408 eV.`
`therefore` Stopping potential ` = 1.408` V.