The frequency of incident light falling on a photosensitive metal plate is doubled, the K.E of the emitted photo-electrons is

Options

A. double the earlier value
B. unchanged
C. more than doubled
D. less than doubled

Answer

C

Explanation

let `E_(1) and E_(2)` be the KE of photoelectron for incident light of frequency v and 2v, respectively.
then, `hv=E_(1)+phi_(0)`
or `h2v=E_(2)+phi`
so, `2(E_(1)+phi_(0))=E_(2)+phi`
or `E_(2)=2E_(1)+phi_(0)`
it means the KE of photoelectron becomes more than double.