For photoelectric emission from certain metal the cut - off frequency is `v`. If radiation of frequency `2 v` incident on the metal plate , the maximum possible velocity of the emitted electron will be (`m` is the electron mass).

Options

A. `sqrt(hv//m)`
B. `sqrt(2hv//m)`
C. `2sqrt(hv//m)`
D. `sqrt(hv//(2m))`

Answer

D

Explanation

`E=phi+K.E._(max)`
`h(2v)=hv+1/2mv_(max)^2`
`v_(max)=sqrt((2hv)/m)`