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

Answer

C

Explanation

As `(1)/(2)mv_(max)^(2)=hvimplies v_(max)^(2)=(2hv)/(m)`
`therefore v_(max)=sqrt((2hv)/(m))`