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

Answer

A

Explanation

Work function , `phi= hupsilon`
According to Einstein.s photoelectric equation
`1/2mv_(max)^2 =h(2upsilon)-hupsilon` `1/2mv_(max)^2 =hupsilon" or " v_(max)^2=(2hupsilon)/m`
`v_(max)=sqrt((2hupsilon)/m)`