Threshold frequency of metal is `f_(0)`. When light of frequency `v = 2f_(0)` is incident on the metal plate, velocity of electron emitted in `V_(1)`. When a plate frequency of incident radiation is `5f_(0), V_(2)` is velocity of emitted electron, then `V_(1):V_(2)` is

Options

A. `1:4`
B. `1:2`
C. `2:1`
D. `4:1`

Answer

B

Explanation

`hv = hv_(0) +(1)/(2)mv^(2), v_(1) sqrt((2f_(0)h)/(m)), v_(2) = sqrt((8f_(0)h)/(m))`