The threshold frequency for certain metal is `v_0`. When light of frequency `2v_0` is incident on it, the maximum velocity of photoelectrons is `4xx10^(6) ms^(-1)`. If the frequency of incident radiation is increaed to `5v_0`, then the maximum velocity of photoelectrons will be

Options

A. `(4)/(5)xx10^(6)ms^(-1)`
B. `2xx10^(6)ms^(-1)`
C. `8xx10^(6)ms^(-1)`
D. `2xx10^(7)ms^(-1)`

Answer

C

Explanation

In the first case,
`(1)/(2)mv_(max)^(2)=2hupsilon_0-hupsilon_0=hupsilon_0`
In the second case,
`(1)/(2)mv_(max)^(2)=5hupsilon_0-hupsilon_0=4hupsilon_0`
Clearly, `v_(max)` is doubled.