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. `4xx10^(6) m//sec`
B. `6xx10^(6) m//sec`
C. `8xx10^(6) m//sec`
D. `16xx10^(6) m//sec`

Answer

C

Explanation

`h.2 upsilon_(0) = h upsilon_(0) +(1)/(2)m(4xx10^(6))^(2)` ….(i)
`h.5 upsilon_(0) = h upsilon_(0) +(1)/(2)m v_(max)^(2)` …..(ii)
`(v_(max)^(2))/((4xx10^(6))^(2)) =4 implies v_(max) =8xx10^(6) m//sec`