The threshold frequency for a certain photosensitive metal is `v_(0)`. When it is illuminated by light of frequency `v=2 v_(0)`, the maximum velocity of photoelectrons is `v_(0)`.The maximum velocity of the photoelectrons when the same metal is illuminated by light of frequency `v=5 v_(0)` is `x v_(0)`. Find `x`

Options

A. `sqrt2v_0`
B. `2v_0`
C. `2sqrt2v_0`
D. `4v_0`

Answer

B

Explanation

`k_("max") = hv - phi_0 `
`1/2 mv_0^2 = h (2v_0) - hv_0 = hv_0" "...(i)`
`1/2 mv_1^2 = h (5v_0) - hv_0 = 4hv_0" "...(ii)`
From (i) and (ii) –
`v_1^2/v_0^2=4/1 impliesv_1^2=4v_0^2impliesv_1 =2v_0`