When the light of frequency `2 v_(0)` (where `v_(0)` is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is `v_(1)`. When the frequency of the incident radiation is increased to `5 v_(0)`, the maximum velocity of electrons emitted from the same plate is `v_(2)`. the ratio of `v_(1)` to `v_(2)` is

Options

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

Answer

D

Explanation

Case`1:h.2v_(0)=hv_(0)+1/2mv_(1)^(2)implies hv_(0)=1/2mv_(1)^(2)...(i)`
Case `2: h.5v_(0)=hv_(0)+1/2mv_(2)^(2)`
`implies h(4v_(0))=1/2mv_(2)^(2).....(ii)`
Dividing eqns. `(i)` and `(ii)` we get
`(v_(1))/(v_(2))=1/2`