When the light of frequency `2v_(0)` (where `v_(0)` is threshold frequency) is incident on a metal plate, the maximum velocity of electrons emitted is `v_(01)`. 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. `1 : 4`
C. `1 : 2`
D. `2 : 1`

Answer

C

Explanation

`E = W_(0) + (1)/(2) mv^(2)`
For the first case, `h (2v_(0)) = hv_(0) + (1)/(2) mv_(1)^(2)`
or, `hv_(0) = (1)/(2) mv_(1)^(2)`
For the second case, `h(5v_(0)) = hv_(0) + (1)/(2) mv_(2)^(2)`
or, `4hv_(0) = (1)/(2) mv_(2)^(2)`
Now dividing (1) by (2), we have
`(v_(1)^(2))/(v_(2)^(2)) = (hv_(0))/(4hv_(0)) = (1)/(4) or, (v_(1))/(v_(2)) = 1 : 2`