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Threshold frequency of metal is f(0). Wh...

Threshold frequency of metal is `f_(0)`. When light of frequency `v = 2f_(0)` is incident on the metal plate, velocity of electron emitted in `V_(1)`. When a plate frequency of incident radiation is `5f_(0), V_(2)` is velocity of emitted electron, then `V_(1):V_(2)` is

A

`1:4`

B

`1:2`

C

`2:1`

D

`4:1`

Text Solution

Verified by Experts

The correct Answer is:
B
`v^2 = (2h(?-?_0))/(m), v_1^2 = (2h(2f_0 - f_0))/(m) ----(1)`
`V_2^2= (2h(f_0- f_0))/(m)-----(2)`
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