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A certain metal when irradiated by light...

A certain metal when irradiated by light `(v=3.2xx10^(16)Hz)` emits photoelectrons with twice of K.E. as did photoelectrons when the same metal is irradiated by light `(v=2.0xx10^(16)Hz)`. The `v_(0)` of the metal is

A

`12 xx 10^(14) Hz`

B

`8 xx 10^(15) Hz`

C

`1.2 xx 10^(16) Hz`

D

`4 xx 10^(12) Hz`

Text Solution

Verified by Experts

The correct Answer is:
D
`KE = hv_(0) - hv_(0)`
`hv_(1) - hv_(0)= 2(hv_(2) - hv_(0))`
`v_(0) = 2(v_(2) - v_(1))`
`= 2(2.0 xx 10^(-16)) - (3.2 xx 10^(16))`
`= 8 xx 10^(15) s^(-1) = 8 xx 10^(-15) Hz`
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