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

Options

A. `1.6xx10^(18)Hz`
B. `0.8xx10^(15)Hz`
C. `8xx10^(15)Hz`
D. `8xx10^(16)Hz`

Answer

C

Explanation

When a certain metal was ……….
`hv = hv_(0)+KE`
`v-v_(0)=KE//h`
`v_(2)-v_(0)=2v_(1)-v_(2)`
`= 4xx10^(16) - 3.2xx10^(18)`
`= 0.8xx10^(16) = 8xx10^(15)Hz`