A photosensitive metallic surface has work funtion `hv_(0)`. If photons of energy `2hv_(0)` fall on this surface the electrons come out with a maximum velocity of `4xx10^(6) m//s`. When the photon energy is increases to `5hv_(0)` then maximum velocity of photo electron will be

Options

A. `2xx10^(7)ms^(-1)`
B. `8xx10^(6)ms^(-1)`
C. `2xx10^(6)ms^(-1)`
D. `8xx10^(5)ms^(-1)`

Answer

B

Explanation

As `(1)/(2)mv_(1)^(2)=2hv_(0)-hv_(0)=hv_(0) and (1)/(2)mv_(2)^(2)=5hv_(0)hv_(0)=4hv_(0)`
So, `(1)/(2) mv_(2)^(2)=4xx(1)/(2)mv_(1)^(2)`
or `v_(2)=2v_(1)=2xx4xx10^(8)`
`=8xx10^(8)ms^(-1)`.