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. `2 xx 10^(7) m//s`
B. `8 xx 10^(6) m//s`
C. `2 xx 10^(6) m//s`
D. `8 xx 10^(5)m//s`

Answer

2

Explanation

`(1)/(2)mv_(max)^(2)=2hv_(0)-hv_(0)`
`(1)/(2)mv_(max)^(2)=5hv_(0)-hv_(0)=4hv_(0)`
So, `v_(max) = 8 xx 10^(6)`