A photosensitive metallic surface has work function, `h v_(0)`. If photons of energy `2hv_(0)` fall on this surface, the electrons come out with a maximum velocity of `4 xx 10^(6) m/s`. When the photon energy is increased to `5hv_(0)`, then photon energy is increased to photoelectrons will be

Options

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

Answer

D

Explanation

`(v_(1))/(v_(2)) = sqrt((2hv_(0) - hv_(0))/(5hv_(0) - hv_(0))) = 1/2 rArr v_(2) = 8 xx 10^(6) ms^(-1)`