A photosensitive surface has work function `W_(0)=hv_(0)`. If photons of energy `2hv_(0)` fall on this surface, the electrons are ejected with a maximum velocity of `4xx10^(6)m//s`. When the energy of the incident photon is increased to `5hv_(0)`, then maximum velocity of the photoelectrons will be

Options

A. `2xx10^(7)`m/s
B. `8xx10^(6)`m/s
C. `8xx10^(5)` m/s
D. `2xx10^(6)` m/s

Answer

B

Explanation

`E=hv=W_(0)+1/2mv_(m)^(2)`
`:.2hv_(0)=hv_(0)+1/2mv_(1)^(2)`
`:.hv_(0)=1/2mv_(1)^(2)` ...(1)
and `5hv_(0)=hv_(0)+1/2mv_(2)^(2)`
`:.4hv_(0)=1/2mv_(2)^(2)` ...(2)
`:.` From (1) and (2) `:.4/1=(v_(2)/v_(1))^(2)`
`:.v_(2)=2v_(1)=2xx4xx10^(6)=8xx10^(6)m//s`