Work function of photo sensitive is `hf_(0)` If surface is incident with photon of `2hf_(0)` energy,then maximum speed of electron emitted is `4xx10^(6)` m/s .If energy of photon incident is changed to `5hf_(0)` then maximum speed of photoelectron will be.....

Options

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

Answer

C

Explanation

We know that,
`hf-phi=(1)/(2)mv_(max)^(2)`
`therefore hf-phi prop v_(max)^(2)`
`therefore (5hf_(0)-hf_(0))/(2hf_(0)-hf_(0))=(v_(2)^(2))/(v_(1)^(2))`
`therefore 4=(v_(2)^(2))/(v_(1)^(2))`
`therefore 2=(v_(2))/(v_(1))`
`therefore v_(2)=2v_(1)=2xx4xx10^(6)=8xx10^(6) m//s`