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. `2 xx 10^6 m//s`
C. `8 xx 10^6 m//s`
D. `8 xx 10^5 m//s`

Answer

C

Explanation

We know that
`hv - phi = K_(max) = 1.2 mv_(max)^2`
According to question
`(5hv_0 - hv_0)/(2hv_0 - hv_0) = (v_2^2)/(v_1^2)`
`v_2 = 2v_1 = 2 xx 4 xx 10^6 = 8 xx 10^6 m//s`