A photo sensitive metallic surface has work funcation `hv_(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 `5 hv_(0)`, then maximum velocity of photo electrons will be-

Options

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

Answer

D

Explanation

`1/2mv_("max")^2 =E-phi_0 =hv-hv_0`
`implies 1/2 mv_1^2 = 2hv_0 -hv_0 =hv_0`
And `1/2 mv_2^2 = 5hv_0 -hv_0 = 4hv_0`
`(v_2/v_1)^2=4/1impliesv_2 =2v_1 = 2xx4 xx10^6`
`=8xx10^6m//s`