By how much would the stopping potential for given photosensitive surface go up if the frequency of the incident radiations were to be increased from `4xx10^(15) Hz` to `8xx10^(15)Hz`? Given `h = 6.4xx10^(-34) J.s, e = 1.6xx10^(-19)C and c = 3xx10^(8) m.s^(-1)`.

Explanation

Given frequencies , `v_1=4xx10^(15) Hz and v_2=8xx10^(15)Hz`
Einstein.s photoelectric equation is
`eV_0 = hv-W`
For `v_1=4xx10^(15) Hz`
`eV_(01)=hv_1-W" "...(i)`
For `v_2=8xx10^(15)Hz`
`eV_(02)=hv_2-W" "...(ii)`
Subtracting (i) from (ii) , we get
`e(V_(0_2)-V_(0_1))=h(v_2-v_1)`
`V_(0_2)-V_(0_1)=(h(v_2-v_1))/e=16.5V`