Question

By how which would the stopping potential for a given photosensitive surface go up if the frequency of the incident radiations were to be increased from `6xx10^(15)Hz` to `16xx10^(15)Hz`? Given, `h=6.6xx10^(-34)Js`, `e=1.6xx10^(-19)C and c=3xx10^8ms^-1`.

Answer 1

Here, `v_1=6xx10^(15)Hz,
v_2=16xx10^(15)Hz , V_(02)-V_(01)=?`
`eV_0=hv-phi_0`
`:. eV_(01)=hv_1-phi_0 and eV_(02)=hv_2-phi_0`
`:. e(V_(02)-V_(01))=h(v_2-v_1)`
or `V_(02)-V_(01)=h/e(v_2-v_1)`
`:. V_(02)-V_(01)=((6.4xx10^(-34)))/(1.6xx10^(-19))`
`x(16xx10^(15)-6xx10^(15))`
`=40V`