If the wavelength of light incident on a photoelectric cell be reduced from `4000Å` to `3600Å`, then what will be the change in the cut off potential? `(h=6.6xx10^(-34)Js , e=1.6xx10^(-19)C)`

Explanation

Let `phi_0` be the work function of the surface of the photoelectric cell to be exposed by light. Maximum KE of emitted photoelectron will be
`K_(max)=(hc)/lambda-phi_0`
If the `V_0` is the cut off potential, then
`eV_0=K_(max)=(hc)/lambda-phi_0`
or `V_0=(hc)/(e lambda)-(phi_0)/e`
When the incident light of wavelength `lambda_1` be changed to wavelength `lambda_(2)`, then the change (increase) in cut off potential will be
`DeltaV_0=((hc)/(elambda_2)-(phi_0)/e)-((hc)/(elambda_1)-(phi_0)/e)`
`=(hc)/e (1/(lambda_1)-1/(lambda_1))`
`=((6.6xx10^(-34))xx(3xx10^8))/((1.6xx10^(-19)))[(10^(10))/3600-(10^(10))/4000]`
`=0.34V`