If the wavelength of the light incident on a photoelectric cell be reduced from `lambda_(1)` to `lambda_(2)` `Å`, then what will be the change in the cut-off potential?

Explanation

Let the work function of the surface be `phi`. If v be the frequency of the light falling on the surface, then according the Einstein's photoelectric equation, the maximum kinetic energy `KE_("max")` of emitted electron is given by
`KE_("max") = hv- phi = ( hc)/(lambda) - phi`
We know that, `KE_("max") = eV_(0)`
Where `V_(0) = ` cut-off potential
or `eV_(0) =( hc)/( lambda) -phi` or `V_(0) = ( hc)/( e lambda) - ( phi)/( e ) `
Now, `Delta V_(0)= V_(02) - V_(01)`
`= ((hc)/( elambda_(2)) - ( phi) /(e )) - ((hc)/( e lambda_(1)) - ( phi) /( e )) `
`= ( hc)/( e ) ((1)/( lambda_(2)) - ( 1)/( lambda_(1))) = ( hc)/( e ) ((lambda_(1)- lambda_(2))/( lambda_(1) lambda_(2)))`