Calculate the velocity of a photo-electron, if the work function of the target material is 1.24 eV and the wavelength of incident light is `4360Å`. What retarding potential is necessary to stop the emission of the electrons?

Explanation

Energy of incident photons in eV on metal surfacce is
`E= ( 12431)/( 4360) eV = 2.85 eV`
According to photo electric effect equation we have `E = phi + ( 1)/( 2) mv_("max")^(2)`
or `(1)/( 2) mv_("max")^(2) = E - phi`
`=2.85 - 1.24 eV = 1.64eV`
The stopping potential for these ejected electrons can be given as
`V_(0) = ( 1//2 mv_("max")^(2))/( e ) = ( 1.61 eV )/( e ) = 1.61 ` volts