question_answer

Ultraviolet light is incident on the two photosensitive materials having work functions \[{{W}_{1}}\]  and \[{{W}_{2}}({{W}_{1}}>{{W}_{2}}).\] In which case win the kinetic energy of the emitted electrons be greater and why?

Answer:

Since, KE of a photoelectron \[=hv-{{W}_{0}}\] Hence, the KE of emitted electrons will be greater for that photo-sensitive material, which has smaller work function, i.e.\[{{W}_{2}}\]. As \[W=h{{v}_{0}}\] (h is constant) where, \[{{v}_{0}}\] is threshold frequency. As work function of material 1 is greater than material 2, material 1 need more energy for ejection of photoelectron and impart less kinetic energy to the photoelectrons. Therefore, kinetic energy of the ejected electron in case 2 is more.