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.
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