When photons of energy hv fall on an aluminium plate (of work function Eo), photoelectrons of maximum kinetic energy K are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be

According to Einstein,s photoelectric effect energy of photon : KE of photoelectron + word function of metal. That is, hc=12mv2+Eo   or  hc=Kmax+Eo   .    .    .   iNow,   we  have  given,   v'=2vTherefore,   Kmax'=h2v−EoKmax'=2hv−Eo               .    .       .   iiFrom  Eq.i   and  ii,  we  have  Kmax'=2Kmax+Eo−Eo=2Kmax+Eo=Kmax+Kmax+Eo=Kmax+hc                    from   Eq.iPutting  Kmax=K⇒Kmax'=K+hv