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

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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

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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'$$=K+hv$$

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

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