In a photoemissive cell with executing wavelength λ, the fastest electron has speed v. If the exciting wavelength is changed to 3λ/4, the speed of the fastest emitted electron will be

hv−Wo=12mvmax2⇒hcλ−hcλo=12mvmax2⇒hcλo−λλλo=12mvmax2⇒vmax2hcmλo−λλλoWhen  wavelength  is  λ  and  velocity  is  v,  thenv=2hcmλo−λλλo    .     .     .    .     iWhen  wavelength  is  3λ4and  velocity  is  v'  thenv'=2hcmλo−3λ/43λ/4×λo     .     .     iiDividing  equation  ii  by  i,  we  getv'v=λo3λ/434λλo×λλoλo−λv'=v431/2λ0−3λ/4λo−λi.e.,   v'>431/2