In a photocell, with excitation wavelength λ, the faster electron has speed v. If the excitation wavelength is changed to 3λ/4, the speed of the fastest electron will be

12mv2=hcλ−W0   .........(i)Let the speed of the fastest electron be V1 when excitation wavelength is changed to 3λ/4∴ 12mv12=4hc3λ−W0⇒ 12mv12=43hcλ−W0+W03⇒ 12mv12=4312mv2+W03     [using Eq.(i)] ⇒ v12=4v23+2W03m∴ v1>43v