Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, successively illuminate a metallic surface whose work function is 0.5 eV. Ratio of maximum speeds of emitted electrons will be :

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is Kmax=hv−ϕ0 where hv is the energy of incident photon and ϕ0 is the work function. But Kmax=12mv2max ∴12mv2max=hv−ϕ0 As per question, 12mv2max1=1 eV−0.5 eV=0.5 eV (i) and 12mv2max2=2.5 eV−0.5 eV=2 eV (ii) Dividing eqn. (i) by eqn. (ii), we get v2max1v2max2=0.5eV2eV=14 vmax1vmax2=√14=12.