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

Step-by-step solution

Kinetic energy, ( K h=f W ) where ( W ) is work function; ( f ) is frequency of incident photon. For photon of energy ( 1 e V, K₁=1 e V-0.5 e V=0.5 e V ) For photon of energy ( 2.5 eV , K₂=2.5 eV -0.5 eV =2.0 eV ) Taking ratio, ( K₁ K₂ = 0.5 eV 2.0 eV ) ( 1 2 m v₁² 1 2 m v₂² = 0.5 2.0 v₁² v₂² = 1 4 v₁ v₂ = 1 2 ) Therefore, ratio of maximum speeds of emitted electrons will be ( 1: 2 ).