Photoelectric emission is observed from a metallic surface for frequencies ₁ and ₂ of the incident light rays ( ₁ > ₂ ). If the ratio of maximum value of kinetic energy of the photoelectron emitted in first case to that in second case 3 : K, then the threshold frequency of the metallic surface is

Step-by-step solution

Using Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by K_ max = h - h ₀ , where ₀ is the threshold frequency. For the first case: K₁ = h ₁ - h ₀ For the second case: K₂ = h ₂ - h ₀ Given the ratio of maximum kinetic energies is 3 : K : K₁ K₂ = 3 K Substituting the expressions for K₁ and K₂ : h ₁ - h ₀ h ₂ - h ₀ = 3 K Canceling h and cross-multiplying: K( ₁ - ₀) = 3( ₂ - ₀) K ₁ - K ₀ = 3 ₂ - 3 ₀ Rearranging to solve for ₀ : 3 ₀ - K ₀ = 3 ₂ - K ₁ ₀(3 - K) = 3 ₂ - K ₁