MHT CET202611 April 2026Morning ShiftPhysicsDual Nature of MatterActual
Photoelectric emission is observed from a metallic surface for frequencies v₁ and v₂ of the incident light rays ( v₁ > v₂ ). If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of k : 1 , then what is the threshold frequency of the metallic surface?
Options
- Av₁ - v₂ k
- Bv₁ - v₂ k - 1
- Ckv₁ - v₂ k - 1
- Dkv₂ - v₁ k - 1
Correct answer
D. kv₂ - v₁ k - 1
Step-by-step solution
Using Einstein's photoelectric equation, the maximum kinetic energy of photoelectrons is given by K = h(v - v₀) , where v₀ is the threshold frequency. For frequency v₁ , K₁ = h(v₁ - v₀) For frequency v₂ , K₂ = h(v₂ - v₀) Given that the ratio of maximum kinetic energies is k : 1 , we have: K₁ K₂ = h(v₁ - v₀) h(v₂ - v₀) = k v₁ - v₀ v₂ - v₀ = k v₁ - v₀ = k(v₂ - v₀) v₁ - v₀ = kv₂ - kv₀ kv₀ - v₀ = kv₂ - v₁ v₀(k - 1) = kv₂ - v₁ v₀ = kv₂ - v₁ k - 1 Answer: kv₂ - v₁ k - 1