MHT CET202521 Apr 2025Morning 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 1: k , then the threshold frequency of metallic surface is
Options
- Ak v₂-v₁ k -1
- Bv₂-v₁ k
- Cv₁-v₂ k -1
- Dk v₁-v₂ k -1
Correct answer
D. k v₁-v₂ k -1
Step-by-step solution
Let h be Planck's constant and v₀ the threshold frequency. The maximum kinetic energy of emitted photoelectrons is given by KE_ = hv - hv₀ , where v is the incident frequency. For frequencies v₁ and v₂ , we have: KE_ 1 = hv₁ - hv₀ KE_ 2 = hv₂ - hv₀ Given that KE_ 1 / KE_ 2 = 1/k , this implies k(hv₁ - hv₀) = hv₂ - hv₀ . Dividing both sides by h gives k(v₁ - v₀) = v₂ - v₀ . Rearranging terms: kv₁ - kv₀ = v₂ - v₀ kv₁ - v₂ = kv₀ - v₀ kv₁ - v₂ = v₀(k - 1) Solving for v₀ yields v₀ = kv₁ - v₂ k - 1 , which corresponds to