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The electric field at a point associated with a light wave is given by E = 200 sin 6 × 10 15 t + sin 9 × 10 15 t Vm - 1 Given: h = 4 . 14 × 10 - 15 eVs If this light falls on a metal surface having a work function of 2 . 50 eV , the maximum kinetic energy of the photoelectrons will be

Options

  1. A1 . 90   eV
  2. B3 . 27   eV
  3. C3 . 60   eV
  4. D3 . 42   eV

Correct answer

D. 3 . 42   eV

Step-by-step solution

Given: E = 200 sin 6 × 10 15 t + sin 9 × 10 15 t   V   m - 1 Here, angular velocity, ω 1 = 6 × 10 15   and ω 2 = 9 × 10 15 Maximum kinetic energy of the photoelectron is KE max . = E - ϕ = h f - ϕ = h ω 2 π - ϕ = 4 . 14 × 10 - 15 × 9 × 10 15 2 × 3 . 14 - 2 . 5 = 5 . 92 - 2 . 50 = 3 . 42   eV

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