JEE Main202229 Jun 2022Evening ShiftPhysicsDual Nature of MatterActual
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
- A1 . 90   eV
- B3 . 27   eV
- C3 . 60   eV
- 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