Concepts Of Physics MCQ Edition [Volume 2]PhysicsPhotoelectric Effect and Wave–Particle Duality
The electric field associated with a light wave at a point is given by E = (100 Vm ⁻¹) ! [(3.0 10¹⁵ s ⁻¹) t ] ! [(6.0 10¹⁵ s ⁻¹) t ] . When this light is incident on a metal surface with a work function of 2.0 eV , determine the maximum kinetic energy of the emitted photoelectrons.
Options
- A5.93 eV
- B3.93 eV
- C1.95 eV
- D35.26 eV
Correct answer
B. 3.93 eV
Step-by-step solution
The given electric field of the light wave is: E = 100 (3.0 10¹⁵ t) (6.0 10¹⁵ t) Using the trigonometric identity 2 A B = (A-B) - (A+B) , the electric field can be written as: E = 50 [ (3.0 10¹⁵ t) - (9.0 10¹⁵ t)] This indicates that the light wave is a superposition of two waves with angular frequencies ₁ = 3.0 10¹⁵ s ⁻¹ and ₂ = 9.0 10¹⁵ s ⁻¹ . The maximum kinetic energy of the emitted photoelectrons is determined by the photon with the higher frequency, which is ₂ = 9.0 10¹⁵ s ⁻¹ . The energy of this incident pho