Concepts Of Physics MCQ Edition [Volume 2]PhysicsPhotoelectric Effect and Wave–Particle Duality
For a monochromatic beam, the associated electric field becomes zero 1.2 10¹⁵ times per second. Determine the maximum kinetic energy of the photoelectrons emitted when this light is incident on a metal surface having a work function of 2.0 eV .
Options
- A2.48 eV
- B0.48 eV
- C4.48 eV
- D2.96 eV
Correct answer
B. 0.48 eV
Step-by-step solution
The electric field of an electromagnetic wave becomes zero twice in each cycle. Frequency of the light, = 1.2 10¹⁵ 2 = 6 10¹⁴ Hz Energy of the incident photon, E = h E = (4.14 10⁻¹⁵ eV s ) (6 10¹⁴ Hz ) = 2.484 eV Using Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by: K_ = E - K_ = 2.484 eV - 2.0 eV = 0.484 eV 0.48 eV