NTA Abhyas JEE Main2020PhysicsDual Nature of MatterPractice
A light of intensity 16 m W and energy of each photon 10 e V incident on a metal plate of work function 5 eV and area 10 - 4 m 2 then find the maximum kinetic energy of emitted electrons and the number of photoelectrons emitted per second if photon efficiency is 10 %.
Options
- A5   eV ,   10 11
- B10   eV ,   10 12
- C5   eV ,   10 13
- D10   eV ,   10 14
Correct answer
A. 5   eV ,   10 11
Step-by-step solution
Maximum K E = h C λ - ϕ = 10 - 5 = 5   eV Intensity I = N p h C λ t A ⇒ N p t = I A h C λ and the number of emitted photoelectrons per second. = N p t × 10 % = I A h C λ × 1 10 = 16 × 10 - 3 × 10 - 4 10 × 1.6 × 10 - 19 × 10 = 10 11