JEE MainPhysicsDual Nature of Matter
Two monochromatic light sources A and B have the same power. When light from these sources is separately used to illuminate a photosensitive surface having a work function of 2.5 eV , the stopping potentials are found to be 1.5 V and 3.5 V respectively. The ratio of the number of photons emitted per second by source A to that by source B is :
Options
- A2:3
- B3:7
- C3:2
- D7:3
Correct answer
C. 3:2
Step-by-step solution
From Einstein's photoelectric equation, the energy of an incident photon is given by E = + eV_s , where is the work function and V_s is the stopping potential. For source A, the photon energy is: E_A = 2.5 eV + 1.5 eV = 4.0 eV For source B, the photon energy is: E_B = 2.5 eV + 3.5 eV = 6.0 eV The power of a light source is related to the number of photons emitted per second n and the energy of each photon E by the relation P = nE . Since both sources have the same power ( P_A = P_B ): n_A E_A = n_B E_B n_A n_B = E_