JEE MainPhysicsDual Nature of Matter
A photosensitive surface with a work function of 2.2 eV is illuminated separately by four different lamps. The lamps are: A. 100 W Infrared lamp ( = 800 nm ) B. 50 W Red lamp ( = 620 nm ) C. 10 W Green lamp ( = 500 nm ) D. 5 W Ultraviolet lamp ( = 310 nm ) Which lamp will produce photoelectrons with a stopping potential of exactly 1.8 V ? (Take hc = 1240 eV nm )
Options
- A100 W Infrared lamp
- B50 W Red lamp
- C5 W Ultraviolet lamp
- D10 W Green lamp
Correct answer
C. 5 W Ultraviolet lamp
Step-by-step solution
The energy of an incident photon is given by E = hc . Calculating the photon energy for each lamp: For the Infrared lamp (A): E_A = 1240 800 = 1.55 eV For the Red lamp (B): E_B = 1240 620 = 2.0 eV For the Green lamp (C): E_C = 1240 500 = 2.48 eV For the Ultraviolet lamp (D): E_D = 1240 310 = 4.0 eV The work function of the surface is = 2.2 eV . Photoelectric emission occurs only if the incident photon energy is greater than the work function ( E > ). Thus, lamps A and B will not cause any emission. According to Ein