NEETPhysicsDual Nature of Matter
Light of wavelength 400 nm is incident on a photosensitive metal surface whose work function is 2.1 eV . The stopping potential required to halt the fastest photoelectrons emitted from the surface is (Take hc = 1240 eV nm )
Options
- A5.2 V
- B3.1 V
- C2.1 V
- D1.0 V
Correct answer
D. 1.0 V
Step-by-step solution
The energy of the incident photons is given by: E = hc Using hc = 1240 eV nm and = 400 nm : E = 1240 400 = 3.1 eV According to Einstein's photoelectric equation, the maximum kinetic energy ( K_ ) of the emitted photoelectrons is: K_ = E - K_ = 3.1 eV - 2.1 eV = 1.0 eV The stopping potential ( V_s ) is related to the maximum kinetic energy by K_ = eV_s . Therefore, the stopping potential is 1.0 V . Answer: 1.0 V