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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

  1. A5.2 V
  2. B3.1 V
  3. C2.1 V
  4. 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

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