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JEE MainPhysicsDual Nature of Matter

Light of wavelength 310 nm is incident on a metal surface whose work function is 2.5 eV . The stopping potential required to halt the fastest emitted photoelectrons is : (Take hc = 1240 eV nm )

Options

  1. A6.5 V
  2. B1.5 V
  3. C4.0 V
  4. D2.5 V

Correct answer

B. 1.5 V

Step-by-step solution

The energy of the incident photons is given by: E = hc E = 1240 eV nm 310 nm = 4.0 eV According to Einstein's photoelectric equation, the maximum kinetic energy ( K_ ) of the emitted photoelectrons is: K_ = E - ₀ K_ = 4.0 eV - 2.5 eV = 1.5 eV The stopping potential V_s is related to the maximum kinetic energy by K_ = eV_s . Therefore, eV_s = 1.5 eV V_s = 1.5 V Answer: 1.5 V

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