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JEE MainPhysicsDual Nature of Matter

Photons of two different energies illuminate a metal surface having a work function . The energy of the first incident photon is 2 and that of the second incident photon is 10 . The ratio of the de Broglie wavelength of the fastest emitted photoelectron in the first case to that in the second case is

Options

  1. A1 : 3
  2. B3 : 1
  3. C9 : 1
  4. D5 : 1

Correct answer

B. 3 : 1

Step-by-step solution

From Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by K_ max = E - For the first incident photon of energy E₁ = 2 : K₁ = 2 - = For the second incident photon of energy E₂ = 10 : K₂ = 10 - = 9 The de Broglie wavelength of a particle is related to its kinetic energy by = h 2mK Therefore, the ratio of the de Broglie wavelengths of the fastest photoelectrons is ₁ ₂ = K₂ K₁ Substituting the values of kinetic energies: ₁ ₂ = 9 = 9 = 3 The ratio is 3 : 1 . Answer: 3 :

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