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JEE MainPhysicsDual Nature of Matter

When a metallic surface is illuminated with radiation of photon energy 5 eV , the minimum de Broglie wavelength of the emitted photoelectrons is ₁ . When the same surface is illuminated with radiation of photon energy 14 eV , the minimum de Broglie wavelength of the emitted photoelectrons is ₂ . If the ratio ₁ : ₂ = 2 : 1 , the work function of the metallic surface is

Options

  1. A4 eV
  2. B17 eV
  3. C2 eV
  4. D23 eV

Correct answer

C. 2 eV

Step-by-step solution

Let the work function of the metal be . The maximum kinetic energy of the emitted photoelectrons is given by Einstein's photoelectric equation: K_ = E - . For the first radiation: K₁ = 5 - For the second radiation: K₂ = 14 - The minimum de Broglie wavelength of an electron is related to its maximum kinetic energy by = h 2mK_ , which implies 1 K_ . Given the ratio ₁ ₂ = 2 , we have: K₂ K₁ = 2 K₂ K₁ = 4 Substituting the expressions for kinetic energies: 14 - 5 - = 4 14 - = 4(5 - ) 14 - = 20 - 4 3 = 6 = 2 eV . Answer:

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