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When photons of energy 4 . 0 eV fall on the surface of a metal A , the ejected photoelectrons have maximum kinetic energy T A (in eV ) and a de-Broglie wavelength λ A . When the same photons fall on the surface of another metal B , the maximum kinetic energy of ejected photoelectrons is T B = T A − 1 .5 eV . If the de-Broglie wavelength of these photoelectrons is λ B = 2 λ A , then the work function of metal B is

Options

  1. A2   eV
  2. B3   eV
  3. C2.5   eV
  4. D3.5   eV

Correct answer

D. 3.5   eV

Step-by-step solution

The relation between de-Broglie wavelength and kinetic energy is λ = h 2 K m e λ A λ B = K B K A ⇒ 1 2 = T A - 1 . 5 T A     ⇒ T A = 2   eV ⇒ K B = 2 - 1 . 5 = 0 . 5   eV So, the work-function of B is ϕ B = 4 . 0 - 0 . 5 = 3 . 5   eV

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