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The difference between kinetic energies of photoelectrons emitted from a surface by the light of wavelength 2500 A ∘ and 5000 A ∘ will be

Options

  1. A1 . 61   eV
  2. B2 . 47 eV
  3. C3 . 96 eV
  4. D3.96 × 10 - 19 e V

Correct answer

B. 2 . 47 eV

Step-by-step solution

Δ E = h c λ 1 - h c λ 2 = h c ( λ 2 - λ 1 ) λ 1 λ 2 ( i n e V ) = 6.62 × 10 - 34 × 3 × 10 8 × ( 5000 - 2500 ) × 10 - 10 2500 × 5000 × 10 - 20 × 1.6 × 10 - 19 = 2.47 e V

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