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The de-Broglie wavelength of an electron moving with kinetic energy of 144   eV is nearly

Options

  1. A102 × 10 - 3   nm
  2. B102 × 10 - 4   nm
  3. C102 × 10 - 5   nm
  4. D102 × 10 - 2   nm

Correct answer

A. 102 × 10 - 3   nm

Step-by-step solution

The formula of de-Broglie wavelength, λ = h 2 m K . E . and we know that, K . E . = e V . Recall the proper relation between wavelength and electric potential, in angstrom, λ = 12 . 27 V Å = 12 . 27 144 × 10 - 10 = 1 . 02 × 10 - 10   m it means the wavelength, λ = 102 × 10 - 3   nm .

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