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The kinetic energy of an electron having de-Broglie wavelength λ is ( h = Planck's constant, m = mass of electron)

Options

  1. Ah 2 m λ
  2. Bh 2 2 m λ 2
  3. Ch 2 2 m 2 λ 2
  4. Dh 2 2 m 2 λ

Correct answer

B. h 2 2 m λ 2

Step-by-step solution

λ = h 2 m K . E λ 2 = h 2 2 m ( K . E ) ( K . E ) = h 2 2 m λ 2

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