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Electrons of mass, m with de-Broglie wavelength, λ fall on the target in an X-ray tube. The cutoff wavelength, λ 0 of the emitted X-ray is

Options

  1. Aλ 0 = 2 m c λ 2 h
  2. Bλ 0 = 2 h m c
  3. Cλ 0 = 2 m 2 c 2 λ 3 h 2
  4. Dλ 0 = λ

Correct answer

A. λ 0 = 2 m c λ 2 h

Step-by-step solution

Momentum, P = h λ ⇒ E = P 2 2 m ⇒ h 2 2 m λ 2 = h c λ 0 ⇒ λ 0 = h c h 2 2 m λ 2 = 2 m c λ 2 h .

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