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MHT CET20228 Aug 2022Morning ShiftPhysicsDual Nature of MatterActual

Electrons of mass m with de-Broglie wavelength fall on the target. The cut-off wavelength ₀ is equal to [ h= Planck's constant, C= velocity of light ]

Options

  1. A2 m c ^2 h
  2. Bm c h
  3. C2 h m c ^2
  4. D2 m c h

Correct answer

A. 2 m c ^2 h

Step-by-step solution

Using de-Broglie equation = h p where p= 2 m E = h 2 m E Energy of the X-ray emitted E= h c ₀ = h 2 m h c ₀ ₀= 2 m c ^2 h

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