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JEE Main 2021 (20 Jul, Shift 2) · Davisson-Germer, Uncertainty, Compton & X-Rays

Davisson-Germer, Uncertainty, Compton & X-Rays question from JEE Main 2021 (20 Jul, Shift 2)

Q8JEE Main 2021 · 20 JUL · SHIFT 2Medium

Davisson-Germer, Uncertainty, Compton & X-Rays

An electron having de-Broglie wavelength λ\lambda is incident on a target in a X-ray tube. Cut-off wavelength of emitted X-ray is:

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Answer: (c)

  1. Kinetic energy of electron: E=h22mλ2E = \frac{h^2}{2m\lambda^2}
  2. All of it can go into one photon: hcλ0=h22mλ2\frac{hc}{\lambda_0} = \frac{h^2}{2m\lambda^2}
  3. λ0=2mcλ2h\lambda_0 = \frac{2mc\lambda^2}{h}
  4. So the answer is (c).
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