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MHT CET 2021 (21 Sep, Shift 2) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from MHT CET 2021 (21 Sep, Shift 2)

Q91MHT CET 2021 · 21 SEP · SHIFT 2Easy

Matter Waves: de Broglie Wavelength

A proton and alpha particle are accelerated through the same potential difference. The ratio of the de-Broglie wavelength of proton to that of alpha particle will be (mass of alpha particle is four times mass of proton.)

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Answer: (b)

  1. λ∝1mq\lambda \propto \dfrac{1}{\sqrt{mq}} for the same accelerating potential.
  2. λpλα=4m⋅2em⋅e=8=22\dfrac{\lambda_p}{\lambda_\alpha} = \sqrt{\dfrac{4m\cdot2e}{m\cdot e}} = \sqrt{8} = 2\sqrt{2}.
  3. So the answer is (b) 22:12\sqrt{2} : 1.
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