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MHT CET 2026 (11 Apr, Shift 1) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from MHT CET 2026 (11 Apr, Shift 1)

Q20MHT CET 2026 · 11 APR · SHIFT 1Medium

Matter Waves: de Broglie Wavelength

A photon and an electron have equal energy E. The ratio of λ\lambda(electron) to λ\lambda(photon) is proportional to

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Answer: (b)

  1. λe=h2mE\lambda_e = \dfrac{h}{\sqrt{2mE}}, λph=hcE\lambda_{ph} = \dfrac{hc}{E}.
  2. λeλph=Ec2mE=Ec2m∝E\dfrac{\lambda_e}{\lambda_{ph}} = \dfrac{E}{c\sqrt{2mE}} = \dfrac{\sqrt{E}}{c\sqrt{2m}} \propto \sqrt{E}
  3. So the answer is (b) E:1\sqrt{E} : 1.
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