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MHT CET 2024 (16 May, Shift 2) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from MHT CET 2024 (16 May, Shift 2)

Q58MHT CET 2024 · 16 MAY · SHIFT 2Medium

Matter Waves: de Broglie Wavelength

Kinetic energy of a proton is equal to energy EE of a photon. Let ' λ1\lambda_1 ' be the de-Broglie wavelength of proton and ' λ2\lambda_2 ' be the wavelength of photon. If (λ1λ2)∝En\left(\frac{\lambda_1}{\lambda_2}\right) \propto E^n then the value of ' nn ' is

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Answer: (d)

  1. Proton: λ1=h2mE\lambda_1 = \dfrac{h}{\sqrt{2mE}}. Photon: λ2=hcE\lambda_2 = \dfrac{hc}{E}.
  2. λ1λ2=Ec2m∝E1/2\dfrac{\lambda_1}{\lambda_2} = \dfrac{\sqrt{E}}{c\sqrt{2m}} \propto E^{1/2}
  3. So n=0.5n = 0.5: the answer is (d).
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