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Matter Waves: de Broglie Wavelength question from NEET 2024

Q49NEET 2024Easy

Matter Waves: de Broglie Wavelength

The graph which shows the variation of (1λ2)\left(\frac{1}{\lambda^2}\right) and its kinetic energy, EE is (where λ\lambda is de Broglie wavelength of a free particle):

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Answer: (d)

  1. λ=h2mE⇒1λ2=2mh2E\lambda = \dfrac{h}{\sqrt{2mE}} \Rightarrow \dfrac{1}{\lambda^2} = \dfrac{2m}{h^2}E.
  2. So 1λ2∝E\dfrac{1}{\lambda^2} \propto E: a straight line through the origin.
  3. So the answer is (d).
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