Question:

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

Updated On: Jan 4, 2025
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  • where λ is de Broglie wavelength of a free particle
  • where λ is de Broglie wavelength of a free particle
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The Correct Option is D

Solution and Explanation

Using $\lambda = \frac{h}{\sqrt{2mE}}$, we find $\frac{1}{\lambda^2} \propto E$, yielding a linear graph.

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