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Question: Answered & Verified by Expert
A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as :
PhysicsDual Nature of MatterJEE MainJEE Main 2024 (09 Apr Shift 1)
Options:
  • A $\lambda_\alpha < \lambda_{\mathrm{p}} < \lambda_{\mathrm{e}}$
  • B $\lambda_{\mathrm{e}}>\lambda_\alpha>\lambda_{\mathrm{p}}$
  • C $\lambda_{\mathrm{p}}>\lambda_{\mathrm{e}}>\lambda_\alpha$
  • D $\lambda_{\mathrm{p}} < \lambda_{\mathrm{e}} < \lambda_\alpha$
Solution:
1860 Upvotes Verified Answer
The correct answer is: $\lambda_\alpha < \lambda_{\mathrm{p}} < \lambda_{\mathrm{e}}$
$\begin{aligned} & \lambda_{\mathrm{DB}}=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}} \\ & \Rightarrow \lambda_{\mathrm{DB}} \alpha \frac{1}{\sqrt{\mathrm{m}}} \\ & \Rightarrow \lambda_{\mathrm{a}} < \lambda_{\mathrm{p}} < \lambda_{\mathrm{e}}\end{aligned}$

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