A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be…

A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as :
  1. $\lambda_\alpha < \lambda_{\mathrm{p}} < \lambda_{\mathrm{e}}$
  2. $\lambda_{\mathrm{e}}>\lambda_\alpha>\lambda_{\mathrm{p}}$
  3. $\lambda_{\mathrm{p}}>\lambda_{\mathrm{e}}>\lambda_\alpha$
  4. $\lambda_{\mathrm{p}} < \lambda_{\mathrm{e}} < \lambda_\alpha$

Solution

$\begin{aligned} & \lambda_{\mathrm{DB}}=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}} \\ & \Rightarrow \lambda_{\mathrm{DB}} \alpha \frac{1}{\sqrt{\mathrm{m}}} \\ & \Rightarrow \lambda_{\mathrm{a}} < \lambda_{\mathrm{p}} < \lambda_{\mathrm{e}}\end{aligned}$

Asked in: JEE Main 2024 (09 Apr Shift 1)

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