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 :
- $\lambda_\alpha < \lambda_{\mathrm{p}} < \lambda_{\mathrm{e}}$
- $\lambda_{\mathrm{e}}>\lambda_\alpha>\lambda_{\mathrm{p}}$
- $\lambda_{\mathrm{p}}>\lambda_{\mathrm{e}}>\lambda_\alpha$
- $\lambda_{\mathrm{p}} < \lambda_{\mathrm{e}} < \lambda_\alpha$
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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