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Proton (p) and electron (e) will have same deBroglie wavelength when the ratio of their momentum is (assume, $m_p=1849 \mathrm{~m}_{\mathrm{e}}$ )
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$1: 1$
De Broglie wavelength is $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$
$\lambda_{\mathrm{p}}=\lambda_{\mathrm{e}} \Rightarrow \frac{\mathrm{h}}{\mathrm{m}_{\mathrm{p}} \mathrm{v}_{\mathrm{p}}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{e}}}$
$\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{e}}=\mathrm{m}_{\mathrm{p}} \mathrm{v}_{\mathrm{p}} \Rightarrow \mathrm{p}_{\mathrm{e}}=\mathrm{p}_{\mathrm{p}}$
$\therefore \frac{\mathrm{p}_{\mathrm{p}}}{\mathrm{p}_{\mathrm{e}}}=\frac{1}{1}$
$\lambda_{\mathrm{p}}=\lambda_{\mathrm{e}} \Rightarrow \frac{\mathrm{h}}{\mathrm{m}_{\mathrm{p}} \mathrm{v}_{\mathrm{p}}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{e}}}$
$\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{e}}=\mathrm{m}_{\mathrm{p}} \mathrm{v}_{\mathrm{p}} \Rightarrow \mathrm{p}_{\mathrm{e}}=\mathrm{p}_{\mathrm{p}}$
$\therefore \frac{\mathrm{p}_{\mathrm{p}}}{\mathrm{p}_{\mathrm{e}}}=\frac{1}{1}$
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