A proton and a $\mathrm{Li}^{3+}$ nucleus are accelerated by the same potential. If $\lambda_{\mathrm{Li}}$ and $\lambda_{\mathrm{P}}$ denote the de Broglie wavelengths of $\mathrm{Li}^{3+}$ and proton respectively, then the value of $\frac{\lambda_{\mathrm{L}}}{\lambda_{\mathrm{p}}}$ is $\times \times 10^{-}$1. The value of $\mathrm{x}$ is
(Rounded off to the nearest integer)
(Mass of $\mathrm{Li}^{3+}=8.3$ mass of proton)
$\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}$
$\frac{\lambda_{\mathrm{Li}}}{\lambda_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}}(\mathrm{e}) \mathrm{V}}{\mathrm{m}_{\mathrm{Li}}(3 \mathrm{e})(\mathrm{V})}} \quad \mathrm{m}_{\mathrm{Li}}=8.3 \mathrm{~m}_{\mathrm{p}}$
$\frac{\lambda_{\mathrm{Li}}}{\lambda_{\mathrm{p}}}=\sqrt{\frac{1}{8.3 \times 3}}=\frac{1}{5}=0.2=2 \times 10^{-1}$
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