# Solve the following

Question:

A proton and $\mathrm{aLi}^{3+}$ 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 $\frac{\lambda_{\mathrm{Li}}}{\lambda_{\mathrm{p}}}$ is $\mathrm{x} \times 10^{-1}$. The value of $x$ is _________

[Rounded off to the nearest integer] [Mass of $\mathrm{Li}^{3+}=8.3$ mass of proton]

Solution:

(2)

De Brogir Davelength $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk} . \mathrm{E}}}$

$\frac{\lambda_{\mathrm{Li}+3}}{\lambda_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}} \times\left(\mathrm{e}^{-} \mathrm{v}\right)_{\mathrm{p}}}{\mathrm{m}_{\mathrm{L}^{+3} \times 3 \mathrm{e}_{\mathrm{p}}} \mathrm{V}}}$

$m_{\mathrm{L}^{+3}}=8.3 \mathrm{mp}$

$\frac{\lambda_{L i+3}}{\lambda_{p}}=\sqrt{\frac{m_{p}}{3 \times 8.3 m_{p}}}=\sqrt{\frac{1}{25}}$

$=\frac{1}{5}=0.2=2 \times 10^{-1}$

$x=2$

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