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A proton and an $\alpha$-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths $\lambda_{\mathrm{p}}$ and $\lambda_\alpha$ related to each other?
Solution:
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Verified Answer
As, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ or $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqv}}}$
$$
\therefore \quad \frac{\lambda_{\mathrm{p}}}{\lambda_\alpha}=\frac{\not}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}} \times v_{\mathrm{p}}}} \times \frac{\sqrt{\not \mathrm{m}_\alpha \mathrm{q}_\alpha v_\alpha}}{K}
$$
Given,
$$
\begin{aligned}
&\mathrm{m}_\alpha=4 \mathrm{~m}_{\mathrm{p}} \\
&\mathrm{q}_\alpha=2 \mathrm{e} ; \\
&\mathrm{q}_{\mathrm{p}}=\mathrm{e} \\
&\mathrm{v}_{\mathrm{p}}=\mathrm{v}_\alpha=\mathrm{v} \\
&\frac{\lambda_{\mathrm{p}}}{\lambda_\alpha}=\frac{\sqrt{\mathrm{m}_\alpha \mathrm{q}_\alpha}}{\sqrt{\mathrm{m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}}}}=\frac{\sqrt{4 \mathrm{~m}_{\mathrm{p}} \times 2 \mathrm{e}}}{\sqrt{\mathrm{m}_{\mathrm{p}} \times \mathrm{e}}}=\sqrt{8} \\
&\lambda_{\mathrm{p}}=\sqrt{8} \lambda_\alpha
\end{aligned}
$$
So, the de-Broglie wavelength of proton is $\sqrt{8}$ times wavelength of $\alpha$-particle.
$$
\therefore \quad \frac{\lambda_{\mathrm{p}}}{\lambda_\alpha}=\frac{\not}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}} \times v_{\mathrm{p}}}} \times \frac{\sqrt{\not \mathrm{m}_\alpha \mathrm{q}_\alpha v_\alpha}}{K}
$$
Given,
$$
\begin{aligned}
&\mathrm{m}_\alpha=4 \mathrm{~m}_{\mathrm{p}} \\
&\mathrm{q}_\alpha=2 \mathrm{e} ; \\
&\mathrm{q}_{\mathrm{p}}=\mathrm{e} \\
&\mathrm{v}_{\mathrm{p}}=\mathrm{v}_\alpha=\mathrm{v} \\
&\frac{\lambda_{\mathrm{p}}}{\lambda_\alpha}=\frac{\sqrt{\mathrm{m}_\alpha \mathrm{q}_\alpha}}{\sqrt{\mathrm{m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}}}}=\frac{\sqrt{4 \mathrm{~m}_{\mathrm{p}} \times 2 \mathrm{e}}}{\sqrt{\mathrm{m}_{\mathrm{p}} \times \mathrm{e}}}=\sqrt{8} \\
&\lambda_{\mathrm{p}}=\sqrt{8} \lambda_\alpha
\end{aligned}
$$
So, the de-Broglie wavelength of proton is $\sqrt{8}$ times wavelength of $\alpha$-particle.
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