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A proton, a neutron, an electron and an $\alpha$-particle have same energy. Then, their de-Broglie wavelengths compare as
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Verified Answer
The correct answer is:
$\lambda_\alpha < \lambda_{\mathrm{p}}=\lambda_{\mathrm{n}}>\lambda_{\mathrm{e}}$
$\lambda_\alpha < \lambda_{\mathrm{p}}=\lambda_{\mathrm{n}}>\lambda_{\mathrm{e}}$
The relation between $\lambda$ and $\mathrm{K}$ is given by
$$
\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}
$$
So, for the given value of kinetic energy $\mathrm{K}$, $\frac{\mathrm{h}}{\sqrt{2 \mathrm{~K}}}$ is a constant.
$$
\begin{aligned}
&\text { Thus, } \lambda \propto \frac{1}{\sqrt{\mathrm{m}}} \\
&\therefore \quad \Rightarrow \lambda_{\mathrm{p}}: \lambda_{\mathrm{n}}: \lambda_{\mathrm{e}}: \lambda_\alpha \\
&\Rightarrow \quad=\frac{1}{\sqrt{\mathrm{m}_{\mathrm{p}}}}: \frac{1}{\sqrt{\mathrm{m}_{\mathrm{n}}}}: \frac{1}{\sqrt{\mathrm{m}_{\mathrm{e}}}}: \frac{1}{\sqrt{\mathrm{m}_\alpha}} \\
&\text { if }\left(\mathrm{m}_{\mathrm{p}}=\mathrm{m}_{\mathrm{n}}\right) \text {, then } \lambda_{\mathrm{p}}=\lambda_{\mathrm{n}} \\
&\text { if }\left(\mathrm{m}_\alpha>\mathrm{m}_{\mathrm{p}}\right) \text {, then } \lambda_\alpha < \lambda_{\mathrm{p}} \\
&\text { if }\left(\mathrm{m}_{\mathrm{e}} < \mathrm{m}_{\mathrm{n}}\right) \text {, then } \lambda_{\mathrm{e}}>\lambda_{\mathrm{n}} \\
&\text { Hence, } \lambda_\alpha < \lambda_{\mathrm{p}}=\lambda_{\mathrm{n}} < \lambda_{\mathrm{e}} .
\end{aligned}
$$
$$
\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}
$$
So, for the given value of kinetic energy $\mathrm{K}$, $\frac{\mathrm{h}}{\sqrt{2 \mathrm{~K}}}$ is a constant.
$$
\begin{aligned}
&\text { Thus, } \lambda \propto \frac{1}{\sqrt{\mathrm{m}}} \\
&\therefore \quad \Rightarrow \lambda_{\mathrm{p}}: \lambda_{\mathrm{n}}: \lambda_{\mathrm{e}}: \lambda_\alpha \\
&\Rightarrow \quad=\frac{1}{\sqrt{\mathrm{m}_{\mathrm{p}}}}: \frac{1}{\sqrt{\mathrm{m}_{\mathrm{n}}}}: \frac{1}{\sqrt{\mathrm{m}_{\mathrm{e}}}}: \frac{1}{\sqrt{\mathrm{m}_\alpha}} \\
&\text { if }\left(\mathrm{m}_{\mathrm{p}}=\mathrm{m}_{\mathrm{n}}\right) \text {, then } \lambda_{\mathrm{p}}=\lambda_{\mathrm{n}} \\
&\text { if }\left(\mathrm{m}_\alpha>\mathrm{m}_{\mathrm{p}}\right) \text {, then } \lambda_\alpha < \lambda_{\mathrm{p}} \\
&\text { if }\left(\mathrm{m}_{\mathrm{e}} < \mathrm{m}_{\mathrm{n}}\right) \text {, then } \lambda_{\mathrm{e}}>\lambda_{\mathrm{n}} \\
&\text { Hence, } \lambda_\alpha < \lambda_{\mathrm{p}}=\lambda_{\mathrm{n}} < \lambda_{\mathrm{e}} .
\end{aligned}
$$
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