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  • A proton, a neutron, an electron and an -particle have same energy. Then, their de-Broglie wavelengths compa

A proton, a neutron, an electron and an \alpha -particle have same energy. Then, their de-Broglie wavelengths compare as:

Option: 1

\lambda_p=\lambda_n>\lambda_e>\lambda_\alpha


Option: 2

\lambda_\alpha<\lambda_{\mathrm{p}}=\lambda_{\mathrm{n}}<\lambda_{\mathrm{e}}


Option: 3

\lambda_e<\lambda_p=\lambda_p>\lambda_e


Option: 4

\lambda_e=\lambda_p=\lambda_n=\lambda_\alpha


Answers (1)

best_answer

We know that the relation between \lambda and \mathrm{K} is given by \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}

Here, for the given value of energy K,  \frac{h}{\sqrt{2 K}} is a constant.

Thus, \lambda \propto \frac{1}{\sqrt{\mathrm{m}}}

\begin{aligned} & \therefore \lambda_{\mathrm{p}}: \lambda_{\mathrm{n}}: \lambda_{\mathrm{e}}: \lambda_\alpha \\ & \Rightarrow=\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}} \end{aligned}

 \begin{aligned} &\text{ Since ,}m_p=m_n, \lambda_p=\lambda_n\\ &\text{As, }\mathrm{m}_\alpha<\mathrm{m}_{\mathrm{p}}, \text{therefore }\lambda_\alpha<\lambda_{\mathrm{p}}\\ &\text{As, } \mathrm{m}_{\mathrm{e}}<\mathrm{m}_{\mathrm{n}}, \text{therefore }\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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Anam Khan

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