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  • The magnitude of the de-Broglie wavelength of electron (e), proton (p), neutron (n) and <img alt="\alph

The magnitude of the de-Broglie wavelength (\lambda) of electron (e), proton (p), neutron (n) and \alpha- particle (\alpha) all having the same energy of 1 \mathrm{Me\nu}, in the increasing order will follow the sequence
 

Option: 1

\lambda_e, \lambda_p, \lambda_n, \lambda_\alpha


Option: 2

\lambda_e, \lambda_n, \lambda_p, \lambda_\alpha


Option: 3

\lambda_\alpha, \lambda_{\mathrm{n}}, \lambda_{\mathrm{p}}, \lambda_{\mathrm{e}}


Option: 4

\lambda_{\mathrm{p}}, \lambda_{\mathrm{e}}, \lambda_\alpha, \lambda_{\mathrm{n}}


Answers (1)

best_answer


\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}

So \mathrm{h} \propto \frac{1}{\sqrt{\mathrm{m}}}

Since  \mathrm{m}_\alpha>\mathrm{m}_{\mathrm{n}}>\mathrm{m}_{\mathrm{p}}>\mathrm{m}_{\mathrm{e}}

So de-Broglie wavelength in increasing order will be \lambda_\alpha, \lambda_{\mathrm{n}}, \lambda_{\mathrm{p}}, \lambda_{\mathrm{e}}

Posted by

sudhir.kumar

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