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  • A proton, a neutron, an electron and -particle have same energy. If <img alt="\m

A proton, a neutron, an electron and \mathrm{\alpha}-particle have same energy. If \mathrm{\lambda _{p},\lambda _{n},\lambda _{e}\: and\: \lambda _{\alpha }} are the de Broglie's wavelengths of prorton, neutron, electron and \mathrm{{\alpha }} particle respectively, then choose the correct relation from the following :

Option: 1

\mathrm{\lambda _{p}=\lambda _{n}> \lambda _{e}> \lambda _{\alpha }}\\


Option: 2

\mathrm{\lambda _{\alpha }< \lambda _{n}< \lambda _{p}< \lambda _{e }}\\


Option: 3

\mathrm{\lambda _{e}< \lambda _{p}= \lambda _{n}> \lambda _{\alpha }}\\


Option: 4

\mathrm{\lambda _{e}=\lambda _{p}= \lambda _{n}= \lambda _{\alpha }}


Answers (1)

\mathrm{K E_{\alpha}=K E_{p}=K E_{e}=K E_{n}}\\

De Broglie wavelength   \mathrm{=\lambda=\frac{h}{\sqrt{2 m(KE)}}}

We know that

\mathrm{m_{\alpha}>m_{n}>m_{p}>m_{e}}

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

Hence the correct answer is option 2.

Posted by

Sumit Saini

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