Dual Nature of Matter

JEE Main level, single correct, medium

85 of 200

Question 85 of 200

Electrons are accelerated through a potential difference V\mathrm{V} and protons are accelerated through a potential difference 4 V4 \mathrm{~V}. The de-Broglie wavelengths are λe\lambda_e and λp\lambda_p for electrons and protons respectively. The ratio of λeλp\frac{\lambda_e}{\lambda_p} is given by: (given mem_e is mass of electron and mpm_p is mass of proton).

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Answer: (D) λeλp=2mpme\frac{\lambda_e}{\lambda_p}=2 \sqrt{\frac{m_p}{m_e}}

Energy in joule (E)
== charge ×\times potential diff. in volt Eelectron =qcV\mathrm{E}_{\text {electron }}=\mathrm{q}_{\mathrm{c}} \mathrm{V} and Eproton =qp4 V\mathrm{E}_{\text {proton }}=\mathrm{q}_{\mathrm{p}} 4 \mathrm{~V}
de-Broglie wavelength λ=hP=h2mE\lambda=\frac{\mathrm{h}}{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}
λe=h2 meeV and λP=h2 mee4 V(∵qe=qP)\begin{array}{r} \lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{eV}}} \text { and } \lambda_{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{e} 4 \mathrm{~V}}} \\ \left(\because \mathrm{q}_{\mathrm{e}}=\mathrm{q}_{\mathrm{P}}\right) \end{array}
∴λeλP=h2 meeVh2 mPe4 V=2 mPe4 V2 meeV=2mPme\begin{aligned} \therefore \frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{P}}} & =\frac{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{eV}}}}{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{P}} \mathrm{e} 4 \mathrm{~V}}}}=\sqrt{\frac{2 \mathrm{~m}_{\mathrm{P}} \mathrm{e} 4 \mathrm{~V}}{2 \mathrm{~m}_{\mathrm{e}} \mathrm{eV}}} \\ & =2 \sqrt{\frac{\mathrm{m}_{\mathrm{P}}}{\mathrm{m}_{\mathrm{e}}}}\end{aligned}

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