Electrons are accelerated through a potential difference $\mathrm{V}$ and protons are accelerated through a…

Electrons are accelerated through a potential difference $\mathrm{V}$ and protons are accelerated through a potential difference $4 \mathrm{~V}$. The de-Broglie wavelengths are $\lambda_e$ and $\lambda_p$ for electrons and protons respectively. The ratio of $\frac{\lambda_e}{\lambda_p}$ is given by: (given $m_e$ is mass of electron and $m_p$ is mass of proton).
  1. $\frac{\lambda_e}{\lambda_p}=\sqrt{\frac{m_p}{m_e}}$
  2. $\frac{\lambda_e}{\lambda_p}=\sqrt{\frac{m_e}{m_p}}$
  3. $\frac{\lambda_e}{\lambda_p}=\frac{1}{2} \sqrt{\frac{m_e}{m_p}}$
  4. $\frac{\lambda_e}{\lambda_p}=2 \sqrt{\frac{m_p}{m_e}}$

Solution

Energy in joule (E) $=$ charge $\times$ potential diff. in volt $\mathrm{E}_{\text {electron }}=\mathrm{q}_{\mathrm{c}} \mathrm{V}$ and $\mathrm{E}_{\text {proton }}=\mathrm{q}_{\mathrm{p}} 4 \mathrm{~V}$ de-Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ $ \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} $ $\begin{aligned} \therefore \frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{P}}} & =\frac{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{eV}}}}{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{P}} \mathrm{e} 4 \mathrm{~V}}}}=\sqrt{\frac{2 \mathrm{~m}_{\mathrm{P}} \mathrm{e} 4 \mathrm{~V}}{2 \mathrm{~m}_{\mathrm{e}} \mathrm{eV}}} \\ & =2 \sqrt{\frac{\mathrm{m}_{\mathrm{P}}}{\mathrm{m}_{\mathrm{e}}}}\end{aligned}$

Asked in: JEE Main 2013 (23 Apr Online)

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