MCQ
Electrons are accelerated through a potential difference $V$ and protons are accelerated through a potential difference $4\, V$. 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).
  • A
    $\frac{{{\lambda _e}}}{{{\lambda _p}}} = \sqrt {\frac{{{m_p}}}{{{m_e}}}} $
  • B
    $\frac{{{\lambda _e}}}{{{\lambda _p}}} = \sqrt {\frac{{{m_e}}}{{{m_p}}}} $
  • C
    $\frac{{{\lambda _e}}}{{{\lambda _p}}} = \frac{1}{2}\sqrt {\frac{{{m_e}}}{{{m_p}}}} $
  • $\frac{{{\lambda _e}}}{{{\lambda _p}}} = 2\sqrt {\frac{{{m_p}}}{{{m_e}}}} $

Answer

Correct option: D.
$\frac{{{\lambda _e}}}{{{\lambda _p}}} = 2\sqrt {\frac{{{m_p}}}{{{m_e}}}} $
d
Energy in joule $(E)$

$=$ charge $\times$ potential diff. in volt

${{\text{E}}_{{\text{electron }}}} = {q_{\text{e}}}{\text{V}}$ and ${{\text{E}}_{{\text{proton }}}} = {q_{\text{p}}}4{\text{V}}$

de-Broglie wavelength

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

$\lambda_{\mathrm{c}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{m}_{\mathrm{e}} \mathrm{e} \mathrm{V}}}$ and $\lambda_{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{m}_{\mathrm{p}} \mathrm{e} 4 \mathrm{V}}}$

${\text{(}}\because {q_c} = {q_p}{\text{)}}$

$\therefore \frac{{{\lambda _{\text{c}}}}}{{{\lambda _{\text{P}}}}} = \frac{{\frac{{\text{h}}}{{\sqrt {2{{\text{m}}_{\text{e}}}{\text{eV}}} }}}}{{\frac{{\text{h}}}{{\sqrt {2{{\text{m}}_p}{\text{e4V}}} }}}}$ $ = \sqrt {\frac{{2{{\text{m}}_p}{\text{e4V}}}}{{2{{\text{m}}_e}{\text{eV}}}}} $

$ = 2\sqrt {\frac{{{{\text{m}}_{\text{p}}}}}{{{{\text{m}}_{\text{e}}}}}} $

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