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}}}}}} $