QJEEM22Ph8363: JEE Main 2022 Physics - Class 12 JEEM22Ph8363 Solution: The ratio of wavelengths of proton and deuteron accelerated by…

$\lambda = {h \over {mv}} = {h \over {\sqrt {2m\,eV} }}$

so ${{{\lambda _p}} \over {{\lambda _d}}} = {{\sqrt {{m_d}{V_d}} } \over {\sqrt {{m_p}{V_p}} }} = {1 \over {\sqrt 2 }}$

${{2{V_d}} \over {{V_p}}} = {1 \over 2}$

${{{V_p}} \over {{V_d}}} = {4 \over 1}$