The ratio of wavelengths of proton and deuteron accelerated by potential Vp and Vd is 1 : $\sqrt2$. Then the ratio of Vp to Vd will be :
Solution
<p>$\lambda = {h \over {mv}} = {h \over {\sqrt {2m\,eV} }}$</p>
<p>so $${{{\lambda _p}} \over {{\lambda _d}}} = {{\sqrt {{m_d}{V_d}} } \over {\sqrt {{m_p}{V_p}} }} = {1 \over {\sqrt 2 }}$$</p>
<p>${{2{V_d}} \over {{V_p}}} = {1 \over 2}$</p>
<p>${{{V_p}} \over {{V_d}}} = {4 \over 1}$</p>
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
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