JEE MAIN - Physics (2022 - 25th July Evening Shift - No. 6)
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 :
1 : 1
$$\sqrt2$$ : 1
2 : 1
4 : 1
Explanation
$$\lambda = {h \over {mv}} = {h \over {\sqrt {2m\,eV} }}$$
so $${{{\lambda _p}} \over {{\lambda _d}}} = {{\sqrt {{m_d}{V_d}} } \over {\sqrt {{m_p}{V_p}} }} = {1 \over {\sqrt 2 }}$$
$${{2{V_d}} \over {{V_p}}} = {1 \over 2}$$
$${{{V_p}} \over {{V_d}}} = {4 \over 1}$$
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