Dual Nature of Matter

JEE Main level, single correct, medium

41 of 200

Question 41 of 200

The ratio of wavelengths of proton and deuteron accelerated by potential Vp{V}_{p} and Vd{V}_{d} is 1:21:\sqrt{2}. Then, the ratio of Vp{V}_{p} to Vd{V}_{d} will be

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Answer: (D) 4:14:1

Kinetic energy gained by a charged particle accelerated by a potential VV is qVqV
KE=qVKE=qV
Using relation between momentum and kinetic energy
⇒p22m=qV⇒p=2mqV\Rightarrow \dfrac{{p}^{2}}{2m}=qV\Rightarrow p=\sqrt{2mqV}
We know that momentum of charged particle is p=hλp=\dfrac{h}{\lambda }.
Thus, λ=h2mqV\lambda =\dfrac{h}{\sqrt{2mqV}}
Now, λpλd=mdVdmpVp\dfrac{{\lambda }_{p}}{{\lambda }_{d}}=\sqrt{\dfrac{{m}_{d}{V}_{d}}{{m}_{p}{V}_{p}}}
⇒12=2VdVp⇒VpVd=4\Rightarrow \dfrac{1}{\sqrt{2}}=\sqrt{\dfrac{2{V}_{d}}{{V}_{p}}} \Rightarrow \dfrac{{V}_{p}}{{V}_{d}}=4

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