Dual Nature of Matter

JEE Main level, single correct, medium

36 of 200

Question 36 of 200

An α\alpha particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particals will be :

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Answer: (B) 22:12\sqrt{2}:1

For α\alpha particle, mα=4{m}_{\alpha }=4 a.m.u and charge qα=2e{q}_{\alpha }=2e.
For proton, mp=1{m}_{p}=1 a.m.u and charge qp=e{q}_{p}=e.
As the charged particle are moving through same potential, the kinetic energy acquired by the particle will be equal to electric potential energy i.e., 12mv2=qV\dfrac{1}{2}m{v}^{2}=qV, where, VV is potential.
Using the relation between momentum (p)\left(p\right) and kinetic energy, we have
p=2mE=2mqVp=\sqrt{2mE}=\sqrt{2mqV}
Then, the ratio of linear momenta of two particles is pαpp=mαqαmpqp=41×21\dfrac{{p}_{\alpha }}{{p}_{p}}=\sqrt{\dfrac{{m}_{\alpha }{q}_{\alpha }}{{m}_{p}{q}_{p}}}=\sqrt{\dfrac{4}{1}\times \dfrac{2}{1}}
=221=22:1=\dfrac{2\sqrt{2}}{1}=2\sqrt{2}:1

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