An alpha  - particle of energy 5 MeV is scattered through 180^∘ by a fixed uranium nucleus. The distance of the closest approach is of the order of

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Explanation:

The potential energy =kinetic energy

⇒1/2 mv2=1/(4πε0 )×(q1 q2)/r

⇒r=(9×109×2×92×(1.6×10-19 )^2) / (5×106×1.6×10-19 )

⇒r=5.3×10-14

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