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An $α$ particles of energy $10 M e V$ is scattered through $180^{0}$ by a fixed uranium nucleus. The distance of closed approach of order of $M e V$

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detailed solution

Correct option is D

KE=PE=14π∈0×2Ze2dd=9×109×2×236×1.6×1.6×10−3810×106×1.6×10−19=18×236×1.6×10−14=6796.8×10−17=6.796×10−14.

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An α particles of energy 10 MeV is scattered through 1800 by a fixed uranium nucleus. The distance of closed approach of order of MeV

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An $α$ particles of energy $10 M e V$ is scattered through $180^{0}$ by a fixed uranium nucleus. The distance of closed approach of order of $M e V$