An $\alpha$-particle of energy $4\  MeV$ is scattered through $180^o $ by a fixed uranium nucleus. The distance of the closest approach is of the order of

  • A
    $1\ \mathring A $
  • B
    ${10^{ - 10}}\ cm$
  • C
    ${10^{ - 12}}\ cm$
  • D
    ${10^{ - 15}}\ cm$

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