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
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
$1\ Å$
- B
${10^{ - 10}}\ cm$
- C
${10^{ - 12}}\ cm$
- D
${10^{ - 15}}\ cm$
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