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An alpha particle of energy 5 MeV is sca...

An alpha particle of energy `5 MeV` is scattered through `180^(@)` by a found uramiam nucleus . The distance of closest approach is of the order of

A

`1Å`

B

`10^(-10)cm`

C

`10^(-12)cm`

D

`10^(-15)cm`

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The correct Answer is:
C
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