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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

`5.3xx10^(-12)`

B

`5.3xx10^(-13)`

C

`5.3xx10^(-14)`

D

`5.3xx10^(-15)`

Text Solution

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The correct Answer is:
C
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