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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

`10^(-15) cm`

B

`10^(-13) cm`

C

`10^(-12) cm`

D

`10^(-19) cm`

Text Solution

Verified by Experts

The correct Answer is:
C
Use `E = (1)/(4 pi epsilon_(0)) ((Ze) (e))/( r_(0))`
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