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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

`1 A^(@)`

B

`10^(-10)cm`

C

`10^(-12)cm`

D

`10^(-16)cm`

Text Solution

Verified by Experts

The correct Answer is:
C
`(1)/(2)mv_(o)^(2)=(q_1 q_2)/(4 pi epsilon_(0) r)`
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