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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 fixed uranium nucleus. The distance of closest approach is of the order of

A

`1 Å`

B

`10^(-11) cm`

C

`10^(-12) cm`

D

`10^(-15) cm`

Text Solution

Verified by Experts

The correct Answer is:
C
Let distance of closet approach be 'd' then
`1/2 mv^(2)=(k(2e)(9)(2e))/d^(2)rArr d=10^(-12) cm`
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