1.

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 ofA. `1Å`B. `10^(-10)cm`C. `10^(-10)cm`D. `10^(-12)cm`

Answer» Correct Answer - C
`1/2mv_(0)=(q_(1)q_(2))/(4piepsilon_(0)r)`


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