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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. `10^(-15) cm`B. `10^(-13) cm`C. `10^(-12) cm`D. `10^(-19) cm` |
Answer» Correct Answer - C Use `E = (1)/(4 pi epsilon_(0)) ((Ze) (e))/( r_(0))` |
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