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. `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))`


Discussion

No Comment Found

Related InterviewSolutions