

InterviewSolution
Saved Bookmarks
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)` |
|