InterviewSolution
Saved Bookmarks
| 1. |
The angular speed of an electron revolving around the H-nucleus is proportional toA. `1//r`B. `1//r^(3//2)`C. `1//r^(2)`D. `r^(3//2)` |
|
Answer» Correct Answer - B We know that, therefore `1/(4pivarepsilon_(0))cdot(Ze^(2))/r^(2)=(mv^(2))/r` therefore`v=sqrt((1/(4pivarepsilon_(0))cdot(Ze^(2))/(mr)))=k/sqrtr` (Let `k=(Ze^(2))/(4pivarepsilon_(0)m)` ) Angular speed ,`omega=v/r=k/(sqrtrcdotr)=k/(r^(3//2))` |
|