InterviewSolution
Saved Bookmarks
| 1. |
The activity of a radioactive substance is `R_(1)` at time `t_(1)` and `R_(2)` at time `t_(2)(gt t_(1))`. Its decay cosntant is `lambda`. Then .A. `R_(1)t_(1)=R_(2)t_(2)`B. `R_(1)=R_(1)e^(-(lambda^(t_(1)-t_(2))))`C. `(R_(1)-R_(2))/(t_(2)-t_(1))`=constantD. `R_(2)=R_(1)e^(lambda(t_(2)-t_(1))` |
|
Answer» Correct Answer - B |
|