InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The half life of radioactive Radon is `3.8 days` . The time at the end of which `(1)/(20) th` of the radon sample will remain undecayed is `(given log e = 0.4343 ) `A. 3.8 daysB. 16.5 daysC. 33 daysD. 76 days |
|
Answer» Correct Answer - B (b) By formula `N = N_0 e^(-lamda t)` Given `(N)/(N_0) = (1)/(20)` and `lamda = (0.6931)/(3.8)` `rArr 20 = e^((0.6931)/(3.8))` Taking log of both sides or `log 20 = (0.6931 xx t)/(3.8) log_10 e` or `1.3010 = (0.6931 xx t xx 0.4343)/(3.8)` `rArr t = 16.5 days`. |
|