InterviewSolution
Saved Bookmarks
| 1. |
The half-life of `4.0mg beta-` emitter of `.^(210)X` is `5` day and the average energy of emitted `beta-` particle is `0.34MeV`. At what rate in watts does the sample emits energy?A. `2.0`B. `0.1`C. `1.5`D. `1.0` |
|
Answer» Correct Answer - `(d)` Power = Energy of `beta` in `J xx "No of" beta-` particles emitted/sec. `= E xx (-(dN)/(dt)) = E xx lambda xx N` `= 0.34xx10^(6)xx1.6xx10^(-19) xx (0.693)/(5xx24xx60xx60)` `xx (4xx10^(-3)xx6.023xx10^(23))/(210) = 1J//sec = 1` watt |
|