InterviewSolution
Saved Bookmarks
| 1. |
The kinetic energy of `alpha`-particle emitted in the `alpha`-decay of `""_(88)Ra^(266)` is [given, mass number of Ra = 222 u]A. `5.201` Me VB. `3.301` Me VC. `6.023` Me VD. `4.871` Me V |
|
Answer» Correct Answer - D (d) the nuclear should be `""_(88)Ra^(226) to ""_(86)Rn ^(222)+ alpha ` Energy corresponding mass defect `Q =[{:("(mass number of Ra )"),("-(Mass number of Ru )"),("-m (4He )"):}]Uxx931.5Me V = 4.871MeV ` |
|