

InterviewSolution
Saved Bookmarks
1. |
Calculate the energy released when three alpha particles combine to form a `^12 C` nucleus. The atomic mass of `_2^4 He` is `4.002603 u`. |
Answer» The mass of `a^(12)C` atom is exactly `12 u`. The energy released in the reaction `3(._(2)^(4)Hr)rarr._(6)^(12)C` is `[3m(._(2)^(4)He)-m(._(6)^(12)C)]c^(2)` `[3xx4.002603u-12 u](931 MeV//u)=7.27 MeV`. |
|