InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
In a fission event of \(^{238}_{92}\,U\) by fast moving neutrons, no neutrons are emitted and final products, after the beta decay of the primary fragments, are \(_ {58}^{140}Ce\) and \(^{99}_{44}Ru\) . Calculate Q for this process. Neglect the masses of electrons / positrons emitted during the intermediate steps.GIven: \(m\left(^{238}_{92}U\right) = 238.05079u;\,m\left(^{140}_{58}Ce=139.90543u\right)\)\(m\left(^{99}_{44}Ru\right)= 98.90594;\, m\left(^1_0n\right) = 1.008665u\) |
|
Answer» \(^{238}_{92}U +\, ^1_0n \longrightarrow \,^{140}_{58}Ce+\,^{99}_{44}Ru\) Then Q value \(Q = \left[m\left(^{238}_{92}U\right)+\,m\left(^1_0n\right) - m\left(^{140}_{58}Ce\right)-m\left(^{99}_{44}Ru\right)\right]\times931.5\) Q = [238.05079 + 1.008665 - 139.90843 - 98.90594] x 931.5 Q = [239.059455 - 238.81137] x 931.5 Q = 0.248085 x 931.5 Q = 231.09 MeV |
|