# ​ In a fission event of 238 92 U  by fast moving neutrons, no neutrons are emitted and final products,

47 views
in Physics
closed

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:

$m\left(^{238}_{92}U\right) = 238.05079u;\,m\left(^{140}_{58}Ce=139.90543u\right)$

$m\left(^{99}_{44}Ru\right)= 98.90594;\, m\left(^1_0n\right) = 1.008665u$

+1 vote
by (30.2k points)
selected

$^{238}_{92}U +\, ^1_0n \longrightarrow \,^{140}_{58}Ce+\,^{99}_{44}Ru$

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