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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: 

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

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

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Best 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

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