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Consider the fission of \(_{92}^{238}U\) by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are \(_{58}^{140}Ce\) and \(_{44}^{99}Ru\). Calculate Q for this fission process. The relevant atomic and particle masses are

\(m(_{92}^{238}U)\) = 238.05079u

\(m(_{58}^{140}Ce)\) = 139.90543u

\(m(_{44}^{99}Ru)\) = 98.90594u

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= [238.05079 + 1.008665 - 139.90543 - 98.90594] x 931.5 MeV

= 0.248085 x 931.5

= 231.09 MeV

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