Consider the fission of 23892U 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 14058Ce and 9944Ru. Calculate Q for this fission process. The relevant atomic and particle masses are m(23892U ) =238.05079 u m(14058Ce ) =139.90543 u m(9944Ru ) = 98.90594 u
The reaction for the fission of 93238U by fast neutrons is as shown 93238U+01n→58140Ce+4499U Q=[m(92238U)+mN−m(58140Ce)−m(4499Ru)]× 931 .5 MeV Q=[238.05079+1.00893−139.90543−98.90594]× 931 .5 MeV Q=0.24835× 931 .5 MeV Q=231.1 MeV
