For the given series in nth step find out number of produced neutrons and energy 238U92--->Ba+Kr+3neutrons +energy Share with your friends Share 0 Yasodharan answered this Dear Student, The computation are as expressed below, 92U238 + 0n1→56Ba141 + 36Kr 92+ 3 on1 + energymass of reactant = 238+1 = 239 amumass of product = 141+92+(3×1) = 236 amumass defect = 239 - 236 = 3 amuEnergy released = 3 amu × 931.5 MeV = 2794.5 MeV is higher energyAns:Neutron = 3Energy = 2794.5 MeV Regards. -2 View Full Answer Kushagra Bedi answered this rfda 0