For the given series in nth step find out number of produced neutrons and energy
​238U92--->Ba+Kr+3neutrons +energy

Dear Student,

The computation are as expressed below,

92U238 + 0n156Ba141 + 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.

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