A nuclear reactor generates power at 50% efficiency, by fusion ""_(92)U^(235) into two canal fragments of ""_(6)Pd^(116) with the emission of two gamma rays and three neutrons. The average B.E. per particle of U^(235) and Pd^(116) is 7.2 MeV and 8.2 MeV respectively. The energy released in one fission event is

A nuclear reactor generates power at 50% efficiency, by fusion ""_(92)

1.	A nuclear reactor generates power at 50% efficiency, by fusion ""_(92)U^(235) into two canal fragments of ""_(6)Pd^(116) with the emission of two gamma rays and three neutrons. The average B.E. per particle of U^(235) and Pd^(116) is 7.2 MeV and 8.2 MeV respectively. The energy released in one fission event is
Answer» 180MeV 190MeV 200MeV 210MeV Solution :B.E. of nucleous of one `U^(235)" NUCLEUS"=7.2 xx 235=1692MeV` FISSION reaction is `""_(92)U^(235)to 2_(46)Pd^(116)+3_(0)n^(1)+2gamma+Q` B.E. of nucleons of 2 `Pd^(116)" NUCLIE" =2 xx 8.2 xx 116=19.02.2MeV` Energy released due to mass defect =(1902.4-1692) =210.4MeV Energy GAINED by `2gamma-"rays"=2 xx 5.2 MeV=10.4 MeV`. Energy released per fission =(210.4-10.4)=200meV

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