Questions
1 kg of undergoes fission process. If energy released per event is 200 MeV, then total energy released is
detailed solution
Correct option is C
Number of nuclei in 1 kg of 235U is N=NA235×(1×103)=6.023×1023×103235=2.56×1024∴ Total energy released 0=N×200MeV=5.12×1026MeVTalk to our academic expert!
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