Question
Question: If 200 MeV energy is released in the fission of a single \(U^{235}\)nucleus, the number of fissions ...
If 200 MeV energy is released in the fission of a single U235nucleus, the number of fissions required per second to produce 1 kilowatt power shall be (Given16mueV=1.6×10−19J).
A
3.125×1013
B
3.125×1014
C
3.125×1015
D
3.125×1016
Answer
3.125×1013
Explanation
Solution
P=n(tE)⇒1000=tn×200×106×1.6×10−19
⇒tn=3.125×1013.