Question
Question: 200 MeV of energy may be obtained per fission of \(U^{235}\). A reactor is generating 1000 kW of pow...
200 MeV of energy may be obtained per fission of U235. A reactor is generating 1000 kW of power. The rate of nuclear fission in the reactor is.
A
1000
B
2×108
C
3.125×1016
D
931
Answer
3.125×1016
Explanation
Solution
Power ⇒(21)4=(21)t/48⇒t=1926muhour.
Rate of nuclear fission
=200×1.6×10−13106= 3.125 × 1016.