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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 U235U^{235}nucleus, the number of fissions required per second to produce 1 kilowatt power shall be (Given16mueV=1.6×1019J1\mspace{6mu} eV = 1.6 \times 10^{- 19}J).

A

3.125×10133.125 \times 10^{13}

B

3.125×10143.125 \times 10^{14}

C

3.125×10153.125 \times 10^{15}

D

3.125×10163.125 \times 10^{16}

Answer

3.125×10133.125 \times 10^{13}

Explanation

Solution

P=n(Et)1000=n×200×106×1.6×1019tP = n\left( \frac{E}{t} \right) \Rightarrow 1000 = \frac{n \times 200 \times 10^{6} \times 1.6 \times 10^{- 19}}{t}

nt=3.125×1013.\Rightarrow \frac{n}{t} = 3.125 \times 10^{13}.