Solveeit Logo

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 U235U^{235}. A reactor is generating 1000 kW of power. The rate of nuclear fission in the reactor is.

A

1000

B

2×1082 \times 10^{8}

C

3.125×10163.125 \times 10^{16}

D

931

Answer

3.125×10163.125 \times 10^{16}

Explanation

Solution

Power (12)4=(12)t/48t=1926muhour.\Rightarrow \left( \frac{1}{2} \right)^{4} = \left( \frac{1}{2} \right)^{t/48} \Rightarrow t = 192\mspace{6mu} hour.

Rate of nuclear fission

=106200×1.6×1013= \frac{10^{6}}{200 \times 1.6 \times 10^{- 13}}= 3.125 × 1016.