`200 MeV` of energy may be obtained per fission of `U^235`. A reactor is generating `1000 kW` of power. The rate of nuclear fission in the reactor is.

Options

A. 1000
B. `2xx10^(8)`
C. `3.125xx10^(16)`
D. 931

Explanation

To solve the problem, we need to find the rate of nuclear fission occurring in a reactor that generates a power of 1000 kW, given that each fission of Uranium-235 releases 200 MeV of energy. Here’s how to approach it step by step:

### Step 1: Convert Power to Watts
The power generated by the reactor is given as 1000 kW. We need to convert this into watts:
\[ \text{Power} = 1000 \, \text{kW} = 1000 \times 10^3 \, \text{W} = 10^6 \, \text{W} \]

### Step 2: Convert Energy per Fission from MeV to Joules
The energy released per fission of Uranium-235 is given as 200 MeV. We need to convert this energy into joules. The conversion factor is:
\[ 1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J} \]
Thus, the energy per fission in joules is:
\[ \text{Energy per fission} = 200 \, \text{MeV} \times 1.6 \times 10^{-13} \, \text{J/MeV} = 3.2 \times 10^{-11} \, \text{J} \]

### Step 3: Calculate the Rate of Nuclear Fission
The rate of nuclear fission can be calculated using the formula:
\[ \text{Rate of fission} = \frac{\text{Power}}{\text{Energy per fission}} \]
Substituting the values we found:
\[ \text{Rate of fission} = \frac{10^6 \, \text{W}}{3.2 \times 10^{-11} \, \text{J}} = \frac{10^6}{3.2 \times 10^{-11}} \approx 3.125 \times 10^{16} \, \text{fissions/second} \]

### Final Answer
The rate of nuclear fission in the reactor is approximately:
\[ \text{Rate of fission} \approx 3.125 \times 10^{16} \, \text{fissions/second} \]