If `200 MeV` energy is released in the fission of a single `U^235` nucleus, the number of fissions required per second to produce `1` kilowatt power shall be (Given `1 eV = 1.6 xx 10^-19 J`).

Options

A. `3.125 xx 10^13`
B. `3.125 xx 10^14`
C. `3.125 xx 10^15`
D. `3.125 xx 10^16`

Explanation

To solve the problem, we need to determine the number of fissions of a single \( U^{235} \) nucleus required to produce 1 kilowatt (kW) of power when each fission releases \( 200 \, \text{MeV} \) of energy.

### Step-by-Step Solution:

1. **Convert kilowatts to watts**:
\[ 1 \, \text{kW} = 1000 \, \text{W} \]

2. **Convert MeV to Joules**:
We know that \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \). Therefore,
\[ 200 \, \text{MeV} = 200 \times 10^6 \, \text{eV} = 200 \times 10^6 \times 1.6 \times 10^{-19} \, \text{J} \]
\[ = 3.2 \times 10^{-11} \, \text{J} \]

3. **Calculate the number of fissions required per second**:
The power \( P \) is given by the energy released per fission multiplied by the number of fissions per second \( N \):
\[ P = N \times \text{Energy per fission} \]
Rearranging gives:
\[ N = \frac{P}{\text{Energy per fission}} \]
Substituting the values:
\[ N = \frac{1000 \, \text{W}}{3.2 \times 10^{-11} \, \text{J}} \]

4. **Calculate \( N \)**:
\[ N = \frac{1000}{3.2 \times 10^{-11}} \approx 3.125 \times 10^{13} \, \text{fissions/second} \]

### Final Answer:
The number of fissions required per second to produce 1 kilowatt of power is approximately \( 3.125 \times 10^{13} \) fissions/second.

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