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. `3.125xx10^13`
B. `1.52xx10^6`
C. `3.125xx10^12`
D. `3.125xx10^14`

Explanation

To solve the problem of how many fissions of a U-235 nucleus are required to produce 1 kilowatt of power, we can follow these steps:

### Step 1: Convert the energy released per fission from MeV to Joules
The energy released in the fission of a single U-235 nucleus is given as 200 MeV. We need to convert this energy into Joules using the conversion factor \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\).

\[ E = 200 \text{ MeV} = 200 \times 10^6 \text{ eV} \]

Now, converting this to Joules:

\[ E = 200 \times 10^6 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} \]

Calculating this gives:

\[ E = 200 \times 10^6 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-11} \text{ J} \]

### Step 2: Determine the energy required per second for 1 kW of power
Power is defined as energy per unit time. To produce 1 kilowatt (kW) of power, we need to produce 1 kJ of energy per second.

\[ \text{Power} = 1 \text{ kW} = 1000 \text{ J/s} \]

### Step 3: Relate the number of fissions to the energy produced
Let \(n\) be the number of fissions occurring per second. The total energy produced by \(n\) fissions in one second is given by:

\[ \text{Total Energy} = n \times E \]

Setting this equal to the energy required per second:

\[ n \times E = 1000 \text{ J} \]

### Step 4: Solve for the number of fissions per second
Substituting the expression for \(E\):

\[ n \times (3.2 \times 10^{-11} \text{ J}) = 1000 \text{ J} \]

Now, solving for \(n\):

\[ n = \frac{1000 \text{ J}}{3.2 \times 10^{-11} \text{ J}} = \frac{1000}{3.2 \times 10^{-11}} \]

Calculating this gives:

\[ n \approx 3.125 \times 10^{13} \]

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