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. `3.125xx10^(14)`
C. `3.125xx10^(15)`
D. `3.125xx10^(16)`

Answer

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

### Step 1: Understand the relationship between power, energy, and time
Power (P) is defined as the energy (E) released per unit time (T):
\[ P = \frac{E}{T} \]
In this case, we want to find the number of fissions (n) per second, so we can rearrange the equation:
\[ n = \frac{P \cdot T}{E} \]

### Step 2: Convert the power from kilowatts to watts
Since 1 kilowatt (kW) is equal to 1000 watts (W), we have:
\[ P = 1000 \, \text{W} \]

### Step 3: Determine the energy released per fission
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:
\[ 200 \, \text{MeV} = 200 \times 10^6 \, \text{eV} \]
Using the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ E = 200 \times 10^6 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} \]
Calculating this gives:
\[ E = 200 \times 10^6 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-11} \, \text{J} \]

### Step 4: Calculate the number of fissions required per second
Now we can substitute the values into the rearranged power equation:
\[ n = \frac{P}{E} = \frac{1000 \, \text{W}}{3.2 \times 10^{-11} \, \text{J}} \]
Calculating this gives:
\[ n = \frac{1000}{3.2 \times 10^{-11}} \approx 3.125 \times 10^{13} \, \text{fissions/second} \]

### Conclusion
Thus, the number of fissions required per second to produce 1 kilowatt of power is approximately:
\[ n \approx 3.125 \times 10^{13} \]