Assuming that 200MeV of energy is released per fission of uranium ato, find the number of fission per second required to release one kilowatt power.

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)`

Answer

To solve the problem of finding the number of fissions per second required to release one kilowatt of power from uranium fission, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Values**:
- Energy released per fission of uranium = 200 MeV
- Power required = 1 kilowatt = 1000 watts

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

3. **Use the Power Formula**:
- Power (P) is defined as the energy per unit time:
\[ P = \frac{N \cdot E}{T} \]
- Rearranging gives us:
\[ N = \frac{P \cdot T}{E} \]
- Since we want the number of fissions per second, we can set \(T = 1\) second, thus:
\[ N = \frac{P}{E} \]

4. **Substitute the Values**:
- Substitute \(P = 1000 \, \text{W}\) and \(E = 3.2 \times 10^{-11} \, \text{J}\):
\[ N = \frac{1000 \, \text{W}}{3.2 \times 10^{-11} \, \text{J}} = 3.125 \times 10^{13} \, \text{fissions/second} \]

5. **Conclusion**:
- The number of fissions per second required to release one kilowatt of power is:
\[ N \approx 3.125 \times 10^{13} \, \text{fissions/second} \]

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

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