If 200 MeV energy is released in the fission of a single nucleus of `._(92)U^(235)`, how many fissions must occur per sec to produce a power of 1 kW?

Options

A. `3.125 xx 10^(13)`
B. `6.250 xx 10^(13)`
C. `1.525 xx 10^(13)`
D. None of these

Answer

To solve the problem, we need to determine how many fissions of Uranium-235 must occur per second to produce a power of 1 kW, given that each fission releases 200 MeV of energy.

### Step-by-Step Solution:

1. **Convert the energy released per fission from MeV to Joules**:
- The energy released in fission is given as 200 MeV.
- We know that 1 MeV = \(1.6 \times 10^{-13}\) Joules.
- Therefore, the energy released per fission in Joules is:
\[ E = 200 \, \text{MeV} \times 1.6 \times 10^{-13} \, \text{J/MeV} = 3.2 \times 10^{-11} \, \text{J} \]

2. **Define the power requirement**:
- The power required is given as 1 kW, which is equivalent to:
\[ P = 1 \, \text{kW} = 1000 \, \text{W} = 1000 \, \text{J/s} \]

3. **Relate power to the number of fissions**:
- If \(n\) is the number of fissions occurring per second, then the total energy produced per second due to fissions is:
\[ \text{Total Energy per second} = n \times E = n \times 3.2 \times 10^{-11} \, \text{J} \]
- Setting this equal to the power requirement gives us:
\[ n \times 3.2 \times 10^{-11} = 1000 \]

4. **Solve for \(n\)**:
- Rearranging the equation to solve for \(n\):
\[ n = \frac{1000}{3.2 \times 10^{-11}} \]
- Calculating \(n\):
\[ n = \frac{1000}{3.2 \times 10^{-11}} \approx 3.125 \times 10^{13} \]

5. **Final Answer**:
- Therefore, the number of fissions that must occur per second to produce a power of 1 kW is approximately:
\[ n \approx 3.125 \times 10^{13} \, \text{fissions/second} \]