A reactor is generating 1000 kW of power and 200 MeV of energy may be obtained per fission of `U^235` The rate of nuclear fission in the reactor is

Options

A. `3.125 xx 10^(16)`
B. `1.253xx10^(16)`
C. `2 xx 10^8`
D. 931

Answer

To find the rate of nuclear fission in the reactor, we can follow these steps:

### Step 1: Identify the given values
- Power generated by the reactor, \( P = 1000 \, \text{kW} = 1000 \times 1000 \, \text{W} = 10^6 \, \text{W} \)
- Energy per fission of \( U^{235} = 200 \, \text{MeV} \)

### Step 2: Convert energy per fission from MeV to Joules
We know that:
1 MeV = \( 1.6 \times 10^{-13} \, \text{J} \)

Thus, the energy per fission in Joules is:
\[ E_{\text{fission}} = 200 \, \text{MeV} \times 1.6 \times 10^{-13} \, \text{J/MeV} = 3.2 \times 10^{-11} \, \text{J} \]

### Step 3: Relate power to energy
Power is defined as the energy generated per unit time. Therefore, the total energy generated in one second is equal to the power:
\[ E_{\text{total}} = P \times t = 10^6 \, \text{W} \times 1 \, \text{s} = 10^6 \, \text{J} \]

### Step 4: Set up the equation for the number of fissions
Let \( x \) be the number of fissions occurring in one second. The total energy generated by \( x \) fissions is:
\[ E_{\text{total}} = x \times E_{\text{fission}} = x \times 3.2 \times 10^{-11} \, \text{J} \]

### Step 5: Equate the two expressions for total energy
Setting the two expressions for total energy equal gives:
\[ 10^6 \, \text{J} = x \times 3.2 \times 10^{-11} \, \text{J} \]

### Step 6: Solve for \( x \)
Rearranging the equation to solve for \( x \):
\[ x = \frac{10^6 \, \text{J}}{3.2 \times 10^{-11} \, \text{J}} = \frac{10^6}{3.2 \times 10^{-11}} \approx 3.125 \times 10^{16} \]

### Conclusion
The rate of nuclear fission in the reactor is approximately:
\[ x \approx 3.125 \times 10^{16} \, \text{fissions/second} \]

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