A nuclear reactor containing `U^(235)` produces 10 MW . Calculate the fission rate assuming that 200 MeV of useful energy is released in one fission ?

Options

A. `3.1xx 10^(17) sec^(-1)`
B. `6 xx 10^8 sec^(-1)`
C. `1.5 xx 10^(17) sec^(-1)`
D. `2 xx 10^(17) sec^(-1)`

Answer

To solve the problem of calculating the fission rate in a nuclear reactor producing 10 MW of power, we can follow these steps:

### Step 1: Understand the relationship between power, energy, and time
Power (P) is defined as the rate of energy produced over time. Mathematically, it can be expressed as:
\[ P = \frac{E}{T} \]
where \( E \) is the total energy produced and \( T \) is the time period.

### Step 2: Relate fission reactions to power
Let \( n \) be the number of fission reactions occurring per unit time (fission rate). If each fission reaction releases an energy \( e \), then the total energy produced in time \( T \) can be expressed as:
\[ E = n \cdot e \cdot T \]
Substituting this into the power equation gives:
\[ P = \frac{n \cdot e \cdot T}{T} \]
This simplifies to:
\[ P = n \cdot e \]

### Step 3: Rearranging to find the fission rate
From the equation \( P = n \cdot e \), we can rearrange it to find the fission rate \( n \):
\[ n = \frac{P}{e} \]

### Step 4: Substitute the known values
We know:
- Power \( P = 10 \, \text{MW} = 10 \times 10^6 \, \text{W} \)
- Energy per fission \( e = 200 \, \text{MeV} \)

First, we need to convert the energy from MeV to Joules. The conversion factor is:
\[ 1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J} \]
Thus,
\[ e = 200 \, \text{MeV} = 200 \times 1.6 \times 10^{-13} \, \text{J} = 3.2 \times 10^{-11} \, \text{J} \]

### Step 5: Calculate the fission rate
Now substitute the values into the equation for \( n \):
\[ n = \frac{10 \times 10^6 \, \text{W}}{3.2 \times 10^{-11} \, \text{J}} \]

Calculating this gives:
\[ n = \frac{10 \times 10^6}{3.2 \times 10^{-11}} \approx 3.125 \times 10^{17} \, \text{fissions/second} \]

### Final Answer
The fission rate is approximately:
\[ n \approx 3.1 \times 10^{17} \, \text{fissions/second} \]