What is the power output of a `._(92) U^(235)` reactor if it is takes `30` days to use up `2 kg` of fuel, and if each fission gives `185 MeV` of usable energy ?.

Options

A. `45` megawatt
B. `58.46` megawatt
C. `72` megawatt
D. `92` megawatt

Explanation

To find the power output of a Uranium-235 reactor, we can follow these steps:

### Step 1: Calculate the total energy produced from the fission of 2 kg of Uranium-235.

1. **Energy per fission**: Each fission of Uranium-235 releases 185 MeV of energy.
2. **Convert MeV to Joules**:
\[ 1 \text{ MeV} = 1.6 \times 10^{-13} \text{ Joules} \]
Therefore,
\[ 185 \text{ MeV} = 185 \times 1.6 \times 10^{-13} \text{ Joules} = 2.96 \times 10^{-11} \text{ Joules} \]

### Step 2: Calculate the number of fissions in 2 kg of Uranium-235.

1. **Molar mass of Uranium-235**: Approximately 235 g/mol.
2. **Convert kg to grams**:
\[ 2 \text{ kg} = 2000 \text{ g} \]
3. **Calculate moles of Uranium-235**:
\[ \text{Moles} = \frac{2000 \text{ g}}{235 \text{ g/mol}} \approx 8.51 \text{ mol} \]
4. **Calculate the number of atoms**:
\[ \text{Number of atoms} = \text{Moles} \times N_A = 8.51 \times 6.022 \times 10^{23} \approx 5.12 \times 10^{24} \text{ atoms} \]

### Step 3: Calculate the total energy produced from all fissions.

1. **Total energy**:
\[ \text{Total energy} = \text{Number of fissions} \times \text{Energy per fission} \]
\[ = 5.12 \times 10^{24} \times 2.96 \times 10^{-11} \approx 1.51 \times 10^{14} \text{ Joules} \]

### Step 4: Convert the time from days to seconds.

1. **Time in seconds**:
\[ 30 \text{ days} = 30 \times 24 \times 60 \times 60 = 2,592,000 \text{ seconds} \]

### Step 5: Calculate the power output.

1. **Power formula**:
\[ P = \frac{E}{T} \]
\[ P = \frac{1.51 \times 10^{14} \text{ Joules}}{2,592,000 \text{ seconds}} \approx 58.3 \text{ Megawatts} \]

### Final Answer:
The power output of the Uranium-235 reactor is approximately **58.3 Megawatts**.

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