The energy liberated on complete fission of `1 kg` of `._92 U^235` is (Assume `200 MeV` energy is liberated on fission of `1` nucleus).

Options

A. `8.2 xx 10^10 J`
B. `8.2 xx 10^9 J`
C. `8.2 xx 10^13 J`
D. `8.2 xx 10^16 J`

Explanation

To calculate the energy liberated on the complete fission of 1 kg of Uranium-235 (\( _{92}^{235}U \)), we will follow these steps:

### Step 1: Determine the number of nuclei in 1 kg of \( _{92}^{235}U \)

1. **Calculate the molar mass of \( _{92}^{235}U \)**:
The atomic mass of \( _{92}^{235}U \) is approximately 235 g/mol.

2. **Convert 1 kg to grams**:
\[ 1 \text{ kg} = 1000 \text{ g} \]

3. **Calculate the number of moles in 1 kg of \( _{92}^{235}U \)**:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{1000 \text{ g}}{235 \text{ g/mol}} \approx 4.255 \text{ mol} \]

4. **Calculate the number of nuclei**:
Using Avogadro's number (\( 6.022 \times 10^{23} \) nuclei/mol):
\[ \text{Number of nuclei} = \text{Number of moles} \times \text{Avogadro's number} = 4.255 \text{ mol} \times 6.022 \times 10^{23} \text{ nuclei/mol} \approx 2.56 \times 10^{24} \text{ nuclei} \]

### Step 2: Calculate the total energy released

1. **Energy released per fission**:
Given that the energy released on fission of one nucleus is \( 200 \text{ MeV} \).

2. **Calculate the total energy released**:
\[ \text{Total energy} = \text{Number of nuclei} \times \text{Energy per nucleus} = 2.56 \times 10^{24} \text{ nuclei} \times 200 \text{ MeV} \]
\[ = 5.12 \times 10^{26} \text{ MeV} \]

### Step 3: Convert energy from MeV to Joules

1. **Conversion factor**:
\( 1 \text{ MeV} = 1.6 \times 10^{-13} \text{ Joules} \)

2. **Convert total energy to Joules**:
\[ \text{Total energy in Joules} = 5.12 \times 10^{26} \text{ MeV} \times 1.6 \times 10^{-13} \text{ Joules/MeV} \]
\[ = 8.192 \times 10^{13} \text{ Joules} \]

### Final Result
The energy liberated on complete fission of 1 kg of \( _{92}^{235}U \) is approximately \( 8.19 \times 10^{13} \text{ Joules} \).

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