In a fission reaction `._92^236 U rarr ^117 X + ^117Y + n + n`, the binding energy per nucleon of `X` and `Y` is `8.5 MeV` whereas of `.^236 U` is `7.6 MeV`. The total energy liberated will be about.

Options

A. 200 KeV
B. 2 MeV
C. 200 MeV
D. 2000 MeV

Explanation

To solve the problem, we need to calculate the total energy liberated during the fission reaction. We will do this by calculating the total binding energy of the products and the original nucleus, and then finding the difference.

### Step-by-Step Solution:

1. **Identify the binding energies:**
- The binding energy per nucleon of \( ^{236}U \) is given as \( 7.6 \, \text{MeV} \).
- The binding energy per nucleon of \( X \) and \( Y \) is given as \( 8.5 \, \text{MeV} \).

2. **Calculate the total number of nucleons:**
- The total number of nucleons in \( X \) and \( Y \) is:
\[ 117 + 117 = 234 \, \text{nucleons} \]

3. **Calculate the total binding energy of \( X \) and \( Y \):**
- The total binding energy for \( X \) and \( Y \) is:
\[ \text{Total Binding Energy}_{X+Y} = \text{Binding Energy per Nucleon} \times \text{Total Nucleons} = 8.5 \, \text{MeV} \times 234 = 1989 \, \text{MeV} \]

4. **Calculate the total binding energy of \( ^{236}U \):**
- The total binding energy for \( ^{236}U \) is:
\[ \text{Total Binding Energy}_{U} = 7.6 \, \text{MeV} \times 236 = 1793.6 \, \text{MeV} \]

5. **Calculate the energy liberated during the fission:**
- The energy liberated (\( \Delta E \)) is the difference between the total binding energy of the products and the original nucleus:
\[ \Delta E = \text{Total Binding Energy}_{X+Y} - \text{Total Binding Energy}_{U} \]
\[ \Delta E = 1989 \, \text{MeV} - 1793.6 \, \text{MeV} = 195.4 \, \text{MeV} \]

6. **Approximate the energy liberated:**
- Since the options provided do not match exactly, we can round \( 195.4 \, \text{MeV} \) to approximately \( 200 \, \text{MeV} \).

### Final Answer:
The total energy liberated will be about \( 200 \, \text{MeV} \).

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