The number of neutrons released when `._92 U^235` undergoes fission by absorbing `._0 n^1` and `(._56 Ba^144 + ._36 Kr^89)` are formed, is.

Options

A. 2
B. 3
C. 1
D. 0

Answer

To solve the problem of determining the number of neutrons released during the fission of Uranium-235, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Fission Reaction**:
The fission of Uranium-235 when it absorbs a neutron can be represented as:
\[ \,^{235}_{92}U + \,^{1}_{0}n \rightarrow \,^{144}_{56}Ba + \,^{89}_{36}Kr + x \,^{1}_{0}n \]
Here, \(x\) represents the number of neutrons released.

2. **Conservation of Mass Number**:
In nuclear reactions, the mass number must be conserved. We will set up an equation based on the mass numbers on both sides of the reaction.

- Left-hand side (LHS) mass number:
\[ 235 + 1 = 236 \]
- Right-hand side (RHS) mass number:
\[ 144 + 89 + x \]

3. **Set Up the Equation**:
Now we equate the total mass numbers from both sides:
\[ 236 = 144 + 89 + x \]

4. **Calculate the Right-Hand Side**:
First, calculate \(144 + 89\):
\[ 144 + 89 = 233 \]
Now, substitute this back into the equation:
\[ 236 = 233 + x \]

5. **Solve for \(x\)**:
Rearranging the equation gives:
\[ x = 236 - 233 \]
Therefore:
\[ x = 3 \]

6. **Conclusion**:
The number of neutrons released during the fission of Uranium-235 is 3.

### Final Answer:
The number of neutrons released is **3**.

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