The number of neutrons accompanying the formation of `._(54)Xe^(139)` and `._(38)Sr^(94)` from the absorption of a slow neutron by `._(92)U^(235)`, followed by nuclear fission is

Options

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

Explanation

To solve the problem of determining the number of neutrons emitted during the nuclear fission of uranium-235, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- The reactant is uranium-235, which can be represented as \( _{92}^{235}U \).
- The products of the fission are xenon-139 and strontium-94, represented as \( _{54}^{139}Xe \) and \( _{38}^{94}Sr \), respectively.

2. **Write the Nuclear Reaction**:
- The nuclear reaction can be represented as:
\[ _{92}^{235}U + n \rightarrow _{54}^{139}Xe + _{38}^{94}Sr + Y \text{ neutrons} \]
- Here, \( n \) represents the absorbed neutron, and \( Y \) is the number of neutrons emitted.

3. **Apply Mass Conservation**:
- According to the law of conservation of mass (or nucleons), the total mass number on the reactant side must equal the total mass number on the product side.
- On the reactant side, the mass number is:
\[ 235 + 1 = 236 \]
- On the product side, the mass number is:
\[ 139 + 94 + Y = 233 + Y \]

4. **Set Up the Equation**:
- Set the total mass numbers equal to each other:
\[ 236 = 233 + Y \]

5. **Solve for Y**:
- Rearranging the equation gives:
\[ Y = 236 - 233 = 3 \]

6. **Conclusion**:
- Therefore, the number of neutrons emitted during the fission process is \( Y = 3 \).

### Final Answer:
The number of neutrons accompanying the formation of \( _{54}^{139}Xe \) and \( _{38}^{94}Sr \) from the absorption of a slow neutron by \( _{92}^{235}U \) is **3**.

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