`92^(U^(235)` mucleus absorbs a slow neutron and undergoes fission into `54^(X^139)` and `38^(sr^94)` nuclie The other particies produced in this fission process are

Options

A. `1beta` and `1alpha`
B. `2 beta` and 1 neutron
C. 2 neutrons
D. 3 neutrons

Answer

To solve the problem, we need to analyze the nuclear fission reaction provided. The fission of Uranium-235 (U-235) is given as follows:

1. **Write the Fission Reaction**:
The fission reaction can be written as:
\[ \text{U}^{235}_{92} + \text{n} \rightarrow \text{X}^{139}_{54} + \text{Sr}^{94}_{38} + \text{n} + \text{n} \]
where \( \text{n} \) represents a neutron.

2. **Conservation of Atomic Mass (A)**:
The atomic mass (A) must be conserved in the reaction.
- Left side: \( 235 \) (from U-235) + \( 1 \) (from the absorbed neutron) = \( 236 \)
- Right side: \( 139 \) (from X) + \( 94 \) (from Sr) + \( 2 \times 1 \) (from 2 neutrons) = \( 139 + 94 + 2 = 235 \)

Therefore, the atomic mass is conserved.

3. **Conservation of Atomic Number (Z)**:
The atomic number (Z) must also be conserved.
- Left side: \( 92 \) (from U-235)
- Right side: \( 54 \) (from X) + \( 38 \) (from Sr) = \( 54 + 38 = 92 \)

Thus, the atomic number is also conserved.

4. **Determine the Other Particles**:
Since the total atomic mass on the left side is \( 236 \) and on the right side is \( 235 \), we have a difference of \( 1 \). This indicates that 2 neutrons were released during the fission process to balance the mass.

5. **Final Answer**:
The other particles produced in this fission process are **2 neutrons**.

### Summary of the Solution:
The fission of Uranium-235 results in the production of X-139, Sr-94, and 2 neutrons.

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