Uranium `U_92^235` on bombardment with slow neutrons produces

Options

A. Deuterons
B. Fusion reaction
C. Fission reaction
D. Endothermic reaction

Answer

To solve the question regarding the reaction of Uranium-235 when bombarded with slow neutrons, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction Type**:
- When Uranium-235 (U-235) is bombarded with slow neutrons, it undergoes a nuclear reaction. This is a type of fission reaction, where the nucleus of U-235 splits into smaller nuclei.

2. **Write the Reaction Equation**:
- The general reaction can be represented as:
\[ \text{U}^{235}_{92} + \text{n} \rightarrow \text{Ba}^{141}_{56} + \text{Kr}^{92}_{36} + 3\text{n} + \text{Energy} \]
- Here, U-235 absorbs a neutron (n) and splits into Barium-141 (Ba-141) and Krypton-92 (Kr-92), along with the release of additional neutrons and a significant amount of energy.

3. **Identify the Products**:
- The primary products of this fission reaction are:
- Barium-141 (Ba-141)
- Krypton-92 (Kr-92)
- Additional neutrons (typically 2 or 3)
- Energy (in the form of kinetic energy of the fission fragments and radiation)

4. **Conclusion**:
- Therefore, the bombardment of Uranium-235 with slow neutrons results in a fission reaction producing Barium-141, Krypton-92, and additional neutrons along with energy.

### Final Answer:
Uranium-235 on bombardment with slow neutrons produces Barium-141, Krypton-92, and additional neutrons, along with energy.

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