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 bombardment of Uranium-235 (`U_92^235`) with slow neutrons, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction Type**:
- When Uranium-235 is bombarded with slow neutrons, it undergoes a nuclear reaction known as fission. In this process, a heavy nucleus splits into smaller nuclei.

2. **Write the Reaction**:
- The fission of Uranium-235 can be represented as follows:
\[ ^{235}_{92}U + ^{1}_{0}n \rightarrow ^{141}_{56}Ba + ^{92}_{36}Kr + 3^{1}_{0}n \]
- Here, `U` represents Uranium, `n` represents a neutron, `Ba` represents Barium, and `Kr` represents Krypton.

3. **Balance the Reaction**:
- Check that the mass numbers and atomic numbers are balanced on both sides of the equation:
- Left side: Mass number = 235 (U) + 1 (n) = 236; Atomic number = 92 (U) + 0 (n) = 92.
- Right side: Mass number = 141 (Ba) + 92 (Kr) + 3 (neutrons) = 236; Atomic number = 56 (Ba) + 36 (Kr) + 0 (neutrons) = 92.
- The reaction is balanced.

4. **Identify the Products**:
- The products of this fission reaction are:
- Barium-141 (`Ba-141`)
- Krypton-92 (`Kr-92`)
- Three neutrons are also produced.

5. **Conclusion**:
- Therefore, the final answer to the question is that Uranium-235 on bombardment with slow neutrons produces Barium-141, Krypton-92, and three neutrons.

### Final Answer:
Uranium-235 (`U_92^235`) on bombardment with slow neutrons produces Barium-141 (`Ba_56^141`), Krypton-92 (`Kr_36^92`), and three neutrons.

---