A neutron strikes a `""_(92)U^(235)` nucleus and as a result `""_(36)Kr^(93) and ""_(56)Ba^(140)` are produced with

Options

A. `alpha`-particle
B. 1-neutron
C. 3-neutron
D. `2-beta`- particle

Answer

To solve the problem, we need to analyze the nuclear reaction that occurs when a neutron strikes a Uranium-235 nucleus. The products of this reaction are Krypton-93 and Barium-140. We need to determine what else is produced in this reaction.

### Step-by-Step Solution:

1. **Identify the Initial and Final Nuclei**:
- The initial nucleus is Uranium-235, which has:
- Atomic number (Z) = 92
- Mass number (A) = 235
- The neutron has:
- Atomic number (Z) = 0
- Mass number (A) = 1

2. **Calculate the Total Initial Mass and Atomic Number**:
- Total initial mass number = Mass of Uranium + Mass of Neutron
\[ A_{initial} = 235 + 1 = 236 \]
- Total initial atomic number = Atomic number of Uranium + Atomic number of Neutron
\[ Z_{initial} = 92 + 0 = 92 \]

3. **Identify the Products**:
- The products of the reaction are:
- Krypton-93 (Z = 36, A = 93)
- Barium-140 (Z = 56, A = 140)

4. **Calculate the Total Final Mass and Atomic Number**:
- Total final mass number = Mass of Krypton + Mass of Barium + Mass of other products
\[ A_{final} = 93 + 140 + A_{other} \]
- Total final atomic number = Atomic number of Krypton + Atomic number of Barium + Atomic number of other products
\[ Z_{final} = 36 + 56 + Z_{other} \]

5. **Set Up the Equations**:
- From mass conservation:
\[ 236 = 93 + 140 + A_{other} \implies A_{other} = 236 - 233 = 3 \]
- From atomic number conservation:
\[ 92 = 36 + 56 + Z_{other} \implies Z_{other} = 92 - 92 = 0 \]

6. **Determine the Other Products**:
- Since the atomic number of the other products is 0 and the mass number is 3, the other products must be 3 neutrons (since neutrons have a mass of 1 and atomic number of 0).

### Final Answer:
The complete reaction can be summarized as:
\[ \text{Uranium-235} + \text{Neutron} \rightarrow \text{Krypton-93} + \text{Barium-140} + 3 \text{ Neutrons} \]