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 \( _{92}^{235}U \) nucleus, resulting in the production of \( _{36}^{93}Kr \) and \( _{56}^{140}Ba \). We will also determine how many neutrons are emitted during this process.

### Step-by-Step Solution:

1. **Write the initial nuclear reaction:**
The initial reaction can be represented as:
\[ _{92}^{235}U + _{0}^{1}n \rightarrow _{36}^{93}Kr + _{56}^{140}Ba + X \]
where \( X \) represents the particles emitted during the reaction.

2. **Conservation of Mass Number:**
The mass number must be conserved in the reaction. The total mass number before the reaction is:
\[ 235 + 1 = 236 \]
The total mass number on the right side is:
\[ 93 + 140 + A_X \]
where \( A_X \) is the mass number of the emitted particles. Setting these equal gives:
\[ 236 = 93 + 140 + A_X \]
Simplifying this:
\[ A_X = 236 - 233 = 3 \]

3. **Conservation of Atomic Number:**
The atomic number must also be conserved. The total atomic number before the reaction is:
\[ 92 + 0 = 92 \]
The total atomic number on the right side is:
\[ 36 + 56 + Z_X \]
where \( Z_X \) is the atomic number of the emitted particles. Setting these equal gives:
\[ 92 = 36 + 56 + Z_X \]
Simplifying this:
\[ Z_X = 92 - 92 = 0 \]

4. **Identify the emitted particles:**
Since \( A_X = 3 \) and \( Z_X = 0 \), the emitted particles must be neutrons. Specifically, we have 3 neutrons emitted during the reaction.

5. **Final Reaction:**
The complete reaction can be written as:
\[ _{92}^{235}U + _{0}^{1}n \rightarrow _{36}^{93}Kr + _{56}^{140}Ba + 3_{0}^{1}n \]

### Conclusion:
Thus, the number of neutrons emitted in this reaction is **3**.