`""_(0)n^(1) + ""_(92)U^(235) to ""_(92)U^(236) to .............. + ""_(36)Kr^(89) + 3""_(0)n^(1)`

Answer

To solve the nuclear reaction given in the question, we need to balance both the atomic numbers and mass numbers on the reactant and product sides. Let's break it down step by step.

### Step 1: Write down the reaction
The reaction can be represented as:
\[ {}_{0}^{1}n + {}_{92}^{235}U \rightarrow {}_{92}^{236}U \rightarrow {}_{36}^{89}Kr + 3{}_{0}^{1}n \]

### Step 2: Identify the reactants and products
- **Reactants**:
- 1 neutron (n)
- Uranium-235 (U-235)

- **Products**:
- Krypton-89 (Kr-89)
- 3 neutrons (n)

### Step 3: Balance the atomic numbers
The atomic number (Z) on the reactant side:
- Neutron: 0
- Uranium-235: 92

Total atomic number on the reactant side = \(0 + 92 = 92\)

The atomic number on the product side:
- Krypton-89: 36
- Neutrons: 0 (3 neutrons contribute 0 to atomic number)

Let \(Z\) be the atomic number of the unknown nucleus \(X\):
\[ Z + 36 + 0 = 92 \]
Thus,
\[ Z = 92 - 36 = 56 \]

### Step 4: Balance the mass numbers
The mass number (A) on the reactant side:
- Neutron: 1
- Uranium-235: 235

Total mass number on the reactant side = \(1 + 235 = 236\)

The mass number on the product side:
- Krypton-89: 89
- Neutrons: 1 (each neutron has a mass number of 1, so 3 neutrons contribute 3)

Let \(A\) be the mass number of the unknown nucleus \(X\):
\[ A + 89 + 3 = 236 \]
Thus,
\[ A = 236 - 89 - 3 = 144 \]

### Step 5: Identify the unknown nucleus
From the calculations:
- Atomic number \(Z = 56\)
- Mass number \(A = 144\)

Looking at the periodic table, the element with atomic number 56 is Barium (Ba).

### Final Answer
The unknown nucleus \(X\) formed in the reaction is:
\[ {}_{56}^{144}Ba \]