for nuclear reaction :
`._(92)U^(235)+._(0)n^(1)to._(56)Ba^(144)+.......+3_(0)n^(1)`

Options

A. `._(26)Kr^(89)`
B. `._(36)Kr^(89)`
C. `._(26)Sr^(90)`
D. `._(38)Sr^(89)`

Explanation

To solve the nuclear reaction problem, we need to identify the missing element in the reaction:

\[ _{92}^{235}\text{U} + _{0}^{1}\text{n} \rightarrow _{56}^{144}\text{Ba} + \ldots + 3_{0}^{1}\text{n} \]

### Step 1: Conservation of Atomic Number
The atomic number (Z) must be conserved in the reaction.

- On the left-hand side (LHS), we have:
- Uranium (U) with atomic number 92
- Neutron (n) with atomic number 0

Therefore, the total atomic number on the LHS is:
\[ Z_{\text{LHS}} = 92 + 0 = 92 \]

- On the right-hand side (RHS), we have:
- Barium (Ba) with atomic number 56
- Let the missing element have atomic number \( Z \)
- Plus 3 neutrons, each with atomic number 0

Therefore, the total atomic number on the RHS is:
\[ Z_{\text{RHS}} = 56 + Z + 0 = 56 + Z \]

Setting both sides equal gives:
\[ 92 = 56 + Z \]

Solving for \( Z \):
\[ Z = 92 - 56 = 36 \]

### Step 2: Conservation of Mass Number
Next, we conserve the mass number (A).

- On the LHS, we have:
- Uranium (U) with mass number 235
- Neutron (n) with mass number 1

Therefore, the total mass number on the LHS is:
\[ A_{\text{LHS}} = 235 + 1 = 236 \]

- On the RHS, we have:
- Barium (Ba) with mass number 144
- Let the missing element have mass number \( A \)
- Plus 3 neutrons, each with mass number 1

Therefore, the total mass number on the RHS is:
\[ A_{\text{RHS}} = 144 + A + 3 = 144 + A + 3 = 147 + A \]

Setting both sides equal gives:
\[ 236 = 147 + A \]

Solving for \( A \):
\[ A = 236 - 147 = 89 \]

### Step 3: Identify the Missing Element
Now we have determined that the missing element has:
- Atomic number \( Z = 36 \)
- Mass number \( A = 89 \)

The element with atomic number 36 is Krypton (Kr).

### Final Answer
The missing element in the nuclear reaction is:
\[ _{36}^{89}\text{Kr} \]