The product p of the nuclear reaction
`._(92)^(235)U + ._(0)^(1) n rarr p + ._(36)^(92)Kr + 3 ._(0)^(1)n` is

Options

A. `._(56)^(141)Sr`
B. `._(56)^(141)La`
C. `._(56)^(141)Ba`
D. `._(56)^(141)Cs`

Explanation

To solve the problem, we will analyze the given nuclear reaction step by step.

### Step 1: Write down the nuclear reaction
The nuclear reaction given is:
\[ _{92}^{235}\text{U} + _{0}^{1}\text{n} \rightarrow P + _{36}^{92}\text{Kr} + 3 \times _{0}^{1}\text{n} \]

### Step 2: Determine the total atomic number and mass number on the left side
- The atomic number of Uranium (U) is 92, and its mass number is 235.
- The atomic number of the neutron (n) is 0, and its mass number is 1.

Calculating the total atomic number and mass number on the left side:
- Total atomic number = \(92 + 0 = 92\)
- Total mass number = \(235 + 1 = 236\)

### Step 3: Determine the total atomic number and mass number on the right side
- The atomic number of Krypton (Kr) is 36, and its mass number is 92.
- There are 3 neutrons, each with an atomic number of 0 and mass number of 1.

Calculating the total atomic number and mass number on the right side:
- Total atomic number = \(36 + 3 \times 0 = 36\)
- Total mass number = \(92 + 3 \times 1 = 95\)

### Step 4: Set up equations for the unknown product P
Let the atomic number of P be \(Z_P\) and the mass number of P be \(A_P\).

From the conservation of atomic number:
\[ Z_P + 36 = 92 \implies Z_P = 92 - 36 = 56 \]

From the conservation of mass number:
\[ A_P + 95 = 236 \implies A_P = 236 - 95 = 141 \]

### Step 5: Identify the element corresponding to \(Z_P = 56\) and \(A_P = 141\)
The element with atomic number 56 is Barium (Ba). Therefore, the product \(P\) is:
\[ P = _{56}^{141}\text{Ba} \]

### Conclusion
The product \(P\) of the nuclear reaction is:
\[ P = _{56}^{141}\text{Ba} \]