In equation `._(92)U^(235) + ._(0)n^1 to ._(56)Ba^(144) + ._(36)Kr^(89) + X` : X is-

Options

A. `._(0)n^(1)`
B. 3H
C. `3._(0)n^(1)`
D. none of these

Answer

To solve the equation \( _{92}^{235}\text{U} + _{0}^{1}\text{n} \rightarrow _{56}^{144}\text{Ba} + _{36}^{89}\text{Kr} + X \), we need to find the particle \( X \).

### Step-by-Step Solution:

1. **Identify Mass Numbers and Atomic Numbers**:
- On the left-hand side (LHS):
- Uranium (U) has a mass number \( A = 235 \) and atomic number \( Z = 92 \).
- Neutron (n) has a mass number \( A = 1 \) and atomic number \( Z = 0 \).
- Therefore, the total mass number on LHS is:
\[ A_{LHS} = 235 + 1 = 236 \]
- The total atomic number on LHS is:
\[ Z_{LHS} = 92 + 0 = 92 \]

2. **Calculate Mass Numbers and Atomic Numbers on the Right-Hand Side (RHS)**:
- On the right-hand side (RHS):
- Barium (Ba) has a mass number \( A = 144 \) and atomic number \( Z = 56 \).
- Krypton (Kr) has a mass number \( A = 89 \) and atomic number \( Z = 36 \).
- Therefore, the total mass number on RHS (excluding \( X \)) is:
\[ A_{RHS} = 144 + 89 + A_X = 233 + A_X \]
- The total atomic number on RHS (excluding \( X \)) is:
\[ Z_{RHS} = 56 + 36 + Z_X = 92 + Z_X \]

3. **Set Up Equations**:
- From the conservation of mass numbers:
\[ A_{LHS} = A_{RHS} \implies 236 = 233 + A_X \implies A_X = 236 - 233 = 3 \]
- From the conservation of atomic numbers:
\[ Z_{LHS} = Z_{RHS} \implies 92 = 92 + Z_X \implies Z_X = 92 - 92 = 0 \]

4. **Determine the Particle \( X \)**:
- Since \( A_X = 3 \) and \( Z_X = 0 \), the particle \( X \) must be a particle with a mass number of 3 and an atomic number of 0.
- This corresponds to 3 neutrons, as neutrons have a mass number of 1 and an atomic number of 0.

### Conclusion:
Thus, the particle \( X \) is \( 3 \) neutrons.