In nuclear reaction `._(4)Be^(9)+._(2)He^(4)rarr._(6)C^(12)+X,X` will be

Options

A. (a)Neutron
B. (b)Proton
C. (c)Positron
D. (d)Electron

Answer

To solve the nuclear reaction given by:

\[ _{4}^{9}\text{Be} + _{2}^{4}\text{He} \rightarrow _{6}^{12}\text{C} + X \]

we need to determine what particle \(X\) is.

### Step-by-Step Solution:

1. **Identify the Reactants and Products:**
- The reactants are Beryllium (\( _{4}^{9}\text{Be} \)) and Helium (\( _{2}^{4}\text{He} \)).
- The product is Carbon (\( _{6}^{12}\text{C} \)) and an unknown particle \(X\).

2. **Calculate the Total Mass Number on the Left-Hand Side (LHS):**
- The mass number of Beryllium is 9.
- The mass number of Helium is 4.
- Total mass number on LHS = \(9 + 4 = 13\).

3. **Calculate the Total Mass Number on the Right-Hand Side (RHS):**
- The mass number of Carbon is 12.
- Let the mass number of \(X\) be \(A_X\).
- Total mass number on RHS = \(12 + A_X\).

4. **Set Up the Equation for Mass Numbers:**
- Since mass numbers must be conserved in a nuclear reaction, we can set up the equation:
\[ 13 = 12 + A_X \]

5. **Solve for \(A_X\):**
- Rearranging the equation gives:
\[ A_X = 13 - 12 = 1 \]

6. **Calculate the Total Atomic Number on the Left-Hand Side (LHS):**
- The atomic number of Beryllium is 4.
- The atomic number of Helium is 2.
- Total atomic number on LHS = \(4 + 2 = 6\).

7. **Calculate the Total Atomic Number on the Right-Hand Side (RHS):**
- The atomic number of Carbon is 6.
- Let the atomic number of \(X\) be \(Z_X\).
- Total atomic number on RHS = \(6 + Z_X\).

8. **Set Up the Equation for Atomic Numbers:**
- Since atomic numbers must also be conserved, we can set up the equation:
\[ 6 = 6 + Z_X \]

9. **Solve for \(Z_X\):**
- Rearranging gives:
\[ Z_X = 6 - 6 = 0 \]

10. **Identify the Particle \(X\):**
- A particle with mass number \(A_X = 1\) and atomic number \(Z_X = 0\) corresponds to a neutron.

Thus, the particle \(X\) is a neutron.

### Final Answer:
\[ X = \text{neutron} \]