In the nuclear reaction `Mg_12^(24) + He_2^4 to n_0^1 + (?)` The product nucleus is

Options

A. `Al_13^27`
B. `Si_14^27`
C. `Al_13^28`
D. `Mg_12^25`

Answer

To solve the nuclear reaction \( \text{Mg}_{12}^{24} + \text{He}_{2}^{4} \to \text{n}_{0}^{1} + (?) \), we need to identify the unknown product nucleus. We can do this by applying the conservation of atomic mass and atomic number.

### Step-by-Step Solution:

1. **Identify the Reactants:**
- Magnesium (\( \text{Mg} \)): Atomic number = 12, Atomic mass = 24
- Helium (\( \text{He} \)): Atomic number = 2, Atomic mass = 4

2. **Identify the Product:**
- Neutron (\( \text{n} \)): Atomic number = 0, Atomic mass = 1
- Let the unknown nucleus be represented as \( X \) with atomic number \( Z \) and atomic mass \( A \).

3. **Write the Conservation Equations:**
- **Conservation of Atomic Mass:**
\[ \text{Total mass on the left} = \text{Total mass on the right} \]
\[ 24 + 4 = 1 + A \]
\[ 28 = 1 + A \implies A = 28 - 1 = 27 \]

- **Conservation of Atomic Number:**
\[ \text{Total atomic number on the left} = \text{Total atomic number on the right} \]
\[ 12 + 2 = 0 + Z \]
\[ 14 = Z \]

4. **Identify the Unknown Nucleus:**
- From the calculations, we have:
- Atomic mass \( A = 27 \)
- Atomic number \( Z = 14 \)
- The nucleus with atomic number 14 and atomic mass 27 is Silicon (\( \text{Si} \)).

5. **Final Answer:**
- The product nucleus is \( \text{Si}_{14}^{27} \).