Which of the following particles is emitted in the nuclear reaction: `._(13)Al^(27) + ._(2)He^(4) rarr ._(14)P^(30 +) ….?`

Options

A. `._(0)n^(1)`
B. `._(-1)e^(0)`
C. `._(1)H^(1)`
D. `._(1)H^(2)`

Explanation

To solve the nuclear reaction given in the question, we need to analyze the reactants and products, ensuring that both the atomic numbers and mass numbers are balanced.

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- Reactants: \( _{13}^{27}Al \) (Aluminum-27) and \( _{2}^{4}He \) (Helium-4)
- Products: \( _{14}^{30}P \) (Phosphorus-30) and an unknown particle \( X \)

2. **Write Down the Mass and Atomic Numbers**:
- For Aluminum-27:
- Atomic number (Z) = 13
- Mass number (A) = 27
- For Helium-4:
- Atomic number (Z) = 2
- Mass number (A) = 4
- For Phosphorus-30:
- Atomic number (Z) = 14
- Mass number (A) = 30

3. **Set Up the Equation**:
The nuclear reaction can be expressed as:
\[ _{13}^{27}Al + _{2}^{4}He \rightarrow _{14}^{30}P + X \]

4. **Balance the Atomic Numbers**:
- Total atomic number on the left side:
\[ 13 (Al) + 2 (He) = 15 \]
- Total atomic number on the right side (including \( X \)):
\[ 14 (P) + Z(X) = 15 \implies Z(X) = 15 - 14 = 1 \]

5. **Balance the Mass Numbers**:
- Total mass number on the left side:
\[ 27 (Al) + 4 (He) = 31 \]
- Total mass number on the right side (including \( X \)):
\[ 30 (P) + A(X) = 31 \implies A(X) = 31 - 30 = 1 \]

6. **Identify the Emitted Particle**:
- From the atomic number and mass number calculated, we find that the emitted particle \( X \) has:
- Atomic number \( Z = 1 \)
- Mass number \( A = 1 \)
- This corresponds to a proton, which is represented as \( _{1}^{1}H \).

### Final Answer:
The particle emitted in the nuclear reaction is \( _{1}^{1}H \) (a proton).

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