`""_(13)^(27) Al +""_(1)^(1)P to X +""_(0)^(0)gamma`, Identify X if reaction is `(p, gamma)` type artificial radioactive reaction :

Options

A. `""_(13)^(28) Al`
B. `""_(14)^(27)Si`
C. `""_(14)^(28)Si`
D. None of these

Explanation

To solve the problem, we need to identify the product \( X \) in the reaction:

\[ _{13}^{27} \text{Al} + _{1}^{1} \text{p} \rightarrow X + _{0}^{0} \gamma \]

### Step 1: Understand the Components of the Reaction
- **Aluminium (Al)** has an atomic number of 13 and a mass number of 27.
- **Proton (p)** has an atomic number of 1 and a mass number of 1.
- **Gamma radiation (\( \gamma \))** has an atomic number of 0 and a mass number of 0.

### Step 2: Apply Conservation of Mass Number
The total mass number on the reactant side must equal the total mass number on the product side.

**Reactants:**
- Mass number of Al = 27
- Mass number of p = 1

Total mass number on the reactant side:
\[ 27 + 1 = 28 \]

**Products:**
Let the mass number of \( X \) be \( A_X \).
Since gamma radiation has a mass number of 0:
\[ A_X + 0 = A_X \]

Setting the total mass number of reactants equal to the total mass number of products:
\[ A_X = 28 \]

### Step 3: Apply Conservation of Atomic Number
The total atomic number on the reactant side must equal the total atomic number on the product side.

**Reactants:**
- Atomic number of Al = 13
- Atomic number of p = 1

Total atomic number on the reactant side:
\[ 13 + 1 = 14 \]

**Products:**
Let the atomic number of \( X \) be \( Z_X \).
Since gamma radiation has an atomic number of 0:
\[ Z_X + 0 = Z_X \]

Setting the total atomic number of reactants equal to the total atomic number of products:
\[ Z_X = 14 \]

### Step 4: Identify Element \( X \)
From the results:
- Mass number \( A_X = 28 \)
- Atomic number \( Z_X = 14 \)

The element with atomic number 14 is Silicon (Si).

### Conclusion
Thus, the product \( X \) in the reaction is:

\[ X = _{14}^{28} \text{Si} \]

### Final Answer
The correct identification of \( X \) is Silicon with atomic number 14 and mass number 28.

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