Bombardment of aluminium of `alpha-` particle leads to its artificial disintegration in two ways `(i)` and `(ii)` as shown below. Product `X,Y,` and `Z`, respectively, are
`._(14)Si^(30)+Xoverset((i))larr._(13)Al^(27)overset((ii))rarr._(15)P^(30)+Y rarr ._(14)Si^(30)+Z`

Options

A. Proton, neutron, positron
B. Neutron, positron, proton
C. Proton, positron, neutron
D. Positron, proton, neutron

Explanation

To solve the problem of the bombardment of aluminum with alpha particles leading to artificial disintegration, we will analyze the reactions step by step.

### Step 1: Analyze the first reaction
The first reaction is:
\[ _{13}^{27}\text{Al} + \alpha \rightarrow _{14}^{30}\text{Si} + X \]
where \(\alpha\) represents the alpha particle, which has a mass number of 4 and an atomic number of 2.

#### Mass Number Calculation:
- Mass number on the left side (reactants):
\[ 27 + 4 = 31 \]
- Mass number on the right side (products):
\[ 30 + \text{mass number of } X \]
- Setting the two sides equal:
\[ 31 = 30 + \text{mass number of } X \]
- Solving for the mass number of \(X\):
\[ \text{mass number of } X = 31 - 30 = 1 \]

#### Atomic Number Calculation:
- Atomic number on the left side (reactants):
\[ 13 + 2 = 15 \]
- Atomic number on the right side (products):
\[ 14 + \text{atomic number of } X \]
- Setting the two sides equal:
\[ 15 = 14 + \text{atomic number of } X \]
- Solving for the atomic number of \(X\):
\[ \text{atomic number of } X = 15 - 14 = 1 \]

Since \(X\) has a mass number of 1 and an atomic number of 1, it is identified as a proton (\(p\)).

### Step 2: Analyze the second reaction
The second reaction is:
\[ _{13}^{27}\text{Al} + \alpha \rightarrow _{15}^{30}\text{P} + Y \]

#### Mass Number Calculation:
- Mass number on the left side (reactants):
\[ 27 + 4 = 31 \]
- Mass number on the right side (products):
\[ 30 + \text{mass number of } Y \]
- Setting the two sides equal:
\[ 31 = 30 + \text{mass number of } Y \]
- Solving for the mass number of \(Y\):
\[ \text{mass number of } Y = 31 - 30 = 1 \]

#### Atomic Number Calculation:
- Atomic number on the left side (reactants):
\[ 13 + 2 = 15 \]
- Atomic number on the right side (products):
\[ 15 + \text{atomic number of } Y \]
- Setting the two sides equal:
\[ 15 = 15 + \text{atomic number of } Y \]
- Solving for the atomic number of \(Y\):
\[ \text{atomic number of } Y = 15 - 15 = 0 \]

Since \(Y\) has a mass number of 1 and an atomic number of 0, it is identified as a neutron (\(n\)).

### Step 3: Analyze the third reaction
The third reaction is:
\[ _{15}^{30}\text{P} \rightarrow _{14}^{30}\text{Si} + Z \]

#### Mass Number Calculation:
- Mass number on the left side (reactants):
\[ 30 \]
- Mass number on the right side (products):
\[ 30 + \text{mass number of } Z \]
- Setting the two sides equal:
\[ 30 = 30 + \text{mass number of } Z \]
- Solving for the mass number of \(Z\):
\[ \text{mass number of } Z = 30 - 30 = 0 \]

#### Atomic Number Calculation:
- Atomic number on the left side (reactants):
\[ 15 \]
- Atomic number on the right side (products):
\[ 14 + \text{atomic number of } Z \]
- Setting the two sides equal:
\[ 15 = 14 + \text{atomic number of } Z \]
- Solving for the atomic number of \(Z\):
\[ \text{atomic number of } Z = 15 - 14 = 1 \]

Since \(Z\) has a mass number of 0 and an atomic number of 1, it is identified as a positron (\(e^+\)).

### Summary of Products
- \(X\) is a proton (\(p\))
- \(Y\) is a neutron (\(n\))
- \(Z\) is a positron (\(e^+\))