`._(23)^(28)Al+_(1)^(1)P rarr X +_(0)^(0)gamma`, type artifical radioactive reaction.

Options

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

Answer

To solve the problem of the artificial radioactive reaction \( _{13}^{28}Al + _{1}^{1}p \rightarrow X + _{0}^{0}\gamma \), we will follow these steps:

### Step 1: Identify the Reactants
The reactants in the equation are:
- Aluminium-28 (\( _{13}^{28}Al \))
- Proton (\( _{1}^{1}p \))

### Step 2: Write Down the Mass and Atomic Numbers
For Aluminium-28:
- Mass number (A) = 28
- Atomic number (Z) = 13

For Proton:
- Mass number (A) = 1
- Atomic number (Z) = 1

### Step 3: Calculate Total Mass and Atomic Numbers of Reactants
Total mass number of reactants:
\[ 28 + 1 = 29 \]

Total atomic number of reactants:
\[ 13 + 1 = 14 \]

### Step 4: Write Down the Products
We have one product \( X \) and gamma radiation (\( _{0}^{0}\gamma \)), which does not affect the mass or atomic number.

### Step 5: Apply Conservation of Mass and Atomic Number
According to the law of conservation of mass and atomic number:
- Total mass number of products = Total mass number of reactants
- Total atomic number of products = Total atomic number of reactants

Let the mass number and atomic number of \( X \) be \( A_X \) and \( Z_X \) respectively.

From the conservation of mass:
\[ A_X = 29 \]

From the conservation of atomic number:
\[ Z_X = 14 \]

### Step 6: Identify Element \( X \)
Now, we need to identify the element with mass number 29 and atomic number 14. Referring to the periodic table:
- The element with atomic number 14 is Silicon (Si).

### Conclusion
Thus, the product \( X \) is \( _{14}^{29}Si \).

### Final Answer
The complete reaction is:
\[ _{13}^{28}Al + _{1}^{1}p \rightarrow _{14}^{29}Si + _{0}^{0}\gamma \]

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