On bombarding `N_7^(14)` with `alpha`- particles, the nuclei of the product formed after the release of a proton will be or In nuclear reaction `N_7^14 + He_2^4 to X_z^A + H_1^1`, the term `X_z^A` represents :

Options

A. `O_8^17`
B. `F_9^18`
C. `F_9^17`
D. `O_8^18`

Answer

To solve the problem, we need to analyze the nuclear reaction given:

\[ N_7^{14} + He_2^{4} \rightarrow X_z^{A} + H_1^{1} \]

### Step 1: Identify the components of the reaction
- **Reactants**: Nitrogen-14 (\(N_7^{14}\)) and Helium-4 (\(He_2^{4}\)).
- **Products**: An unknown element \(X_z^{A}\) and a proton (\(H_1^{1}\)).

### Step 2: Write down the conservation laws
In nuclear reactions, both the atomic number (number of protons) and the mass number (total number of protons and neutrons) must be conserved.

### Step 3: Calculate the total atomic number and mass number of reactants
- **Atomic number of reactants**:
- Nitrogen (\(N\)): 7
- Helium (\(He\)): 2
- Total: \(7 + 2 = 9\)

- **Mass number of reactants**:
- Nitrogen: 14
- Helium: 4
- Total: \(14 + 4 = 18\)

### Step 4: Set up equations for the products
Let \(Z\) be the atomic number and \(A\) be the mass number of the unknown element \(X\).

- From the atomic number conservation:
\[ 9 = Z + 1 \quad \text{(since one proton is released)} \]
Solving for \(Z\):
\[ Z = 9 - 1 = 8 \]

- From the mass number conservation:
\[ 18 = A + 1 \quad \text{(since one proton is released)} \]
Solving for \(A\):
\[ A = 18 - 1 = 17 \]

### Step 5: Identify the element \(X\)
Now we have:
- Atomic number \(Z = 8\)
- Mass number \(A = 17\)

The element with atomic number 8 is Oxygen (O).

### Final Answer
Thus, the product formed after the release of a proton when \(N_7^{14}\) is bombarded with \(\alpha\)-particles is:
\[ X_8^{17} \text{ (Oxygen-17)} \]