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 of determining the product nucleus formed when \( N_7^{14} \) is bombarded with an alpha particle \( He_2^{4} \) and releases a proton \( H_1^{1} \), we can follow these steps:

### Step 1: Write the nuclear reaction
The given nuclear reaction can be expressed as:
\[ N_7^{14} + He_2^{4} \rightarrow X_z^{A} + H_1^{1} \]

### Step 2: Identify the atomic and mass numbers
- The atomic number (Z) of nitrogen \( N \) is 7, and its mass number (A) is 14.
- The atomic number (Z) of the alpha particle \( He \) is 2, and its mass number (A) is 4.
- The atomic number (Z) of the proton \( H \) is 1, and its mass number (A) is 1.

### Step 3: Apply conservation of atomic numbers
According to the conservation of atomic numbers in nuclear reactions:
\[ Z_{initial} = Z_{final} \]
The initial atomic numbers are:
\[ Z_{initial} = Z(N) + Z(He) = 7 + 2 = 9 \]
The final atomic numbers are:
\[ Z_{final} = Z(X) + Z(H) = Z + 1 \]
Setting these equal gives:
\[ 9 = Z + 1 \]
Thus, solving for Z:
\[ Z = 9 - 1 = 8 \]

### Step 4: Apply conservation of mass numbers
Now, applying conservation of mass numbers:
\[ A_{initial} = A_{final} \]
The initial mass numbers are:
\[ A_{initial} = A(N) + A(He) = 14 + 4 = 18 \]
The final mass numbers are:
\[ A_{final} = A(X) + A(H) = A + 1 \]
Setting these equal gives:
\[ 18 = A + 1 \]
Thus, solving for A:
\[ A = 18 - 1 = 17 \]

### Step 5: Identify the product nucleus
Now we have determined that:
- The atomic number \( Z \) of the product nucleus \( X \) is 8.
- The mass number \( A \) of the product nucleus \( X \) is 17.

The nucleus with atomic number 8 and mass number 17 is:
\[ X = O_8^{17} \]

### Final Answer
Thus, the product formed after the reaction is:
\[ \boxed{O_8^{17}} \]

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