In the reaction ltBrgt `""_(7)N^(14) + ""_(2)He^(4) to ""_(a)X^(b) +""_(1)H^(1)` the nucleus X is

Options

A. N-16
B. N-17
C. O-16
D. O-17

Explanation

To solve the nuclear reaction given in the question, we need to identify the nucleus \( X \) in the reaction:

\[ _{7}^{14}N + _{2}^{4}He \rightarrow _{a}^{b}X + _{1}^{1}H \]

### Step 1: Write down the known values
- For nitrogen (\( N \)): Atomic number = 7, Mass number = 14
- For helium (\( He \)): Atomic number = 2, Mass number = 4
- For hydrogen (\( H \)): Atomic number = 1, Mass number = 1

### Step 2: Calculate the total mass number and atomic number on the left side
- Total mass number on the left side = \( 14 + 4 = 18 \)
- Total atomic number on the left side = \( 7 + 2 = 9 \)

### Step 3: Set up equations for the right side
Let the mass number of nucleus \( X \) be \( b \) and its atomic number be \( a \).

From the reaction, we can write two equations based on the conservation of mass number and atomic number:

1. For mass numbers:
\[ b + 1 = 18 \quad \text{(since the mass number of } H \text{ is 1)} \]
Therefore,
\[ b = 18 - 1 = 17 \]

2. For atomic numbers:
\[ a + 1 = 9 \quad \text{(since the atomic number of } H \text{ is 1)} \]
Therefore,
\[ a = 9 - 1 = 8 \]

### Step 4: Identify the nucleus \( X \)
Now we have \( a = 8 \) and \( b = 17 \). The nucleus \( X \) can be identified as:
\[ _{8}^{17}O \]
This indicates that nucleus \( X \) is an isotope of oxygen.

### Final Answer
Thus, the nucleus \( X \) in the reaction is \( _{8}^{17}O \) (Oxygen-17).

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