In the nuclear raction given by `._2He^4 +._7N^(14) rarr ._1H^1 +X` the nucleus X is

Options

A. Nitrogen of mass 16 .
B. Nitrogen of mass 17
C. Oxygen of mass 16
D. Oxygen of mass 17

Answer

To determine the nucleus X in the nuclear reaction given by:

\[ _2He^4 + _7N^{14} \rightarrow _1H^1 + X \]

we will follow these steps:

### Step 1: Write down the atomic and mass numbers of the reactants and products.
- For the reactants:
- Helium-4 (\( _2He^4 \)): Atomic number (Z) = 2, Mass number (A) = 4
- Nitrogen-14 (\( _7N^{14} \)): Atomic number (Z) = 7, Mass number (A) = 14

- For the products:
- Hydrogen-1 (\( _1H^1 \)): Atomic number (Z) = 1, Mass number (A) = 1
- Nucleus X: Atomic number (Z) = B, Mass number (A) = A

### Step 2: Balance the mass numbers.
The total mass number on the left side (reactants) is:
\[ 4 + 14 = 18 \]

On the right side, we have:
\[ 1 + A \]

Setting these equal gives:
\[ 18 = 1 + A \]
\[ A = 18 - 1 = 17 \]

### Step 3: Balance the atomic numbers.
The total atomic number on the left side (reactants) is:
\[ 2 + 7 = 9 \]

On the right side, we have:
\[ 1 + B \]

Setting these equal gives:
\[ 9 = 1 + B \]
\[ B = 9 - 1 = 8 \]

### Step 4: Identify the nucleus X.
Now we know:
- Mass number (A) = 17
- Atomic number (Z) = 8

From the periodic table, the element with atomic number 8 is Oxygen (O). Therefore, nucleus X is:

\[ X = _8O^{17} \]

### Final Answer:
The nucleus X is Oxygen-17 (\( _8O^{17} \)).

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