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

Explanation

To solve the nuclear reaction given by \( _2He^4 + _7N^{14} \rightarrow _1H^1 + X \), we need to identify the nucleus \( X \).

### Step 1: Conservation of Mass Number
The first step is to conserve the mass number (A). The mass number is the total number of protons and neutrons in the nucleus.

- On the left side (reactants):
- Mass number of \( _2He^4 \) = 4
- Mass number of \( _7N^{14} \) = 14
- Total mass number on the left = \( 4 + 14 = 18 \)

- On the right side (products):
- Mass number of \( _1H^1 \) = 1
- Let the mass number of \( X \) be \( A_X \).
- Total mass number on the right = \( 1 + A_X \)

Setting the total mass numbers equal gives us:
\[ 18 = 1 + A_X \]
Solving for \( A_X \):
\[ A_X = 18 - 1 = 17 \]

### Step 2: Conservation of Atomic Number
Next, we conserve the atomic number (Z). The atomic number is the number of protons in the nucleus.

- On the left side (reactants):
- Atomic number of \( _2He^4 \) = 2
- Atomic number of \( _7N^{14} \) = 7
- Total atomic number on the left = \( 2 + 7 = 9 \)

- On the right side (products):
- Atomic number of \( _1H^1 \) = 1
- Let the atomic number of \( X \) be \( Z_X \).
- Total atomic number on the right = \( 1 + Z_X \)

Setting the total atomic numbers equal gives us:
\[ 9 = 1 + Z_X \]
Solving for \( Z_X \):
\[ Z_X = 9 - 1 = 8 \]

### Step 3: Identifying the Element
Now we have determined:
- Mass number \( A_X = 17 \)
- Atomic number \( Z_X = 8 \)

From the periodic table, the element with atomic number 8 is oxygen (O). Therefore, the nucleus \( X \) is \( _8O^{17} \).

### Final Answer
The nucleus \( X \) is \( _8O^{17} \) (oxygen of mass 17).

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