What is X in the following nuclear reaction? `N_7^14 + H_1^1 to O_8^15 +X`

Options

A. `e_(+1)^@`
B. `n_0^1`
C. `gamma`
D. `e_(-1)^@`

Answer

To determine what X is in the nuclear reaction \( N_7^{14} + H_1^1 \to O_8^{15} + X \), we will follow these steps:

### Step 1: Write down the reaction and identify the components
The nuclear reaction given is:
\[ N_7^{14} + H_1^1 \to O_8^{15} + X \]
Here, we have:
- Nitrogen (\( N \)) with atomic number 7 and mass number 14.
- Hydrogen (\( H \)) with atomic number 1 and mass number 1.
- Oxygen (\( O \)) with atomic number 8 and mass number 15.
- We need to find \( X \).

### Step 2: Apply conservation of atomic mass
In nuclear reactions, the total atomic mass on the reactant side must equal the total atomic mass on the product side.

Calculating the total mass on the reactant side:
\[ 14 \, (N) + 1 \, (H) = 15 \]

Calculating the total mass on the product side:
\[ 15 \, (O) + A \, (X) \]

Setting the two sides equal:
\[ 15 = 15 + A \]
This simplifies to:
\[ A = 0 \]

### Step 3: Apply conservation of atomic number
Next, we apply the conservation of atomic number.

Calculating the total atomic number on the reactant side:
\[ 7 \, (N) + 1 \, (H) = 8 \]

Calculating the total atomic number on the product side:
\[ 8 \, (O) + Z \, (X) \]

Setting the two sides equal:
\[ 8 = 8 + Z \]
This simplifies to:
\[ Z = 0 \]

### Step 4: Identify the particle X
From the calculations, we found that:
- The atomic mass \( A \) of \( X \) is 0.
- The atomic number \( Z \) of \( X \) is 0.

A particle with an atomic mass of 0 and an atomic number of 0 is a gamma particle (\( \gamma \)), which is a form of electromagnetic radiation.

### Conclusion
Thus, the particle \( X \) in the nuclear reaction is a gamma particle.