The bombarment of `alpha` - particle on `""_7N^14`, emits proton then new atom will be :

Options

A. `""_8O^17`
B. `""_8O^16`
C. `""_6C^14`
D. Ne

Answer

To solve the problem of determining the new atom formed when an alpha particle bombards nitrogen-14 (\( _7N^{14} \)) and emits a proton, we can follow these steps:

### Step 1: Write the reaction equation
When an alpha particle (\( _2He^4 \)) bombards nitrogen-14 (\( _7N^{14} \)), the reaction can be represented as:
\[ _7N^{14} + _2He^4 \rightarrow X + _1H^1 \]
Here, \( X \) is the new atom formed, and \( _1H^1 \) represents the emitted proton.

### Step 2: Determine the total atomic number and mass number on the reactant side
- The atomic number of nitrogen (\( _7N^{14} \)) is 7.
- The atomic number of the alpha particle (\( _2He^4 \)) is 2.
- Therefore, the total atomic number on the reactant side is:
\[ 7 + 2 = 9 \]
- The mass number of nitrogen is 14, and the mass number of the alpha particle is 4. Thus, the total mass number on the reactant side is:
\[ 14 + 4 = 18 \]

### Step 3: Set up the equations for the product side
Let the atomic number of the new atom \( X \) be \( Z \) and its mass number be \( A \). The emitted proton has an atomic number of 1 and a mass number of 1. Therefore, we can write:
- For atomic numbers:
\[ Z + 1 = 9 \implies Z = 8 \]
- For mass numbers:
\[ A + 1 = 18 \implies A = 17 \]

### Step 4: Identify the new atom
The new atom \( X \) has an atomic number of 8 and a mass number of 17. This corresponds to oxygen, which is represented as:
\[ _8O^{17} \]

### Conclusion
The new atom formed after the bombardment of an alpha particle on nitrogen-14 that emits a proton is:
\[ \boxed{_8O^{17}} \]