A deutron is bombarded on `._8 O^16` nucleus and `alpha-`particle is emitted. The product nucleus is.

Options

A. `._7 N^13`
B. `._5 B^10`
C. `._4Bc^9`
D. `._7 N^14`

Explanation

To solve the problem of identifying the product nucleus when a deuteron is bombarded on an oxygen-16 nucleus resulting in the emission of an alpha particle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants and Products:**
- The reactants are:
- Oxygen-16 nucleus: \( _8^{16}O \)
- Deuteron: \( _1^{2}H \) (which can also be written as \( D \))
- The product of the reaction includes:
- An alpha particle: \( _2^{4}He \)
- A product nucleus, which we will denote as \( _Z^{A}X \).

2. **Write the Nuclear Reaction:**
The reaction can be written as:
\[ _8^{16}O + _1^{2}H \rightarrow _2^{4}He + _Z^{A}X \]

3. **Apply Conservation of Mass Number (A):**
The mass number (A) must be conserved in the reaction:
\[ 16 + 2 = 4 + A \]
Simplifying this gives:
\[ 18 = 4 + A \implies A = 18 - 4 = 14 \]

4. **Apply Conservation of Atomic Number (Z):**
The atomic number (Z) must also be conserved:
\[ 8 + 1 = 2 + Z \]
Simplifying this gives:
\[ 9 = 2 + Z \implies Z = 9 - 2 = 7 \]

5. **Identify the Product Nucleus:**
From the values obtained:
- The atomic number \( Z = 7 \)
- The mass number \( A = 14 \)
Therefore, the product nucleus is \( _7^{14}N \) (Nitrogen-14).

### Final Answer:
The product nucleus formed when a deuteron is bombarded on an oxygen-16 nucleus, emitting an alpha particle, is \( _7^{14}N \).

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