When a deuterium is bombarded on `._(8)O^(16)` nucleus, an `alpha`-particle is emitted, then the product nucleus is

Options

A. `._(7)N^(13)`
B. `._(5)B^(10)`
C. `._(4)Be^(9)`
D. `._(7)N^(14)`

Answer

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

### Step 1: Identify the initial and final particles
- The initial particle is the oxygen-16 nucleus, denoted as \( _{8}^{16}O \).
- The deuterium nucleus is denoted as \( _{1}^{2}H \).
- The emitted alpha particle is denoted as \( _{2}^{4}\alpha \) or simply \( _{2}^{4}He \).

### Step 2: Write the nuclear reaction
The nuclear reaction can be represented as:
\[ _{1}^{2}H + _{8}^{16}O \rightarrow _{2}^{4}He + X \]
where \( X \) is the unknown product nucleus we need to determine.

### Step 3: Apply conservation of mass number
The mass number (A) is conserved in nuclear reactions. Therefore, we can set up the equation:
\[ A_{\text{initial}} = A_{\text{final}} \]
Calculating the mass numbers:
- Initial mass number: \( 2 + 16 = 18 \)
- Final mass number: \( 4 + A_X \) (where \( A_X \) is the mass number of the unknown nucleus)

Setting them equal gives:
\[ 18 = 4 + A_X \]
Solving for \( A_X \):
\[ A_X = 18 - 4 = 14 \]

### Step 4: Apply conservation of atomic number
Next, we apply the conservation of atomic number (Z):
\[ Z_{\text{initial}} = Z_{\text{final}} \]
Calculating the atomic numbers:
- Initial atomic number: \( 1 + 8 = 9 \)
- Final atomic number: \( 2 + Z_X \) (where \( Z_X \) is the atomic number of the unknown nucleus)

Setting them equal gives:
\[ 9 = 2 + Z_X \]
Solving for \( Z_X \):
\[ Z_X = 9 - 2 = 7 \]

### Step 5: Identify the product nucleus
Now we have determined that the unknown nucleus has a mass number \( A_X = 14 \) and an atomic number \( Z_X = 7 \). The element with atomic number 7 is nitrogen (N).

Thus, the product nucleus is:
\[ _{7}^{14}N \]

### Final Answer
The product nucleus formed when deuterium is bombarded on an oxygen-16 nucleus, emitting an alpha particle, is \( _{7}^{14}N \) (Nitrogen-14).

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