The reaction `""_(7)N^(14) + ""_(0)n^(1) rarr ""_(6)C^(14)+X`
requires emission of

Options

A. `""_(2)^(4)He`
B. `""_(1)^(1)H`
C. `""_(0)^(1)n`
D. `""_(-1)e^(0)`

Explanation

To solve the reaction \( _{7}^{14}N + _{0}^{1}n \rightarrow _{6}^{14}C + X \), we need to determine what particle \( X \) is emitted during this nuclear reaction.

### Step 1: Identify the initial and final particles
- The initial particle is nitrogen-14 (\( _{7}^{14}N \)) and a neutron (\( _{0}^{1}n \)).
- The final particle is carbon-14 (\( _{6}^{14}C \)) and an unknown particle \( X \).

### Step 2: Write down the conservation laws
In nuclear reactions, both the mass number (A) and atomic number (Z) must be conserved.

- **Mass number conservation**:
\[ A_{initial} = A_{final} \]
For our reaction:
\[ 14 + 1 = 14 + A_X \implies 15 = 14 + A_X \implies A_X = 1 \]

- **Atomic number conservation**:
\[ Z_{initial} = Z_{final} \]
For our reaction:
\[ 7 + 0 = 6 + Z_X \implies 7 = 6 + Z_X \implies Z_X = 1 \]

### Step 3: Identify particle \( X \)
From the above calculations, we found:
- The mass number \( A_X = 1 \)
- The atomic number \( Z_X = 1 \)

The particle with mass number 1 and atomic number 1 is a proton (\( _{1}^{1}H \)).

### Conclusion
Thus, the emitted particle \( X \) is a proton.

### Final Answer
The reaction requires the emission of a proton (\( _{1}^{1}H \)).

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