`._(7)N^(13)` changes to `._(6)C^(13)` by the emission of

Options

A. `_(8)^(13)O+`_(-1)^(0)e`
B. `_(6)^(13)C+`_(+1)^(0)e`
C. `_(6)^(13)C +K` electron capture
D. `_(5)^(9)Be+`_(2)^(4)He`

Explanation

To solve the problem of how \( _{7}^{13}N \) (Nitrogen-13) changes to \( _{6}^{13}C \) (Carbon-13) by the emission of a particle, we can follow these steps:

### Step 1: Identify the Reactants and Products
The given reaction is:
\[ _{7}^{13}N \rightarrow _{6}^{13}C + ? \]

Here, Nitrogen-13 is the reactant, and Carbon-13 is one of the products. We need to determine what particle is emitted during this transformation.

### Step 2: Analyze the Mass and Charge
- The mass number (A) of Nitrogen-13 is 13, and its atomic number (Z) is 7.
- The mass number of Carbon-13 is also 13, and its atomic number is 6.

### Step 3: Balance the Mass and Charge
To balance the equation, we need to ensure that both the mass numbers and the atomic numbers are conserved.

1. **Mass Number**:
- Left side: 13 (from Nitrogen)
- Right side: 13 (from Carbon) + mass of emitted particle
- Therefore, the mass of the emitted particle must be 0.

2. **Atomic Number**:
- Left side: 7 (from Nitrogen)
- Right side: 6 (from Carbon) + atomic number of emitted particle
- Therefore, the atomic number of the emitted particle must be +1 (since 7 - 6 = 1).

### Step 4: Identify the Emitted Particle
The particle that has a mass number of 0 and an atomic number of +1 is a positron (\( e^+ \)). This is because:
- A positron is the antimatter counterpart of an electron and has a charge of +1.

### Step 5: Write the Final Balanced Equation
Thus, the complete balanced equation is:
\[ _{7}^{13}N \rightarrow _{6}^{13}C + e^+ \]

### Conclusion
The emission that occurs is a positron emission.

### Final Answer
The answer is: \( _{6}^{13}C + e^+ \) (positron emission).

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