`._6^12 C` absorbs an energenic neutron and emits beta particles. The resulting nucleus is.

Options

A. `._(7)N^(14)`
B. `._(7)N^(13)`
C. `._(5)B^(13)`
D. `._(6)C^(13)`

Answer

To solve the problem, we will analyze the nuclear reaction step by step.

### Step 1: Identify the initial nucleus
The given nucleus is Carbon-12, represented as \( _6^{12}C \). Here, the atomic number (Z) is 6, and the mass number (A) is 12.

### Step 2: Absorption of a neutron
When Carbon-12 absorbs an energetic neutron, the reaction can be represented as follows:
\[ _6^{12}C + _0^{1}n \rightarrow _6^{13}C \]
After absorption, the mass number increases by 1 (from 12 to 13), while the atomic number remains the same (still 6). Therefore, we have:
- New mass number (A) = 13
- New atomic number (Z) = 6

### Step 3: Emission of a beta particle
Next, the nucleus emits a beta particle. A beta particle (beta minus, \( \beta^- \)) is represented as \( _{-1}^{0}e \). The emission of a beta particle results in the following changes:
- The atomic number (Z) increases by 1 (since a neutron is converted into a proton).
- The mass number (A) remains unchanged.

So, the reaction can be represented as:
\[ _6^{13}C \rightarrow _7^{13}N + _{-1}^{0}e \]
After the emission of the beta particle:
- New mass number (A) = 13
- New atomic number (Z) = 7

### Step 4: Identify the resulting nucleus
The resulting nucleus with an atomic number of 7 and a mass number of 13 corresponds to Nitrogen-13, represented as \( _7^{13}N \).

### Conclusion
Thus, the resulting nucleus after the absorption of a neutron and the emission of a beta particle is Nitrogen-13.

### Final Answer
The resulting nucleus is \( _7^{13}N \) (Nitrogen-13).

---