`._(13)Al^(27)` is a stable isotope. `._(13)Al^(29)` is expected to disintegrate by

Options

A. `alph-` emission
B. `beta-` emission
C. Positron emission
D. Proton emission

Explanation

To determine how the isotope aluminum-29 (\( _{13}^{29}\text{Al} \)) disintegrates, we will analyze its neutron-to-proton (N/P) ratio and identify the appropriate decay process.

### Step-by-Step Solution:

1. **Identify the Isotope Information**:
- The mass number of aluminum-29 is 29.
- The atomic number of aluminum is 13, which indicates it has 13 protons.

2. **Calculate the Number of Neutrons**:
- The number of neutrons (N) can be calculated using the formula:
\[ \text{Number of Neutrons} = \text{Mass Number} - \text{Atomic Number} \]
- For aluminum-29:
\[ N = 29 - 13 = 16 \]

3. **Calculate the Neutron-to-Proton Ratio (N/P)**:
- The neutron-to-proton ratio is given by:
\[ \text{N/P} = \frac{\text{Number of Neutrons}}{\text{Number of Protons}} = \frac{N}{P} \]
- For aluminum-29:
\[ \text{N/P} = \frac{16}{13} \approx 1.23 \]

4. **Analyze Stability**:
- A stable nucleus typically has an N/P ratio close to 1 for lighter elements. Since the N/P ratio of aluminum-29 is greater than 1 (1.23), it indicates that the isotope is unstable.

5. **Determine the Decay Process**:
- To achieve stability, the N/P ratio must decrease. This can occur through the conversion of a neutron into a proton.
- This process is characteristic of **beta emission**, where a neutron decays into a proton and emits a beta particle (an electron).

6. **Conclusion**:
- Therefore, aluminum-29 is expected to disintegrate by **beta emission**.

### Final Answer:
Aluminum-29 (\( _{13}^{29}\text{Al} \)) is expected to disintegrate by **beta emission**.

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