`(Ho)_(67)^(165)` is stable isotope. `(Ho)_(67)^(150)` is expected to distegrate by:

Options

A. `alpha`-emission
B. `beta`-emission
C. Positron emission
D. `gamma` -emission

Answer

To determine how the isotope \( \text{Ho}_{67}^{150} \) is expected to disintegrate, we will analyze the neutron-to-proton ratio and its implications for stability.

### Step-by-Step Solution:

1. **Identify the components of the isotope**:
- The isotope is \( \text{Ho}_{67}^{150} \).
- Here, the atomic number (number of protons) is 67, and the mass number (total number of protons and neutrons) is 150.

2. **Calculate the number of neutrons**:
- The number of neutrons can be calculated using the formula:
\[ \text{Number of Neutrons} = \text{Mass Number} - \text{Atomic Number} \]
- Substituting the values:
\[ \text{Number of Neutrons} = 150 - 67 = 83 \]

3. **Determine the neutron-to-proton ratio**:
- The neutron-to-proton ratio is given by:
\[ \text{Neutron-to-Proton Ratio} = \frac{\text{Number of Neutrons}}{\text{Number of Protons}} = \frac{83}{67} \]
- Calculating this ratio:
\[ \text{Neutron-to-Proton Ratio} \approx 1.24 \]

4. **Analyze the stability based on the neutron-to-proton ratio**:
- A neutron-to-proton ratio greater than 1 indicates that the isotope is neutron-rich.
- Neutron-rich isotopes tend to undergo beta decay, where a neutron is converted into a proton, emitting a beta particle (electron) and an antineutrino.

5. **Conclusion**:
- Since the neutron-to-proton ratio of \( \text{Ho}_{67}^{150} \) is greater than 1, it is expected to disintegrate by **beta decay**.

### Final Answer:
\( \text{Ho}_{67}^{150} \) is expected to disintegrate by **beta decay**.

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