`""_(27)Co^(60)` is radioactive because

Options

A. its atomic number is high
B. it has high `p//n` ratio
C. it has high `n//p` ratio
D. none of these

Answer

To determine why the isotope \( _{27}^{60}\text{Co} \) (Cobalt-60) is radioactive, we can follow these steps:

### Step 1: Identify the Components of the Isotope
- The atomic number (Z) of cobalt is 27, which indicates the number of protons in the nucleus.
- The atomic mass (A) is 60, which is the total number of protons and neutrons in the nucleus.

### Step 2: Calculate the Number of Neutrons
- The number of neutrons (N) can be calculated using the formula:
\[ N = A - Z \]
Substituting the values:
\[ N = 60 - 27 = 33 \]

### Step 3: Determine the Number of Protons
- The number of protons (P) is equal to the atomic number:
\[ P = Z = 27 \]

### Step 4: Calculate the Neutron-to-Proton Ratio
- The neutron-to-proton ratio (N/P) is calculated as follows:
\[ \frac{N}{P} = \frac{33}{27} \approx 1.22 \]

### Step 5: Analyze the Stability Criteria
- For a stable nucleus, the neutron-to-proton ratio should be approximately 1 or slightly greater for lighter elements. As the ratio we calculated (1.22) is greater than 1, this indicates instability.

### Step 6: Conclusion
- Since the neutron-to-proton ratio is greater than 1, \( _{27}^{60}\text{Co} \) is considered radioactive. This is because nuclei with a high neutron-to-proton ratio tend to undergo radioactive decay to achieve stability.

### Final Answer
Cobalt-60 is radioactive because its neutron-to-proton ratio (1.22) is greater than 1, indicating instability.

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