The radioactive isotope `CO^(60)` which is used in the treatment of cancer can be made by (n.p) reaction for this reaction the target nucleus is

Options

A. `_28Ni^(59)`
B. `_27Co^(59)`
C. `_28Ni^(60)`
D. `_27Co^(60)`

Explanation

To determine the target nucleus used to produce the radioactive isotope \( \text{Co}^{60} \) through a neutron-proton reaction, we can follow these steps:

### Step 1: Understand the Isotope
Cobalt-60 (\( \text{Co}^{60} \)) is a radioactive isotope of cobalt that is commonly used in cancer treatment.

### Step 2: Identify the Reaction
The question specifies that the isotope is produced via a neutron-proton reaction. In nuclear reactions, a neutron is typically bombarded onto a target nucleus.

### Step 3: Determine the Target Nucleus
To produce \( \text{Co}^{60} \), we need to identify which nucleus can transform into cobalt-60 when a neutron is added.

### Step 4: Analyze the Transformation
The atomic number of cobalt (\( \text{Co} \)) is 27. The atomic number of nickel (\( \text{Ni} \)) is 28. The reaction can be represented as follows:

\[ \text{Ni}^{60} + n \rightarrow \text{Co}^{60} + \text{p} \]

Here, \( n \) represents the neutron and \( p \) represents the proton emitted during the reaction.

### Step 5: Conclusion
Thus, the target nucleus for the production of \( \text{Co}^{60} \) in this neutron-proton reaction is \( \text{Ni}^{60} \).

### Final Answer
The target nucleus is \( \text{Ni}^{60} \).

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