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Step-by-Step Solution
Step 1: Identify the Desired Product
The problem states that we want to produce the radioactive isotope $\,_{27}^{60}\text{Co}\,$, which is used in cancer treatment. This is formed by an $(n, p)$ reaction, meaning that a neutron enters the target nucleus and a proton is emitted.
Step 2: Recall What an $(n, p)$ Reaction Entails
In an $(n, p)$ nuclear reaction, the nucleus absorbs a neutron $(\,_{0}^{1}\text{n}\,)$ and emits a proton $(\,_{1}^{1}\text{p}\,)$, effectively decreasing its atomic number by one (because a neutron is gained and a proton is lost), but keeping the mass number the same (one nucleon goes in, one nucleon goes out).
Step 3: Determine the Target Nucleus
Because we end with $\,_{27}^{60}\text{Co}\,$ after an $(n, p)$ process, the original isotope must have an atomic number higher by 1 (i.e., 28) with the same mass number (i.e., 60). This identifies the target nucleus as $\,_{28}^{60}\text{Ni}\,$ (Nickel-60).
Step 4: Write the Nuclear Reaction
The nuclear reaction can be expressed as:
$\,_{28}^{60}\text{Ni} + \,_{0}^{1}\text{n} \;\to\; \,_{27}^{60}\text{Co} + \,_{1}^{1}\text{p}\,$
Your original information showed it as emitting another neutron, but for an $(n, p)$ reaction specifically, a proton is the emitted particle. The crucial idea is that Nickel-60 effectively loses a proton and converts into Cobalt-60.
Step 5: Conclusion
The correct target nucleus for producing $\,_{27}^{60}\text{Co}\,$ by an $(n, p)$ reaction is $\,_{28}^{60}\text{Ni}\,$.
