In the equation provided, you have a cobalt-60 (\(^{60}_{27}Co\)) nucleus undergoing beta decay (positron emission). The reaction can be represented as follows:
\[ ^{60}{27}Co \rightarrow ^{60}{26}Ni + \beta^+ + \nu_e \]
In beta decay, a proton is transformed into a neutron and a positron (\(\beta^+\)) is emitted along with a neutrino (\(\nu_e\)). When cobalt-60 emits a positron, its atomic number decreases by 1, converting it into nickel-60 (\(^{60}_{28}Ni\)).
So, the daughter nuclide formed in this beta decay reaction is:
Nickel - 60 (\(^{60}_{28}Ni\)).
Therefore, among the options provided, the correct response is:
- Nickel - 60.