Write balanced equations for (a) ${ }_{92}^{238} \mathrm{U}(\alpha, \mathrm{n}){ }^{241} \mathrm{Pu},$ (b) ${ }^{14} \mathrm{~N}(\alpha, \mathrm{p}){ }^{17} \mathrm{O},(\mathbf{c}){ }_{26}^{56} \mathrm{Fe}\left(\alpha, \beta^{-}\right)_{29}^{60} \mathrm{Cu} .$

We start with the Uranium-238 nucleus and the alpha particle: \(_{92}^{238} U + _{2}^{4} \alpha\) After the alpha particle is absorbed, a neutron is emitted: \(_{92}^{238} U + _{2}^{4} \alpha \rightarrow_{94}^{241} Pu + _{0}^{1} n\) Hence, the balanced equation for the reaction is: \(_{92}^{238} U + _{2}^{4} \alpha \rightarrow_{94}^{241} Pu + _{0}^{1} n\) (b)

We start with the Nitrogen-14 nucleus and the alpha particle: \(_{7}^{14} N + _{2}^{4} \alpha\) After the alpha particle is absorbed, a proton is emitted: \(_{7}^{14} N + _{2}^{4} \alpha \rightarrow_{8}^{17} O + _{1}^{1} p\) Hence, the balanced equation for the reaction is: \(_{7}^{14} N + _{2}^{4} \alpha \rightarrow_{8}^{17} O + _{1}^{1} p\) (c)

We start with the Iron-56 nucleus and the alpha particle: \(_{26}^{56} Fe + _{2}^{4} \alpha\) After the alpha particle is absorbed, a beta-minus particle is emitted: \(_{26}^{56} Fe + _{2}^{4} \alpha \rightarrow_{28}^{60} Ni + _{0}^{-1} \beta^{-}\) Finally, the beta-minus decay takes place, converting a neutron in the Nickel-60 nucleus into a proton and emitting an electron, forming Copper-60: \(_{28}^{60} Ni \rightarrow_{29}^{60} Cu + _{0}^{-1} \beta^{-}\) So, combining these reactions we get the overall equation for the reaction: \(_{26}^{56} Fe + _{2}^{4} \alpha \rightarrow_{28}^{60} Ni + _{0}^{-1} \beta^{-} \rightarrow_{29}^{60} Cu + _{0}^{-1} \beta^{-}\) Balancing these equations ensures that mass and atomic numbers are conserved throughout each nuclear reaction.