Write balanced equations for (a) \({ }_{92}^{238} \mathrm{U}(\alpha, \mathrm{n}){ }_{94}^{241} \mathrm{Pu},\) (b) \({ }_{7}^{14} \mathrm{~N}(\alpha, \mathrm{p}){ }^{17} \mathrm{O}\) (c) \({ }_{26}^{56} \mathrm{Fe}(\alpha, \beta)_{29}^{60} \mathrm{Cu}\).

For \({}_{92}^{238}\mathrm{U}(\alpha, \mathrm{n}){}_{94}^{241}\mathrm{Pu}\), we need to balance the atomic numbers and mass numbers on both sides of the reaction. 1. Identify the alpha particle, which has an atomic number of 2 and a mass number of 4: \({}_{2}^{4}\mathrm{\alpha}\). 2. Identify the neutron, which has an atomic number of 0 and a mass number of 1: \({}_{0}^{1}\mathrm{n}\). 3. Write down the reaction with atomic and mass numbers: \({}_{92}^{238}\mathrm{U} + {}_{2}^{4}\mathrm{\alpha} \rightarrow {}_{94}^{241}\mathrm{Pu} + {}_{0}^{1}\mathrm{n}\). 4. Check if the atomic numbers and mass numbers are balanced: - Atomic numbers: 92 + 2 = 94 - Mass numbers: 238 + 4 = 241 + 1 The reaction is balanced.

For \({}_{7}^{14}\mathrm{N}(\alpha, \mathrm{p}){}^{17}\mathrm{O}\), we need to balance the atomic numbers and mass numbers on both sides of the reaction. 1. Identify the alpha particle, which has an atomic number of 2 and a mass number of 4: \({}_{2}^{4}\mathrm{\alpha}\). 2. Identify the proton, which has an atomic number of 1 and a mass number of 1: \({}_{1}^{1}\mathrm{p}\). 3. Write down the reaction with atomic and mass numbers: \({}_{7}^{14}\mathrm{N} + {}_{2}^{4}\mathrm{\alpha} \rightarrow {}_{8}^{17}\mathrm{O} + {}_{1}^{1}\mathrm{p}\). 4. Check if the atomic numbers and mass numbers are balanced: - Atomic numbers: 7 + 2 = 8 + 1 - Mass numbers: 14 + 4 = 17 + 1 The reaction is balanced.

For \({}_{26}^{56}\mathrm{Fe}(\alpha, \beta)_{29}^{60}\mathrm{Cu}\), we need to balance the atomic numbers and mass numbers on both sides of the reaction. 1. Identify the alpha particle, which has an atomic number of 2 and a mass number of 4: \({}_{2}^{4}\mathrm{\alpha}\). 2. Identify the beta particle (electron), which has an atomic number of -1 and a mass number of 0: \({}_{-1}^{0}\mathrm{\beta}\). 3. Write down the reaction with atomic and mass numbers: \({}_{26}^{56}\mathrm{Fe} + {}_{2}^{4}\mathrm{\alpha} \rightarrow {}_{29}^{60}\mathrm{Cu} + {}_{-1}^{0}\mathrm{\beta}\). 4. Check if the atomic numbers and mass numbers are balanced: - Atomic numbers: 26 + 2 = 29 - 1 - Mass numbers: 56 + 4 = 60 The reaction is balanced.