In 1994 it was proposed (and eventually accepted) that element 106 be named seaborgium, \(\mathrm{Sg}\), in honor of Glenn \(\mathrm{T}\). Seaborg, discoverer of the transuranium elements. a. \({ }^{263} \mathrm{Sg}\) was produced by the bombardment of \({ }^{249} \mathrm{Cf}\) with a beam of \({ }^{18} \mathrm{O}\) nuclei. Complete and balance an equation for this reaction. b. \({ }^{263} \mathrm{Sg}\) decays by \(\alpha\) emission. What is the other product resulting from the \(\alpha\) decay of \({ }^{263} \mathrm{Sg}\) ?

For part (a), it is given that \({ }^{263} \mathrm{Sg}\) is produced by bombarding \({ }^{249}\mathrm{Cf}\) with a beam of \({ }^{18}\mathrm{O}\) nuclei. So the reactants are \({ }^{249}\mathrm{Cf}\) and \({ }^{18}\mathrm{O}\), and the product is \({ }^{263}\mathrm{Sg}\). Now we need to balance the equation and find any other product(s).

To balance a nuclear reaction, we need to ensure that the total mass numbers and atomic numbers of the reactants are equal to the sum of the mass numbers and atomic numbers of the products. For this reaction, we have the following equation: \({ }^{249}\mathrm{Cf} + { }^{18}\mathrm{O} \rightarrow { }^{263}\mathrm{Sg} + X\) We need to determine the mass number and atomic number of X. To balance the mass numbers, we have: \(249 + 18 = 263 + A_X\), where \(A_X\) is the mass number of X. Solving for \(A_X\), we get: \(A_X = (249 + 18) - 263 = 4\), Now, let's balance the atomic numbers. The atomic numbers of Cf, O, and Sg are 98, 8, and 106, respectively. Therefore, we have: \(98 + 8 = 106 + Z_X\), where \(Z_X\) is the atomic number of X. Solving for \(Z_X\), we get: \(Z_X = (98 + 8) - 106 = 0\), So, the unknown particle X is a neutron with a mass number of 4. Thus, the balanced nuclear equation is: \({ }^{249}\mathrm{Cf} + { }^{18}\mathrm{O} \rightarrow { }^{263}\mathrm{Sg} + 4{ }^{1}\mathrm{n}\)

For part (b), we're given that \({ }^{263} \mathrm{Sg}\) decays via alpha emission, and we need to find the other product resulting from this decay. An alpha particle has a mass number of 4 and an atomic number of 2, and is represented as \({ }^{4}\mathrm{He}\). The decay is represented by the following equation: \({ }^{263}\mathrm{Sg} \rightarrow { }^{4}\mathrm{He} + Y\) Again, we need to balance the mass numbers and atomic numbers. To balance the mass numbers, we have: \(263 = 4 + A_Y\), where \(A_Y\) is the mass number of Y. Solving for \(A_Y\), we get: \(A_Y = 263 - 4 = 259\) Now, let's balance the atomic numbers: \(106 = 2 + Z_Y\), where \(Z_Y\) is the atomic number of Y. Solving for \(Z_Y\), we get: \(Z_Y = 106 - 2 = 104\) As a result, the other product Y is \({ }^{259}\mathrm{Rf}\), since Rf (Rutherfordium) has an atomic number of 104. The alpha decay equation for \({ }^{263}\mathrm{Sg}\) is: \({ }^{263}\mathrm{Sg} \rightarrow { }^{4}\mathrm{He} + { }^{259}\mathrm{Rf}\)