In 1994 it was proposed (and eventually accepted) that element 106 be named seaborgium, Sg, in honor of Glenn T. Seaborg, discoverer of the transuranium elements. a. \(^{263}\) 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. \(^{263}\) g decays by \(\alpha\) emission. What is the other product resulting from the \(\alpha\) decay of \(^{263} \mathrm{Sg}\) ?

In this reaction, we're given that Californium-249 (\(^{249}\mathrm{Cf}\)) reacts with Oxygen-18 (\(^{18}\mathrm{O}\)) to produce Seaborgium-263 (\(^{263}\mathrm{Sg}\)). We need to find the balanced equation for this reaction.

In a balanced nuclear reaction, the total number of protons and neutrons before the reaction (on the left side) should equal the total number of protons and neutrons after the reaction (on the right side). For the given reaction: \(^{249}\mathrm{Cf} + ^{18}\mathrm{O} \longrightarrow ^{263}\mathrm{Sg} + X\) We know that Californium has 98 protons, Oxygen has 8 protons, and Seaborgium has 106 protons. Now we can calculate the number of neutrons in the missing product X, which can be a nucleus that results from the reaction. The total number of protons and neutrons before the reaction: \(249 + 18\) = 267 The total number of protons and neutrons after the reaction (assuming one Seaborgium nucleus and one X nucleus): \(263 + A_X\) To balance the reaction, the sum of A values should be equal on both sides: \(267 = 263 + A_X\) Solving for \(A_X\): \(A_X = 267 - 263\) \(A_X = 4\) Since X has 4 nucleons, and an alpha particle consists of 2 protons and 2 neutrons, the missing product X in the reaction is an alpha particle (\(\alpha\)). Now we can write the balanced nuclear reaction equation: \(^{249}\mathrm{Cf} + ^{18}\mathrm{O} \longrightarrow ^{263}\mathrm{Sg} + ^4\alpha\) For Part b:

We are given that Seaborgium-263 (\(^{263}\mathrm{Sg}\)) decays by alpha emission.

During alpha decay, the decaying nucleus loses 2 protons and 2 neutrons as an alpha particle (\(^4\alpha\)). Let's denote the mass number of the other product as \(A_Y\) and its atomic number (number of protons) as \(Z_Y\). Since Seaborgium-263 has a mass number of 263 and an atomic number of 106, after losing an alpha particle: \(A_Y = A_{Sg} - A_\alpha = 263 - 4 = 259\) \(Z_Y = Z_{Sg} - Z_\alpha = 106 - 2 = 104\) The other product resulting from the alpha decay of \(^{263}\mathrm{Sg}\) is a nucleus with a mass number of 259 and an atomic number of 104, which is Rutherfordium, denoted as \(^{259}\mathrm{Rf}\).