Thorium-232 decays by the following sequence of emissions: \(\alpha, \beta, \beta, \alpha, \alpha, \alpha, \beta, \alpha, \beta, \alpha\). What is the final product of this sequence of emissions (give its full atomic symbol)?

First, let's write down the starting element, Thorium-232, in terms of its atomic number Z and its mass number A. Thorium: \(Z_0 = 90\), \(A_0 = 232\)

Now, we need to apply the given sequence of emissions and update the values of Z and A accordingly. The sequence of emissions: \(\alpha, \beta, \beta, \alpha, \alpha, \alpha, \beta, \alpha, \beta, \alpha\) Alpha emission 1 (\(\alpha_1\)): \(Z_1 = 88\), \(A_1 = 228\) Beta emission 1 (\(\beta_1\)): \(Z_2 = 89\), \(A_2 = 228\) Beta emission 2 (\(\beta_2\)): \(Z_3 = 90\), \(A_3 = 228\) Alpha emission 2 (\(\alpha_2\)): \(Z_4 = 88\), \(A_4 = 224\) Alpha emission 3 (\(\alpha_3\)): \(Z_5 = 86\), \(A_5 = 220\) Alpha emission 4 (\(\alpha_4\)): \(Z_6 = 84\), \(A_6 = 216\) Beta emission 3 (\(\beta_3\)): \(Z_7 = 85\), \(A_7 = 216\) Alpha emission 5 (\(\alpha_5\)): \(Z_8 = 83\), \(A_8 = 212\) Beta emission 4 (\(\beta_4\)): \(Z_9 = 84\), \(A_9 = 212\) Alpha emission 6 (\(\alpha_6\)): \(Z_{10} = 82\), \(A_{10} = 208\)

The final element has an atomic number \(Z_{10} = 82\) and a mass number \(A_{10} = 208\). This element is Lead-208 (Pb-208). Our final product is: Pb-208