An ore of uranium is found to contain _92U^238 and _82Pb^206 in the mass ratio of 1:0.1. The half-li — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Let mass of U^238 = 1 g, Pb^206 = 0.1 g. Moles N_U = 1/238 = 0.004202 mol; N_Pb = 0.1/206 = 0.000485 mol. Initial moles of U^238 is N_0 = N_U + N_Pb = 0.004687 mol. Ratio N_0/N = 0.004687 / 0.004202 = 1.1154. Using first-order rate equation: t = (t_1/2 / log 2) * log(N_0/N) = (4.5 * 10^9 / 0.3) * log(1.1154) = (1.5 * 10^10) * 0.048 = 7.2 * 10^8 years.