²³⁸ U decays to ²⁰⁶ Pb with a half-life of 4.47 10⁹ y through a chain of processes. A rock sample is discovered to contain 2.00 mg of ²³⁸ U and 0.600 mg of ²⁰⁶ Pb . Assuming that all the lead originated from the uranium, determine the life of the rock.

Step-by-step solution

Let N be the present number of ²³⁸ U atoms and N_ Pb be the present number of ²⁰⁶ Pb atoms. The number of moles of ²³⁸ U is n_U = 2.00 10⁻³ 238 The number of moles of ²⁰⁶ Pb is n_ Pb = 0.600 10⁻³ 206 Since all the lead originated from the uranium, the initial number of moles of ²³⁸ U was: n₀ = n_U + n_ Pb The ratio of the initial amount of uranium to the present amount is: n₀ n_U = 1 + n_ Pb n_U = 1 + 0.600 206 2.00 238 n₀ n_U = 1 + 0.600 238 2.00 206 = 1 + 142.8 412 = 1 + 0.3466 = 1.3466 According to the radioacti