Samarium 147 decays to Neodymium 143 with a decay constant A = 0.006539 Ga', with Ga being a fancy term for "billions of years" (note that the units of a here are per billion years). This means that if the initial amount of radioactive 147Sm is labeled 47Sm(t 0), then the amount at some later time t is given by 14"Sm(1) = 14"Sm(t = 0) e = 14"Sm(t = 0) exp(-A). a) [10 pts] Show that this means the half-life (t2) is given by 0.5 = exp(-0.006539 tu2), which is the same as t2 In(2)/0.006539 b) [10 pts] What is the value of the half-life of 147Sm in billions of years? c) [10 pts] How is it possible to measure such an incredibly long half life? The answer comes from the large number of atoms in a visible amount of material. Specifically, 147 grams of 147Sm contains 1 mole 6.022 x 1023 of atoms. There more than 10 billion decays in a day from such a quantity! For this problem, show that a microgram would give more than 70 decays in a day.