In a samarskite sample discovered recently, there was 3 grams of Thorium (232~~)

Chapter 1, Problem 6

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In a samarskite sample discovered recently, there was 3 grams of Thorium (232~~). Thorium decays to lead-208 (208Pb) through the reaction 232Th+Z08~b +6(44He). It was determined that 0.0376 of a gram of lead-208 was produced by the disintegration of the original Thorium in the sample. Given that the half-life of Thorium is 13.9 billion years, derive the age of this samarskite sample.(Hint: 0.0376 grams of 208Pb is the product of the decay of (232/208)x0.0376 grams of Thorium.)One of the most accurate ways of dating archaeological finds is the methodof carbon- 14 (I4C) dating discovered by Willard Libby around 1949. The basisof this method is delightfully simple: The atmosphere of the earth is continuouslybombarded by cosmic rays. These cosmic rays produce neutrons in theearth's atmosphere, and these neutrons combine with nitrogen to produce I4C,which is usually called radiocarbon, since it decays radioactively. Now, this radiocarbonis incorporated in carbon dioxide and thus moves through the atmosphereto be absorbed by plants. Animals, in turn, build radiocarbon into theirtissues by eating the plants. In living tissue, the rate of ingestion of I4C exactlybalances the rate of disintegration of I4C. When an organism dies, though, itceases to ingest carbon-14 and thus its I4C concentration begins to decreasethrough disintegration of the I4C present. Now, it is a fundamental assumptionof physics that the rate of bombardment of the earth's atmosphere by cosmicrays has always been constant. This implies that the original rate of disintegrationof the- I4C in a sample such as charcoal is the same as the rate measuredtoday." This assumption enables us to determine the age of a sample of charcoal.Let N(t) denote the amount of carbon-14 present in a sample at time t,and No the amount present at time t =0 when the sample was formed. If A denotesthe decay constant of I4C (the half-life of carbon-14 is 5568 years) thendN (t)/dt = -AN (t), N (0) = N,. Consequently, N (t) = No=-". Now the presentrate R (t) of disintegration of the I4C in the sample is given by R (t) =AN (t) =ANoe -" and the original rate of disintegration is R (O)= AN,. Thus. R (I)/ R (0)= -AI so that r = (1 /A)ln[R (O)/ R (t)]. Hence if we measure R (t), the presentrate of disintegration of the 14C in the charcoal, and observe that R(0) mustequal the rate of disintegration of the I4C in a comparable amount of livingwood, then we can compute the age t of the charcoal. The following two problemsare real life illustrations of this method.