The amount of ^{14}C in an old, isolated sample of vegetation is 0.809 that expected from a present day sample with an equivalent carbon content. Calculate the age of the old sample in years.

k = (ln2) / t(1/2) You calculate k in reciprocal years Then you use [A] / [A]o = exp(-kt) with t = 8740 The fraction on the left is the fraction asked for in the first question. for the next one, 0.809 = exp(-kt) and you solve this time for t. take the log of both sides ln(0.666) = -kt

N=Ao*(1/2)^t/T

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