An archaeologist finds the ¹⁴C in a sample of 3.10 g of material to be decaying at 107 counts per second. A modern 1.00-g sample of the same material decays at 151 counts per second. The half-life of ¹⁴C is 5730 y. How old is the sample? A) 12,200 y B) 8460 y C) 25,100 y D) 12,600 y

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The decay rate is proportional to the amount of radioactive material present. The ratio of the decay rates is equal to the ratio of the amounts of radioactive material present.$$\frac{107 	ext{ counts/s}}{151 	ext{ counts/s}} = \frac{3.10 	ext{ g}}{1.00 	ext{ g}}$$Solving for the age of the sample, we get:$$t = \frac{5730 	ext{ y}}{ln(2)} ln(\frac{151 	ext{ counts/s}}{107 	ext{ counts/s}}) = 12,200 	ext{ y}$$

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