A fossilized bone contains 3.875g of Carbon-14 and 120.125g of its daughter isotope (Nitrogen-14). How old is the fossilized bone if Carbon-14 has a half-life of 5730 years?

Approximately 28650 years. Assumptions. 1. All of the Nitrogen-14 came from decay of Carbon-14 2. The lifespan of the animal which the bone came from is insignificant compared to the age of the bone. This is because there will be carbon-14 decay happening during the lifetime of the animal. So the original amount of carbon-14 will be the sum of the masses of carbon-14 and nitrogen-14. So that is 3.875 g + 120.125 g = 124 g originally Divide the original amount of Carbon-14 by the amount of carbon-14 remaining, getting 124 / 3.875 g = 32 Calculate the base 2 logarithm of the ratio to determine how many half lives expired. ln (32) / ln (2) = 3.46574 / 0.693147 = 5 So 5 half lives have expired. Now simply multiply by the duration of a half life. So 5 * 5730 years = 28650 years