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After a plant or animal dies, its $^{14} \mathrm{C}$ content decreases with a half - life of 5 730 yr. If an archaeologist finds an ancient fire pit containing partially consumed firewood and the $^{14} \mathrm{C}$ Content of the wood is only 12.5$\%$ that of an equal carbon sample from a present-day tree, what is the age of the ancient site?

1.72 \times 10^{4} y r

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Cornell University

Rutgers, The State University of New Jersey

Hope College

University of Winnipeg

in number 22. Ah, Carbon 14 is used to find out how old piece of wood is. We know the half life of carbon 14 is 57 30 years, and this piece of wood only had 12.5% of the carbon 14 that originally had well, open fire. That's easy enough. You can tell us divide and half. 30 times 50 than 25. Then half of 25 is this. But we're gonna go through because it won't always be that easy. Well, so I'm going to use the equation. We're the remaining number of nuclei. Is the original number of nuclei time 1/2 raised to the number of half life that you have. And if this has 25% So it's 0.25 of what the original was is what has left. So this would be 0.1 to 5 of the original is what the new mount is. So you gonna tell my that cancels out? And here for my this is the number of half lifes. Um, I can think of this. That would be, however many years. It was divided by the half life, so first, twice as many would be to half laces, three times as many. Like this would be 3/2 lives. So now I'm just gonna take the log of both sides. So I have the log of 0.1 to 5 over here. I have the log of 1/2 raised to that. Remember how long's work? I could bring that out in front. You know, the exponents turns into go what? You multiply in the front. I'm gonna change it. So desperate here. So now if I want to solve for that X, I'm just gonna divide by local five. It's a local 0.1 to 5, divided by log, appoint fire. And then I would motor point by this so multiplied by the 57. 30 and I get 70 miles in 190 years.

University of Virginia