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Problem 62 Hard Difficulty

A piece of charcoal used for cooking is found at the remains of an ancient campsite. A 1.00-kg sample of carbon from the wood has an activity equal to $5.00 \times 10^{2}$ decays per minute. Find the age of the charcoal. Hint: Living material has an activity equal to 15.0 decays/min per gram of carbon present.

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Top Physics 103 Educators
Elyse G.

Cornell University

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Hope College

Video Transcript

for number 62. There's a one kilogram piece of ancient wood charcoal that is measured have 500 decays per minute and we want to hold. That is how old that piece of ancient wood is. Hold that charcoal is and were given. That living stuff has the Corbin is the cable rate of 15 decays per minute for each gram. So I just need to find how old this is. I'm gonna use the equation I always use with this half life kind of stuff. Where you have the number of nuclei is equal to the original number of nuclei times on half, raised to the number of half lives. I don't have the number of nuclear. I have the activity. But if you take number nuclei and multiply it by activity by the decay constant Now this is activity. So this is my new activity. My original activity 1/2 raised with number of half lifes. And if I can find how many half lifes I have, I can figure out how old is. So I just need to make sure that these are in the same unit, whatever that is. Um, so I know that my current is 500 is my activity. Originally, it was 15 2 K's per minute per gram, but I have a whole kilogram. I have a whole kilogram, so I'm gonna multiply this by 1000. So this is gonna be 15,000 decays per minute for the whole kilogram. Well, this is gonna be one here to the end, and I'm trying to find, so I'm gonna divide both sides by this 1500. I'm taking this to the side of the equation, and I get it. 0.0, 433 is gonna equal. We'll have him would already this 0.5 0.5 race to the end. So I'm gonna just take the natural log of both side, not the natural. You take the regular log of both sides, so I'm gonna have a log of went. 0333 Over here. It's gonna be log 0.5 and then for logs. If something's raised, the exponents could just bring it out here in front. So now I'm just going to divide this side by 1.5. So I've logged 0.3333 divided by local 0.5. And that I get was equal to four point. Remember what that is? It's four point 9/2 lives. So now if I just look up this carbon carbon 14 I find that the half life of that is 5000 730 years. So I have four point 9/2 lives. It's going to 4.9 times 5000 730 and these are in years where this is the half life off carbon. So I get 20 1000 77 years. That's an awful lot of 66 We call it 40,000 100 years.

University of Virginia
Top Physics 103 Educators
Elyse G.

Cornell University

LB
Liev B.

Numerade Educator

Farnaz M.

Other Schools

Zachary M.

Hope College