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An old campfire is uncovered during an archaeological dig. Its charcoal is found to contain less than 1$/ 1000$ the normal amount of $^{14} \mathrm{C}$ . Estimate the minimum age of the charcoal, noting that $2^{10}=1024$

57300 years

01:37

Salamat A.

Physics 103

Chapter 31

Radioactivity and Nuclear Physics

Nuclear Physics

Cornell University

University of Michigan - Ann Arbor

Simon Fraser University

University of Sheffield

Lectures

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Okay. So this is a half flat problem. So, we're going to need two equations. The first one is one of the basic half life equations. The number of decays at some point in time equals the initial number of decays times E to the negative lambda T. Where T. Is the time lambdas the decay constant. Okay. And we're also going to need the formula for the decay constant constant. Lambda Is equal to the natural log of two divided by half life. Okay, So let's get the decay constant first. So, we know, I'm assuming we can get this from looking it up. Right? The half life of carbon 14. And if you look it up, it's a fairly common one. They use It's 5730. Here's Okay, So, our decay constant let's find our decay constant first. So our decay constant is the natural log of two divided by the half life. So that's Alan to over 5730. And if we do that, we get 1.21 Times 10 to the -4. All right, So now we can come solve our equation. Right? So, let's uh that's right this out again in as a function of T. Right? So in at some time equals and not E to the negative lambda T. So, let's do some algebra. We need to get T by itself. Right? We're trying to solve for T. So, I'm going to do in divided by and not Right, I'm gonna divide that over and that equals E. To the negative lambda T. And then if I take the natural log on both sides, that will get rid of the E. So the natural log of in over and not equals negative lambda, etc. And so now if I want t let's go up here. So t is going to equal the natural log of in over in not over negative landau. Okay. So if we plug in natural log that the ratio of into and not the problem, it says the charcoal contains Less than 1 1,000th of the normal amount. Right? So it's saying one 1000th of the normal. Well, that is the ratio in over and not right? Because what this number means in is the number of counts at some time in not as your initial Number of counts a number of decades, a number of atoms or whatever it is. Right? So if you're saying 1,000th is remaining, that is the ratio in over and not. It's telling you how much is left. So the natural log of one over a 1000 over negative lambda, right and Negative Lambda is negative 1.21 Times 10 of -4. And so if we do that, we get a negative number for that, right? And a negative number for this. And so it all works out becomes positive. So it's 50,057,000 and about 89 Years. If you do that. So if you want to use significant figures, you can just run that to 57,000 years.

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