Ar ar dating technique
In this video, I want to go through a concrete example.
Ar age data that appeared in more than 195 peer-reviewed publications and in 31 Senior, MSc and Ph D theses.Negative k is the negative of this over the negative natural log of 2 over 1.25 times 10 to the ninth. You would be able to do that to figure out this is a 157-million-year-old sample of volcanic rock.And now, we can get our calculator out and just solve for what this time is. So this is 1 divided by 1 plus 0.01 divided by 0.11. And we go into more depth and kind of prove it in other Khan Academy videos.But we know that the amount as a function of time-- so if we say N is the amount of a radioactive sample we have at some time-- we know that's equal to the initial amount we have.So how can we use this information-- in what we just figured out here, which is derived from the half-life-- to figure out how old this sample right over here? So we need to figure out what our initial amount is. So if you want to think about the total number of potassium-40s that have decayed since this was kind of stuck in the lava.
How do we figure out how old this sample is right over there? And we learned that anything that was there before, any argon-40 that was there before would have been able to get out of the liquid lava before it froze or before it hardened. Let's see how many-- this is thousands, so it's 3,000-- so we get 156 million or 156.9 million years if we round.
Well, what we need to figure out-- we know that n, the amount we were left with, is this thing right over here. And that's going to be equal to some initial amount-- when we use both of this information to figure that initial amount out-- times e to the negative kt. So to figure out how much potassium-40 this is derived from, we just divide it by 11%. And this isn't the exact number, but it'll get the general idea. So this is approximately a 157-million-year-old sample.
So maybe I could say k initial-- the potassium-40 initial-- is going to be equal to the amount of potassium 40 we have today-- 1 milligram-- plus the amount of potassium-40 we needed to get this amount of argon-40. And that number of milligrams there, it's really just 11% of the original potassium-40 that it had to come from. And so our initial-- which is really this thing right over here. This is going to be equal to-- and I won't do any of the math-- so we have 1 milligram we have left is equal to 1 milligram-- which is what we found-- plus 0.01 milligram over 0.11. And what you see here is, when we want to solve for t-- assuming we know k, and we do know k now-- that really, the absolute amount doesn't matter. Because if we're solving for t, you want to divide both sides of this equation by this quantity right over here. We're going to divide that by the negative-- I'll use parentheses carefully-- the negative natural log of 2-- that's that there-- divided by 1.25 times 10 to the ninth. So the whole point of this-- I know the math was a little bit involved, but it's something that you would actually see in a pre-calculus class or an algebra 2 class when you're studying exponential growth and decay.
So you get 1 over this quantity, which is 1 plus 0.01 over the 11%. And then, if you want to solve for t, you want to take the natural log of both sides. And then, to solve for t, you divide both sides by negative k. And you can see, this a little bit cumbersome mathematically, but we're getting to the answer. The mathematics really is something that you would see in high school.
So we got the natural log of 1 over 1 plus 0.01 over 0.11 over negative k. We're just dividing both sides of this equation by negative k. So let's take the natural log of our previous answer. If you saw a sample that had this ratio of argon-40 to potassium-40, you would actually be able to do that high school mathematics.
Using the argon-argon dating technique, by which scientists measure the decay of an isotope called Argon-40 into Argon-39 in order to find the age of crystals, they came up with a rough approximation of the footprints' age: 19,000 years at the oldest, 10,000 or 12,000 years at the youngest.