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Video transcript
In the last video, we give a bit of an overview of potassium-argon dating. In this video, I want to go through a concrete example. And it'll get a little bit mathy, usually involving a little bit of algebra or a little bit of exponential decay, but to really show you how you can actually figure out the age of some volcanic rock using this technique, using a little bit of mathematics. So we know that anything that is experiencing radioactive decay, it's experiencing exponential decay. And we know that there's a generalized way to describe that. 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. We'll call that N sub 0, times e to the negative kt-- where this constant is particular to that thing's half-life. In order to do this for the example of potassium-40, we know that when time is 1.25 billion years, that the amount we have left is half of our initial amount. So let's write it that way. So let's say we start with N0, whatever that might be. It might be 1 gram, kilogram, 5 grams-- whatever it might be-- whatever we start with, we take e to the negative k times 1.25 billion years. That's the half-life of potassium-40. So 1.25 billion years. We know, after that long, that half of the sample will be left. So we will have 1/2 N0 left. Whatever we started with, we're going to have half left after 1.25 billion years. Divide both sides by N0. And then to solve for k, we can take the natural log of both sides. So you get the natural log of 1/2-- we don't have that N0 there anymore-- is equal to the natural log of this thing. The natural log is just saying-- to what power do I have to raise e to get e to the negative k times 1.25 billion? So the natural log of this-- the power they'd have to raise e to to get to e to the negative k times 1.25 billion-- is just negative k times 1.25 billion. Or I could write it as negative 1.25-- let me write times-- 10 to the ninth k. That's the same thing as 1.25 billion. We have our negative sign, and we have our k. And then, to solve for k, we can divide both sides by negative 1.25 billion. And so we get k. And I'll just flip the sides here. k is equal to the natural log of 1/2 times negative 1.25 times 10 to the ninth power. And what we can do is we can multiply the negative times the top. Or you could view it as multiplying the numerator and the denominator by a negative so that a negative shows up at the top. And so we could make this as over 1.25 times 10 to the ninth. It's just 1.25 billion. Let me write it over here in a different color. The negative natural log-- well, I could just write it this way. If I have a natural log of b-- we know from our logarithm properties, this is the same thing as the natural log of b to the a power. So the negative natural log of 1/2 is the same thing as the natural log of 1/2 to the negative 1 power. And so this is the same thing. Anything to the negative power is just its multiplicative inverse. So this is just the natural log of 2. So negative natural log of 1 half is just the natural log of 2 over here. So we were able to figure out our k. It's essentially the natural log of 2 over the half-life of the substance. So we could actually generalize this if we were talking about some other radioactive substance. And now let's think about a situation-- now that we've figured out a k-- let's think about a situation where we find in some sample-- so let's say the potassium that we find is 1 milligram. I'm just going to make up these numbers. And usually, these aren't measured directly, and you really care about the relative amounts. But let's say you were able to figure out the potassium is 1 milligram. And let's say that the argon-- actually, I'm going to say the potassium-40 found, and let's say the argon-40 found-- let's say it is 0.01 milligram. 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? How do we figure out how old this sample is right over there? Well, what we need to figure out-- we know that n, the amount we were left with, is this thing right over here. So we know that we're left with 1 milligram. 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. And we know what k is. And we'll figure it out later. So k is this thing right over here. So we need to figure out what our initial amount is. We know what k is, and then we can solve for t. How old is this sample? And to figure out our initial amount, we just have to remember that for every argon-40 we see, that must have decayed from-- when you have potassium-40, when it decays, 11% decays into argon-40 and the rest-- 89%-- decays into calcium-40. We saw that in the last video. So however much argon-40, that is 11% of the decay product. 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. 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. So to figure out how much potassium-40 this is derived from, we just divide it by 11%. 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. We have this amount of argon-40-- 0.01 milligrams. And that number of milligrams there, it's really just 11% of the original potassium-40 that it had to come from. The rest of it turned into calcium-40. So we divide it by 11% or 0.11. And this isn't the exact number, but it'll get the general idea. And so our initial-- which is really this thing right over here. I could call this N0. 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 then, all of that times e to the negative kt. 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. What actually matters is the ratio. Because if we're solving for t, you want to divide both sides of this equation by this quantity right over here. So you get this side-- the left-hand side-- divide both sides. You get 1 milligram over this quantity-- I'll write it in blue-- over this quantity is going to be 1 plus-- I'm just going to assume, actually, that the units here are milligrams. So you get 1 over this quantity, which is 1 plus 0.01 over the 11%. That is equal to e to the negative kt. And then, if you want to solve for t, you want to take the natural log of both sides. This is equal right over here. You want to take the natural log of both sides. So you get the natural log of 1 over 1 plus 0.01 over 0.11 or 11% is equal to negative kt. And then, to solve for t, you divide both sides by negative k. So I'll write it over here. And you can see, this a little bit cumbersome mathematically, but we're getting to the answer. So we got the natural log of 1 over 1 plus 0.01 over 0.11 over negative k. Well, what is negative k? We're just dividing both sides of this equation by negative k. Negative k is the negative of this over the negative natural log of 2 over 1.25 times 10 to the ninth. And now, we can get our calculator out and just solve for what this time is. And it's going to be in years because that's how we figured out this constant. So let's get my handy TI-85. First, I'll do this part. So this is 1 divided by 1 plus 0.01 divided by 0.11. So that's this part right over here. That gives us that number. And then, we want to take the natural log of that. So let's take the natural log of our previous answer. So it's the natural log of 0.9166667. It gives us negative 0.087. So that's this numerator over here. And we're going to divide that. So this number is our numerator 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 it's negative natural log of 2 divided by 1.25. E9 means times 10 to the ninth. And I closed both parentheses. And now, we need our drum roll. So this should give us our t in years. 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. So this is approximately a 157-million-year-old sample. 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. But the whole point I wanted to do this is to show you that it's not some crazy voodoo here. And, you know, Sal, gave this very high-level explanation, and then, you say, oh, well, there must be some super difficult mathematics after that. The mathematics really is something that you would see in high school. 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. You would be able to do that to figure out this is a 157-million-year-old sample of volcanic rock.