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# 钾-氩定年法计算

钾-氩定年法实例（需要指数函数与对数函数的知识）. 由 Sal Khan 创建

## 视频字幕

in the last video we gave a bit of an overview of potassium-argon dating in this video I want to go through a concrete example it'll get a little bit Matthew 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 we can 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 the amount as a function of time so if we say n is the amount of a radioactive sample we have at some at some time we know that's equal to the initial amount we have we'll call that Eden Sub Zero times e to the negative KT where this constant is particular to that things half-life and to figure it out and we're going to do this for the example of potassium-40 we know that after 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 when we we start with n naught we start with that not at whatever that might be might be one gram kilogram five 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 1/2 and naught left whatever we started with we're going to have 1/2 left after 1.25 billion years divide both sides by n not divide both sides by n not 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 end not 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 that have to raise e to to get e get 2 e to the negative K times 1.25 billion is just negative K times 1.25 billion I could write it as negative 1.25 let me write it times 10 to the 9th times 10 to the 9th 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 one point two five times ten 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 negative so the negative shows up at the top and so we could make this as over one point two five times ten to the ninth it's just one point two five billion and negative let me write it over here in a different color negative the negative natural log well I could just write it this way if I have a a natural log of B we know from our logarithm properties this is the same thing as a 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 one power and so this is the same thing anything to the negative 1 power is just this multiplicative inverse so this is just the natural log of 2 so negative natural log of 1/2 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 this substance so we could actually generalize this if we're talking about some other radioactive substance and now let's think about a situation now that we 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 let's say it is one milligram I'm just gonna make up these numbers and let's say and usually these aren't measured directly and you really care about the relative amounts but let's say you're able to figure out the potassium is one milligram and let's say that the argon next let me say the potassium 40 found let's say the argon-40 found let's say it is 0.01 0.01 milligram so how do we how can we use this information what we just figured out here which is derived from the half-life to figure out how old how old this sample right over here how do we figure out how old this sample is right over there well what we what we need to figure out we know that n we know that n the amount we were left with is this thing right over here so we know that we're left with one milligram so we know that what we have left is one milligram and that's going to be equal to some initial amount it's going to be some initial amount women will use both of this information to figure that initial amount out times e to the negative K T 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 for every argon-40 we see that must have decayed from when you have potassium-40 11% after 11% decays when it decays 11% decays into argon-40 and the rest 89% decays into calcium for T 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-40 s that have decayed since this was since this was kind of stuck in the in the lava and we learned that anything that was there before any argon-40 those there before would have would have been able to get out of the liquid lava before it froze or before it hardened so to figure out how much how much potassium-40 this is derived from we just derive it we divide it by 11% so maybe I could say okay K initial the potassium-40 initial is going to be equal to the amount of potassium-40 we have today one milligram plus the amount of potassium-40 we needed to get this amount of argon-40 so we have this amount of argon forty point zero one milligrams and that the number of milligrams there it's really just 11% of the original potassium potassium-40 that had to come from the rest of it turned into calcium for T so we divide it by eleven percent or 0.1 one 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 and not I could call this and not this is going to be equal to and I won't do any of the math so we have one milligram we have left is equal to 1 milligram which is what we found plus 0.01 milligram over 0.1 one and then all of that times e to the negative 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 then it really the absolute amount a mountain 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 one milligram over this quantity I'll write it in blue over this quantity is going to be one plus I'm just going to assume actually that the unit's here are milligrams so you get 1 over this quantity which is 1 plus 0.01 over the 11 over the 11% that is equal to e to the negative KT and then you want to take if you want to solve for T you want to take the natural log of both sides so then you get so 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 over point 1 1 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 you can see this is 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 point 1 1 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 point 2 5 times 10 to the 9th and now we can get our calculator around and just solve for what this I'm is and it's going to be in years because that's how we figured out this constant so let's get out my handy ti-85 and so first I'll do this part so this is one divided by one plus point zero one divided by point one one 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 R this is just our previous answer so it's the natural log of point nine one six six six six seven gives us negative point zero eight seven 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 let me make I'll use parentheses carefully the negative natural log of 2 the negative natural log of two that's that they're divided by divided by one point two five times 10 to the ninth times 10 to the ninth so divided by so it's negative natural log of 2 divided by one point 2 5 e 9 means times 10 to the ninth and I closed both parentheses and now we need our drumroll so this should give us our T in years and we get let's see how many this is this is thousand so it's three thousand so we get 156 million or 156 point nine million years if we round so this is approximately I could just say approximately 157 million years 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 kind of a precalculus 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 you know there must be some super difficult mathematics after that the mathematics really is something that you would see in high school and if you saw if you saw a sample that had this ratio of argon-40 to potassium-40 you would actually be able to do that fairly you know that that high school mathematics you would able to do that to figure out this is a hundred and fifty seven million year old sample a volcanic rock