Half life equation radiometric dating

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.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.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 the natural log of 1 over 1 plus 0.01 over 0.11 or 11% is equal to negative kt. 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.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.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 say we start with N0, whatever that might be. We know, after that long, that half of the sample will be left. Whatever we started with, we're going to have half left after 1.25 billion years. And then to solve for k, we can take the natural log of both sides.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. 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.

Negative k is the negative of this over the negative natural log of 2 over 1.25 times 10 to the ninth. If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.and *.are unblocked. So the negative natural log of 1/2 is the same thing as the natural log of 1/2 to the negative 1 power. Anything to the negative power is just its multiplicative inverse. So negative natural log of 1 half is just the natural log of 2 over here. 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.

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