# Half life equation radiometric dating dating rebound still love ex wife

Rated 4.68/5 based on 512 customer reviews

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 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. 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. So however much argon-40, that is 11% of the decay product.

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.

Or I could write it as negative 1.25-- let me write times-- 10 to the ninth k. Or you could view it as multiplying the numerator and the denominator by a negative so that a negative shows up at the top. The negative natural log-- well, I could just write it this way.

And so we could make this as over 1.25 times 10 to the ninth. 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.

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.

So you might get a question like, I start with, oh I don't know, let's say I start with 80 grams of something with, let's just call it x, and it has a half-life of two years.

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.

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 this side-- the left-hand side-- divide both sides. So it's negative natural log of 2 divided by 1.25. But the whole point I wanted to do this is to show you that it's not some crazy voodoo here.