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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.
So what we do is we come up with terms that help us get our head around this. So I wrote a decay reaction right here, where you have carbon-14. So now you have, after one half-life-- So let's ignore this. I don't know which half, but half of them will turn into it. And then let's say we go into a time machine and we look back at our sample, and let's say we only have 10 grams of our sample left.
Now you could say, OK, what's the probability of any given molecule reacting in one second? But we're used to dealing with things on the macro level, on dealing with, you know, huge amounts of atoms. So I have a description, and we're going to hopefully get an intuition of what half-life means. And how does this half know that it must stay as carbon? So if you go back after a half-life, half of the atoms will now be nitrogen. Then all of a sudden you can use the law of large numbers and say, OK, on average, if each of those atoms must have had a 50% chance, and if I have gazillions of them, half of them will have turned into nitrogen. How much time, you know, x is decaying the whole time, how much time has passed?
Let's say I have a bunch of, let's say these are all atoms. And let's say we're talking about the type of decay where an atom turns into another atom. Or maybe positron emission turning protons into neutrons. And we've talked about moles and, you know, one gram of carbon-12-- I'm sorry, 12 grams-- 12 grams of carbon-12 has one mole of carbon-12 in it.
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.
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.
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.
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.Tags: Adult Dating, affair dating, sex dating