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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.

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So we know that anything that is experiencing radioactive decay, it's experiencing exponential decay.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. Nov 2009 - 5 min - Uploaded by drlerocks Calculate the age of a sample using radiometric dating. When carbon-14 is used the process is called radiocarbon dating, but.

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