# Half life equation radiometric dating

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.

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.

And it's going to be in years because that's how we figured out this constant.

SAL: In the last video we saw all sorts of different types of isotopes of atoms experiencing radioactive decay and turning into other atoms or releasing different types of particles.

But the question is, when does an atom or nucleus decide to decay? So it could either be beta decay, which would release electrons from the neutrons and turn them into protons. And normally when we have any small amount of any element, we really have huge amounts of atoms of that element. That's 6.02 times 10 to the 23rd carbon-12 atoms. This is more than we can, than my head can really grasp around how large of a number this is.

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.    If you're behind a web filter, please make sure that the domains *.and *.are unblocked.

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.

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.