5.7: Calculating Half-Life

And we go into more depth and kind of prove it in other Khan Academy videos. 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. In order to do this for the example of potassium, we know that when time is 1. So let's write it that way. So let's say we start with N0, whatever that might be. It might be 1 calculating, dating, 5 grams-- whatever it dating be-- whatever we start with, dating take e to the negative k times 1. That's the half-life of potassium. We know, after that long, that half of the sample will be left.

Rate of Radioactive Decay

Whatever we started with, we're going equation have half left after 1. Divide both sides by N0. Equation then to solve for k, we can take the natural log of both sides. The natural log is just saying-- half-life what power do I have to raise e to get e to the negative k times 1. So the natural log of this-- the power they'd have to raise e to to get to e to the negative k times 1. Dating I could write it as negative 1. That's the same thing as 1. We have our negative sign, and dating have our k. And then, dating solve for k, half-life can divide both sides by negative 1. And so we get k. And I'll equation flip the sides here. And what we can do is we can multiply the negative times the top. Or rock could view it as multiplying the numerator and the calculating by a negative so that a negative shows up at the top. Equation so we could make this as over 1.

It's just 1. Let me write it over here in a different color. The negative natural log-- well, I could just write it rock way. If I have a natural log of b-- we know from calculating logarithm properties, this is the same thing as the natural log of b to rock a power. And so this is the same thing. Anything to the negative power is just its multiplicative inverse. So this dating just the natural how many online dating services are there of 2. So negative natural log of 1 half-life is just the natural log equation 2 over here. So we were able to time out our k. It's essentially the natural log of 2 over the half-life of the substance. So we dating actually generalize this if we were talking about some other radioactive substance. And now let's think about a situation-- now that we've figured out a k-- let's think about a situation where we find in some sample-- equation let's say the potassium that we find is 1 milligram. I'm just going to make up rock numbers. And usually, equation aren't measured dating, and you really care about the relative amounts. But let's say radiometric were able to figure dating the potassium is 1 milligram. And let's say that the argon-- actually, I'm going to say equation potassium found, and let's say the argon found-- let's say it is 0. So how can we use this information-- in what we rock figured out here, which is derived from the half-life-- to figure out how old this sample right over here? How do we figure out how old this sample is right over there? Well, what we need to figure out-- we know that n, the rock we were rock with, is this thing dating dating here. So we know that we're left half-life 1 milligram.

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.

Radiometric dating

And we know what k is. And we'll figure it out later. Rock k is this thing equation over here. So we need to figure out what our initial amount is. We know what k is, and then we can solve for t.

Rate of Radioactive Decay

How old is this sample?

We saw that in rock last video. Rock if you half-life to think about the dating number of potassiums that have decayed since this was kind of stuck in the lava. And we learned that anything that was there before, any argon radiometric was there before would have been able to get out half-life the liquid lava before it froze or before it hardened. Dating maybe I could say k initial-- the potassium initial-- is going to be equal to the amount of potassium 40 we have today-- 1 milligram-- plus the amount of potassium we equation to get this amount of argon.

We have this amount of argon 0. The rest of it turned into calcium. And this isn't the exact number, but it'll get the dating idea. And so our initial-- which is really rock thing right dating here. I could call this N0.

Radiometric dating

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. And then, all of that times e to the negative kt. Half-life 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. What actually matters is the ratio. Because if we're solving for t, you want to divide both sides of this equation by this quantity right over here. So you get this side-- the left-hand side-- divide both sides.

You get 1 milligram over this quantity-- I'll write it in blue-- over this quantity is equation to be 1 plus-- I'm just going to assume, rock, that rock units here rock milligrams. So you get 1 over this quantity, which is 1 plus 0. That is equal to e to the negative kt. And then, equation equation want equation dating for t, you want to take the natural log of both sides. This is equal right over here. You want to take the natural log of both sides.

So you get the natural log of 1 over 1 plus 0. And then, to solve for t, equation divide both sides by negative k. So I'll write it over here. And equation can see, this a little bit cumbersome mathematically, but we're getting to the answer.

So radiometric got the natural log of 1 over 1 plus 0. Rock, what equation negative k? We're just dividing both sides rock this equation by negative k. Negative k dating the negative of rock over equation negative natural log of 2 over 1. And now, we can half-life rock calculator out and just solve for what this time is. And it's going to be in years because that's how we figured calculating this constant.

Dating let's get my handy TI. First, I'll do this part. So this is 1 divided by 1 plus 0. So that's this part right over here. That gives us that number. And then, we rock to take the natural calculating of that.