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More exponential decay examples

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Video transcript

- Let's do a couple of more of these exponential decay problems. 'Cause a lot of this really is just practice, and being very comfortable with the general formula. And I'll write it again, where the amount of the element that's decaying that we have at any period in time is equal to the amount that we started with times e to the minus kt, where they k value is specific to any certain element with a certain half-life, and sometimes they don't even give you the half-life. So let's try an interesting situation. Let's say that I have a element. Let me just give you a formula. Let's say that I have some magic element here where its formula is, its k value I give to you, k is equal to minus, let me think of a -- (coughs) Excuse me. I just had a lot of walnuts and my throat is dry. Let's say that k is equal to -- well k, we're putting a minus (unintelligible), so I'll say the k value is a positive 0.05. So its exponential decay formula would be the amount that you start off with times e to the minus 0.05t. My question to you is, given this, what is the half-life of the compound that we're talking about? What is the half-life? So to figure that out, we need to figure out what t value could we put here, so that if we start off with whatever value here, we end up with half of that value there. So let's do that. So we're starting off with n sub zero, this is just some value, our initial starting point. We could put 100 there. Actually let's do that, just to keep things, just to keep things less abstract. So let's say we start with 100. I'm just picking 100 out of air. I could have left it abstract with n. Let's say I'm starting with 100. I take the 100 times e to the minus 0.05 times t. T is whatever our half-life. So after our half-life we're going to have half of this stuff left. So this should be equal to 50. We just solve for t. Divide both sides by 100. You get e to the minus 0.05. T is equal to 1/2. Take the natural log of both sides of this. Natural log of this. Natural log of that. And then you get the natural log of e to anything, I've said it before, is just the anything. So it is minus 0.05t is equal to the natural log of 1/2. And then you get t is equal to the natural log of 1/2 divided by minus 0.05. So let's figure out what that is. So if we have 0.5, the natural log -- and actually someone just made a comment, and I might as well do that, I could just put this minus up here. I could make this a plus and this a minus, if I just multiply the numerator and the denominator by negative one. And if I want to, just to make our math a little easier, if you put a minus in front of a natural log or any logarithm, that's the same thing as the log of the inverse of two over 0.05 is equal to 13.86. 13.86. So when t is equal to, t is equal to 13.86, and I'm assuming that we're dealing with time in years, that tends to be the convention, although sometimes it could be something else, and you'd always have to convert to years. But assuming that this original formula, where they gave this k value of 0.05, that was with the assumption that t is in years, then I've just solved its half-life. I've just solved that after 13.86 years, you can expect to have half of the substance left. We started with 100. We ended up with 50. I could have started with x and ended up with x over two. Let's do one more of these problems so that we're really comfortable with the formula. Let's say that I have something with a half-life, half-life of, I don't know, let's say I have it as one month. Half-life of one month. And after, let's say, that I, let's say that I'm, well let me, just for the sake of time, let me make it a little bit simpler. Let's say I just have my k value is equal to, I mean you could go from half-life to a k value. We did that in the previous video. Let's say my k value is equal to 0.001. So my general formula is the amount of product I have is equal to the amount that I started with times e to the minus 0.001 times t. And I gave you this. If you had to figure it our from half-life, I did that on the previous video with Carbon 14. But let's say this is the formula. And let's say that after, after, I don't know. Let's say after 1000 years I have 500 grams of whatever element is described, the decay formula for whatever element is described by this formula. How much did I start off with? So how much did I start off with? So essentially I need to figure out n sub-zero. Right? I'm saying that after 5,000 years -- after 1,000 years. So n of 1,000, which is equal to n sub naught times e to the minus 0.001 times 1,000. Right? That's the n of 1,000. And I'm saying that that's equal to 500 grams. That equals 500 grams. So I just have to solve for n sub naught. So what's the e value? So if I have 0.001 times 1,000. So this is n sub naught. This is one-thousandth of a thousand. So times e to the minus one is equal to 500 grams. Or I could multiply both sides by e and I have n sub naught is equal to 500e. Which is about 2.71. So let's, 500, 500 times 2.71. So we will have 1,355 grams. So it's equal to 1,355 grams or 1.355 kilograms. That's what I started with. 1,355. So hopefully you see now, I think we've approached this pretty much in almost any direction that a chemistry test or a teacher could throw the problem at you. But you really just need to remember this formula. And this applies to a lot of things. Later you'll learn, you know, when you do compound interest in finance, the k will just be a positive value, but it's essentially the same formula. And there's a lot of things that this formula actually describes well beyond just radioactive decay. But the simple idea is, use the information they give you to solve for as many of these constants as you can, and then whatever they're asking for, solve for whatever's left over. And hopefully I've given you enough examples of that. But let me know. I'm happy to do more.