4 days ago Radiocarbon dating (usually referred to simply as carbon dating) is fix atmospheric carbon dioxide (CO 2) into organic compounds during. This talk will outline how radiocarbon dating is performed using AMS, of 14C measurements, including compound-specific radiocarbon.

Let's say we have a collection of atoms.

## Exponential decay formula proof (can skip, involves calculus)

And normally when we have any small amount of any element, we really have huge amounts of atoms of that element. And we've talked about moles and, you know, one gram of carbon I'm sorry, 12 grams-- 12 grams of carbon has one mole of carbon in it.

One mole of carbon And what is one mole of carbon? This is a ginormous number. This is more than we can, than my head can really grasp around how large of a number this is. And this is only when we have 12 grams. For example, one kilogram is about two pounds. So this is about, what? I want to say [? And pounds is obviously force. You get the idea.

### Half-life and carbon dating (video) | Nuclei | Khan Academy

On Earth, well anywhere, mass is invariant. This is not a tremendous amount.

So with that said, let's go back to the question of how do we know if one of these guys are going to decay in some way. And maybe not carbon, maybe we're talking about carbon or something.

How do we know that they're going to decay? And the answer is, you don't. They all have some probability of the decaying. At any given moment, for a certain type of element or a certain type of isotope of an element, there's some probability that one of them will decay. That, you know, maybe this guy will decay this second. And then nothing happens for a long time, a long time, and all of a sudden two more guys decay.

And so, like everything in chemistry, and a lot of what we're starting to deal with in physics and quantum mechanics, everything is probabilistic. I mean, maybe if we really got in detail on the configurations of the nucleus, maybe we could get a little bit better in terms of our probabilities, but we don't know what's going on inside of the nucleus, so all we can do is ascribe some probabilities to something reacting.

Now you could say, OK, what's the probability of any given molecule reacting in one second? Or you could define it that way. But we're used to dealing with things on the macro level, on dealing with, you know, huge amounts of atoms. So what we do is we come up with terms that help us get our head around this. And one of those terms is the term half-life. And let me erase this stuff down here. So I have a description, and we're going to hopefully get an intuition of what half-life means.

So I wrote a decay reaction right here, where you have carbon It decays into nitrogen And we could just do a little bit of review.

You go from six protons to seven protons. Your mass changes the same. So one of the neutrons must have turned into a proton and that is what happened.

And it does that by releasing an electron, which is also call a beta particle. We could have written this as minus 1 charge. It does have some mass, but they write zero.

This is kind of notation. So this is beta decay.

Beta decay, this is just a review. But the way we think about half-life is, people have studied carbon and they said, look, if I start off with 10 grams-- if I have just a block of carbon that's 10 grams. If I wait carbon's half-life-- this is a specific isotope of carbon. Remember, isotopes, if there's carbon, can come in 12, with an atomic mass number of 12, or with 14, or I mean, there's different isotopes of different elements.

And the atomic number defines the carbon, because it has six protons. Carbon has six protons. But they have a different number of neutrons. So when you have the same element with varying number of neutrons, that's an isotope. So the carbon version, or this isotope of carbon, let's say we start with 10 grams. If they say that it's half-life is 5, years, that means that if on day one we start off with 10 grams of pure carbon, after 5, years, half of this will have turned into nitrogen, by beta decay.

And you might say, oh OK, so maybe-- let's see, let me make nitrogen magenta, right there-- so you might say, OK, maybe that half turns into nitrogen. And I've actually seen this drawn this way in some chemistry classes or physics classes, and my immediate question is how does this half know that it must turn into nitrogen?

And how does this half know that it must stay as carbon? And the answer is they don't know. And it really shouldn't be drawn this way.

So let me redraw it. So this is our original block of our carbon What happens over that 5, years is that, probabilistically, some of these guys just start turning into nitrogen randomly, at random points. So if you go back after a half-life, half of the atoms will now be nitrogen.

So now you have, after one half-life-- So let's ignore this. So we started with this. All 10 grams were carbon. This is after one half-life. And now we have five grams of c And we have five grams of nitrogen Let's think about what happens after another half-life. So if we go to another half-life, if we go another half-life from there, I had five grams of carbon So let me actually copy and paste this one.

This is what I started with. Now after another half-life-- you can ignore all my little, actually let me erase some of this up here. So clearly the amount you lose is dependent on the amount you started with, right? Over any fraction of time, and here it's a very small fraction. So what I set up here is really fairly simple, but it doesn't sound so simple to a lot of people if you say it's a differential equation. We can actually solve this using pretty straightforward techniques.

This is actually a separation of variables problem. And so, what can we do?

Let's divide both sides by N. We want to get all the N's on this side and all the t stuff on the other side. So if we have 1 over N, dN over dt is equal to minus lambda. I just divided both sides of this by N. And then I can multiply both sides of this by dt, and I get 1 over N dN is equal to minus lambda dt. Now I can take the integral of both sides of this equation.

And what do I get? I'm taking the indefinite integral or the antiderivative. What's the antiderivative of 1 over N? Well that's the natural log of N plus some constant-- I'll just do that in blue-- plus some constant. And then that equals-- What's the antiderivative of just some constant?

Well it's just that constant times the derivative, the variable. We're taking the antiderivative with respect to. So minus lambda, times t, plus some constant. These are different constants, but they're arbitrary.

So if we want, we can just subtract that constant from that constant, and put them all on one side and then we just get another constant.

So this boils down to our solution to our differential equation is the natural log of N is equal to minus lambda-t, plus some other constant, I call it c3, it doesn't matter. And now if we want to just make this a function of N in terms of t, let's take both of these, or both take e to the power of both sides of this.

You can view that as kind of the inverse natural log. So e to the power of ln of N, ln of N is just saying what power do you raise e to to get to N? So if you raise e to that power, you get N.

So I'm just raising both sides of this equation. I'm raising e to both sides of this equation. And that is equal to e to the minus lambda-t, plus c3. And now this can be rewritten as, N is equal to e to the minus lambda-t, times e to the c3. And now once again this is an arbitrary constant, so we can just really rename that as, I don't know, let me rename it as c4. So, our solution to our differential equation, N, as a function of t, is equal to our c4 constant, c4e to the minus lambda-t.

Now let's say, even better, let's say is N equals 0. Let's say that N equals 0. We have N sub 0 of our sample. That's how much we're starting off with. So let's see if we can substitute that into our equation to solve for c4. So we said N sub-0 is equal to, let's put 0 in here, so let's see, that's equal to N sub naught. And that's equal to c4 times e to the minus lambda, times 0. Well, minus anything times 0 is 0. So it's e to the 0.

So that's just 1. So c4 is equal to N naught, our starting amount for the sample. So we've actually got an expression. We have the number of particles, or the amount as a function of t, is equal to the amount that we start off with, at time is equal to 0, times e to the minus lambda, times time.

### Exponential decay formula proof (can skip, involves calculus) (video) | Khan Academy

And we just have to be careful that we're always using the time constant when we solve for the different coefficients. So this seems all abstract. How does this relate to half-life? Well let's try to figure out this equation for carbon. This'll be true for anything where we have radioactive decay.

## 17.6: Radiocarbon Dating: Using Radioactivity to Measure the Age of Fossils and Other Artifacts

If we actually had a plus sign here it'd be exponential growth as well. We know that carbon, c, has a 5,year half-life. So the way you could think about it, is if at time equals 0 you start off with t-- So time equals 0. If at N of 0 is equal to-- and we could write there if we want. Actually why don't we do that? If N of 0 we start off with And then at N of 5, years-- so we're going to take t to be in years, you just have to be consistent with your units-- how much will we have left?

We'll have 50 left.