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In my previous question I asked Please explain C14 half-life The OP mentioned that I was thinking of linear decay and C14 was measured in exponential decay.

As I understand it, C14 is always in a state of decay. If we know the exact rate of decay then shouldn't it be linear?

How do we know that C14 decay's exponentially compared to linear and have there been any studies to verify this?

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atomic transmutation are one of the few left out areas in physics that couldn't be answered by any theory, notable among them is QED. –  Vineet Menon Sep 2 '11 at 10:03
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@Vineet: that's not true, "transmutation" (a.k.a. radioactive decay, fission, and fusion) is explained quite well by the standard model. (Also, I deleted an inappropriate comment here.) –  David Z Sep 2 '11 at 15:41
    
@David: I'm telling you by the book I read by Feynman titled QED. –  Vineet Menon Sep 4 '11 at 18:40
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@Vineet: yes, QED itself doesn't explain nuclear interactions. You need the weak force for that. If you'd like to discuss this further, let's take it to Physics Chat. –  David Z Sep 4 '11 at 21:30
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6 Answers

up vote 22 down vote accepted

It's also worth noting that there is nothing special about atoms.

If you have any system where in every period of time an event has a certain chance of happening which only depends on internal effects of the object and no memory or communications with others - you will get the same decay curve. It's purely a matter of the statistics.

If you have a handful of coins and every minute toss them all and remove all the heads into a separate pile - the number of coins remaining in the hand will decay with a half-life of 1 minute.

What is special about carbon14 - and why it is useful for archeaology is that new carbon14 is being made all the time in the atmosphere, and while you are alive you take in this new carbon so the decays don't have any effect until you die. It's like tossing the coins, but while you are alive adding new random coins after each toss - but then when you die have somebody else start to remove the heads. If you assume you died with an equal number of heads and tails, you can work out how many tosses have happened since you died - and so how long ago the sample died.

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+1 for the handful of coins analogy. The entire reason for exponential decay is that there is less stuff left that can decay. –  JollyJoker Sep 2 '11 at 5:32
    
Out of all of the explanations, yours makes the most sense :) Thank you. –  Jonathon Byrd Sep 2 '11 at 15:37
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AC's answer below is better - if the atoms/coins are independant and have no way of effecting or signaling each other to decay then exponential is the only possible function. It takes a bit of thinking about to believe it though! –  Martin Beckett Sep 2 '11 at 15:47
    
I still desire a little bit more comprehensiveness to establish why a random event would have a pdf of occurring at time t of $p(t) = \lambda e^{-\lambda t}$. The coin analog works well as an example with discrete math, but coming up with an intuitive example of a continuous process seems a little bit harder. The examples I can think of are unsavory, like 18th century warfare when they fired in the general direction of the other army but couldn't aim at an individual. In some war examples like that your chance of dying could follow that p(t). –  AlanSE Sep 2 '11 at 16:09
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The derivations of exponential decay rates above can also be rendered into a very intuitive physical idea: "The rate of decay directly depends upon the number of atoms present at the very moment"

Hence $r$ is directly proportional to amount. That is, if $a$ is the amount of carbon and $k$ is some constant,, $\frac{da}{dt}$ is proportional to $a$

Or

$r= \frac{da}{dt} = k\cdot a$

$\frac{1}{a}da = k\cdot dt$

Performing integration (indefinite, just for simplicity)

$\log(a) = kt$

Or

$a = e^{kt}$

Which is exponential decay (generally with a negative $k$)

So, the fact that decay of carbon (or for that matter any element) can be modeled using exponential equation is just common sense. Rate depends upon amount so the decay is exponential. However, this is a statistical model and applies to a huge sample of atoms. It cannot predict the behavior of an individual atom and it's decay pattern. That is still considered VERY random. And hence Schrodinger's cat experiment is born.

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How do we know that C14 decay's exponentially compared to linear

Here's an argument that might help: suppose, temporarily, that radioactive decay was linear. Let's say you started out with a sample, call it sample #1, of a billion atoms in a box, 5700 years ago (that's one half-life). By the current day, half of them would have decayed, so you'd have 500 million atoms left.

Now, let's say you take a different sample (sample #2) of 500 million atoms and put it in another box. Then wait 5700 years. According to the linear decay model, sample #1 would be entirely gone, but sample #2 would still have 250 million atoms. But if you think about it, that doesn't make sense, because if you jump back to the present, there is nothing to distinguish sample #1 from sample #2. Each of them consists of 500 million radioactive ${}^{14}\mathrm{C}$ atoms in a box. So there's no reason that sample #1 should behave any differently from sample #2. It doesn't "remember" that it came from an earlier sample of a billion atoms.

and have there been any studies to verify this?

I'm not sure if it has been explicitly verified for carbon 14, but using other radioactive elements with shorter half-lives, it has been verified probably hundreds of thousands of times over the past century or so that they decay exponentially. It's a pretty simple experiment: you just measure the number of atoms that decay in a short interval of time $\Delta t$ using a Geiger counter or something similar, then wait some time $T > \Delta t$, then again measure the number of atoms that decay in the interval $\Delta t$. The second measurement will give you fewer decays than the first, which is a sign of nonlinear decay. Making more measurements of this kind will reveal that the decay is exponential.

To see this quantitatively, you can basically just reverse the math in Chris's answer. Suppose that the number of atoms remaining after time $t$ is $N(t)$. The number of atoms which decay in a short time interval $\Delta t \ll t_{\frac{1}{2}}$ is, on average, $N(t) - N(t + \Delta t)$, which means that the rate of atoms decaying (i.e decays per minute) is

$$R(t) = \frac{N(t) - N(t + \Delta t)}{\Delta t} \approx -\frac{\mathrm{d}N(t)}{\mathrm{d}t}$$

For exponential decay, this means that

$$R(t) = -\frac{\mathrm{d}}{\mathrm{d}t}N_0 e^{-\lambda t} = \lambda N_0 e^{-\lambda t}$$

and for linear decay, it would be

$$R(t) = -\frac{\mathrm{d}}{\mathrm{d}t}(N_0 - \beta t) = \beta$$

So if radioactive elements underwent linear decay, the rate of decay would be constant, which is definitely not what is observed.

It's possible (and easy) to calculate that $t_{\frac{1}{2}} = \frac{\ln 2}{\lambda}$, so for elements that have a long half-life, $\lambda$ is pretty small. This means in turn that the decay rate $R(t)$ (what you measure) is low and can be difficult to detect. Furthermore, not only is $R(t)$ small, but the change in $R(t)$ over a medium-length period of time is even smaller. This is why direct measurements of the half life of ${}^{14}\mathrm{C}$ are somewhat tricky: the decay rate drops by only a hundredth of a percent per year. But I wouldn't be too surprised if it's been done. I can try to look for a reference if you like. (In practice, there are other ways to calculate the half-life indirectly.)

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Nice explanation with the boxes. –  David Sep 2 '11 at 2:13
    
I think your analogy with the boxes is wrong. If you assume a linear decay, then for C14 it might be that 500 million atoms decay in 5700 years. That is about 88 thousand per year. This would then be a property of C14, so if you create an other box with 500 million atoms, the two boxes would both be decayed after another 5700 years. –  Ruud v A Sep 7 '11 at 15:44
    
@Ruud: it's not an analogy, it's a demonstration of why the linear decay model is inconsistent with the idea of a half-life. But yes, if you don't assume that radioactive decay has a constant half-life, then the explanation with the boxes doesn't require exponential decay. (I realized that after posting but I forgot to come back and clarify it) –  David Z Sep 7 '11 at 17:13
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Galileo would be proud! @Ruud: So 2 separate boxes with 500M atoms would decay entirely in 5700y, but if you put them together half would remain. That seems a bit strange, to say the least... –  yatima2975 Sep 8 '11 at 10:55
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Re: How do we know that C14 decay's exponentially compared to linear?

Experiment.

Re: Have there been any studies to verify this?

Yes. You can read a nice writeup of some of the early studies of C14 at the nobel prize website: http://static.nobelprize.org/nobel_prizes/chemistry/laureates/1960/libby-lecture.pdf


But to expand a bit on the underlying model as to why all radioactive decay (and spontaneous emission in general) is exponential in nature:

Radioactive decay is exponential is because the nuclei have no memory. Nuclei don't grow old like people - an 80-year-old man is very different than a 2-year-old boy, and their expected remaining lifetimes are very different. But all nuclei (until the moment they decay) are perfectly identical: they have no memory of when they were born. You can't distinguish a C14 atom that was created 10 seconds ago from one lucky atom that has managed to survive for 10000 years.

Consequently, every C14 nucleus has the same probability of decaying at a given moment in time. Thus, if $N$ is the number of remaining C14 atoms in your sample, $$\frac{\partial N}{\partial t} = -\Gamma \cdot N$$ where $\Gamma$ is some constant that describes how fast the atoms tend to decay. This equation, of course, gives rise to $N(t) = N_0 \cdot e^{-\Gamma t}$, which is just exponential decay.

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Because the half-life of C-14 is long compared to a human life time it takes careful, sophisticated measurements to demonstrate the exponential decay. However, it's probably worth noting that there are many other radioactive isotopes with very short half-lives. It a standard exercise in 1-year college physics labs to measure the exponential decay of radioactive isotopes of Silver. See for example carleton.edu/departments/PHAS/P128/lab/AgDecay04.pdf –  Charles E. Grant Sep 2 '11 at 1:24
    
@Charles E. Grant: I certainly agree: if all you wanted to do was demonstrate that radioactive decay is exponential in nature, other isotopes would be a much better choice. But because the lifetime of C-14 is so important (carbon dating, etc.), folks have gone to considerable effort to nail down the lifetime. And since the original questioner seemed fixated on C14 for some reason, I figured I'd put in a reference to data that showed the decay of C14, and tell the story in terms of that isotope. –  Anonymous Coward Sep 2 '11 at 2:08
    
The only big experimental difficulties are insuring that you don't have gain and acceptance drifts over the experimental time scale and getting enough C-14 in one place to let you ignore the counting statistics limits (i.e. get $10^{10}$--$10^{11}$ total counts and you're good to go). Not that this is trivial or that short-lived isotopes aren't easier, of course. –  dmckee Sep 3 '11 at 21:41
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We can show this by thinking about what is happening.

Suppose we have a set of $N$ nuclei that are all radioactive. Each of these nuclei has a chance of decaying, $\lambda$. In people lifetimes, some people live longer and some live shorter than others, but there is an average lifetime; this is what $\lambda^{-1}$ represents for nuclei.

Now how many nuclei, $\Delta N$, decay over some interval of time, $\Delta t$, can be represented mathematically by (remember $\Delta N < 0$ because nuclei are decaying), $$\Delta N = - \lambda N \Delta t$$

We can rewrite this in differential form as, $$\frac{d N}{N} = -\lambda dt$$

Using standard calculus, $$\int \frac{d N}{N} =- \int \lambda dt$$

We will find that, $$\ln N =- \lambda t + C$$

Standard algebraic manipulation gives,

$$ N = e^{-\lambda t}e^{C} $$

At $t=0$ we say that $N=e^C=N_0$, thus we usually write the equation: $$N = N_0 e^{-\lambda t}$$

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I would say the Activity reduces exponentially, which is a mathematical way of saying the amount of actvity is proportional to the amount of material left. Decay is a process that only involves individual atoms, and when something decays is a matter decided only by that atom.

If we have an ensemble of many atoms of the same type then it has been noted that the activity reduces over time, and follows a decreasing exponential.

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So if you would allow questions for me to better understand: The atoms are not always in a state of decay? But rather you cannot determine when they will decide to start decaying. Thus when you have a hand full of these atoms you can assume that you can average the results of their decay? sorry if that doesn't make sense. –  Jonathon Byrd Sep 1 '11 at 16:36
    
That sounds about right! Except I would say that they $are$ always in a state of decay, but different atoms have different half lives, so its only the shorter ones that we 'notice'. The activity is a function of both the number and half life. –  Nic Sep 1 '11 at 16:49
    
It doesn't seem like this would be a very solid method over a greater stretch of time. It seems like there's a lot of assumptions that have to be made. no? –  Jonathon Byrd Sep 1 '11 at 17:00
    
The 'Law of large numbers' somewhat guarantees that this 'averaging', or probabalistic approach is actually very accurate. Remeber we might be talking in excess of $10^20$ atoms. that is a huge number. –  Nic Sep 1 '11 at 17:05
    
Interesting, that is a huge number. I can see with that amount of excess and the fact that the atoms are always in a state of decay that you can begin to determine their age. I guess now I'm confused back at the beginning again. If all of the atoms are in a state of constant decay, did you just say that some C14 atoms have a longer rate of decay? I wouldn't mind talking more in chat either. –  Jonathon Byrd Sep 1 '11 at 17:09
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