Carbon-14 has a half-life of 5,730 years. That means that after 5,730 years, half of that sample decays. After another 5,730 years, a quarter of the original sample decays (and the cycle goes on and on, and one could use virtually any radioactive isotope). Why is this so? Logically, shouldn't it take 2,865 years for the quarter to decay, rather than 5,730?

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    $\begingroup$ If you like this question you might also enjoy reading physics.stackexchange.com/q/7584/2451 $\endgroup$
    – Qmechanic
    Commented Sep 1, 2011 at 15:53
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    $\begingroup$ "Logically, shouldn't it take 2,865 years for the quarter to decay, rather than 5,730?" Why? $\endgroup$ Commented Mar 6, 2014 at 3:21
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    $\begingroup$ 2,865 is half of 5,730. If it takes 5,730 years for half the sample to decay, then half of a half (one quarter) should take half the time of half $\endgroup$
    – user37390
    Commented Mar 6, 2014 at 3:37
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    $\begingroup$ After the first half-life, half of the original sample is the new whole sample. It still has the original half-life. $\endgroup$ Commented Mar 6, 2014 at 3:40
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    $\begingroup$ Related: physics.stackexchange.com/q/30185 $\endgroup$ Commented Mar 6, 2014 at 3:51

14 Answers 14


The right way to think about this is that, over 5,730 years, each single carbon-14 atom has a 50% chance of decaying. Since a typical sample has a huge number of atoms1, and since they decay more or less independently2, we can statistically say, with a very high accuracy, that after 5,730 years half of all the original carbon-14 atoms will have decayed, while the rest still remain.

To answer your next natural question, no, this does not mean that the remaining carbon-14 atoms would be "just about to decay". Generally speaking, atomic nuclei do not have a memory3: as long as it has not decayed, a carbon-14 nucleus created yesterday is exactly identical to one created a year ago or 10,000 years ago or even a million years ago. All those nuclei, if they're still around today, have the same 50% probability of decaying within the next 5,730 years.

If you like, you could imagine each carbon-14 nucleus repeatedly tossing a very biased imaginary coin very fast (faster than we could possibly measure): on each toss, with a very, very tiny chance, the coin comes up heads and the nucleus decays; otherwise, it comes up tails, and the nucleus stays together for now. Over a period of, say, a second or a day, the odds of any of the coin tosses coming up heads are still tiny — but, over 5,730 years, the many, many tiny odds gradually add up to a cumulative decay probability of about 50%.

1 A gram of carbon contains about 0.08 moles, or about 5 × 1022 atoms. In a typical natural sample, about one in a trillion (1 / 1012) of these will be carbon-14, giving us about 50 billion (5 × 1010) carbon-14 atoms in each gram of carbon.

2 Induced radioactive decay does occur, most notably in fission chain reactions. Carbon-14, however, undergoes spontaneous β decay, whose rate is not normally affected by external influences to any significant degree.

3 Nuclear isomers and other excited nuclear states do exist, so it's not quite right to say that all nuclei of a given isotope are always identical. Still, even these can, in practice, be effectively modeled as discrete states, with spontaneous transitions between different states occurring randomly with a fixed rate over time, just as nuclear decay events do.

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    $\begingroup$ Dammit, accepted. And just when I was this close to getting a Populist badge... ;-) $\endgroup$ Commented Mar 7, 2014 at 0:07
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    $\begingroup$ My homeland elected populists and it's not in good shape. Great answer, though. There is also the question physics.stackexchange.com/q/69448 . My answer gives examples of "partilces" with internal states that do have memory. $\endgroup$ Commented Mar 31, 2014 at 23:11

I know exactly where you're coming from. If I can put it into my own words: If it takes a sample some amount of time to decay, shouldn't a sample of half the size take half the time to decay? I have fallen into this seemingly sensical but somehow incorrect belief more than once.

Here's a graph that shows what I believe you're currently thinking.

enter image description here

The horizontal axis is time. On the vertical I graph amount of sample left. This graph would be true if half the sample took half the time to decay. (Can you see this in the graph? Look at $t=T/2$ where time $T$ is when the sample is gone.) I think this does make sense in a way, but this isn't how nature works.

Now here's a graph of what actually happens.

enter image description here

This graph is "exponentially decaying" It is a consequence of the following: A sample half the size will decay at half the rate. This also makes sense in a way (thankfully): If you have half the sample size, you'll have half the decay rate. In contrast, note that the first graph has the constant rate of decay, no matter the size of the sample (that is, a constant slope).

So these two possibilities are mutually exclusive: Either the rate of decay is constant regardless of the size (first graph), or the rate of decay is proportional to the sample size (second graph). Observation shows the second graph is correct.

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    $\begingroup$ Right. Also, the lower graph is not exclusive to radioactive decay; it occurs all over the place. The temperature of cooling soup, the loudness of a tuning fork, you name it. It's just how nature works. $\endgroup$
    – Mr Lister
    Commented Mar 6, 2014 at 14:34

Logically, shouldn't it take 2,865 years for the quarter to decay, rather than 5,730?

Imagine that the quantity $q(n)$ of something decays as

$$q(n) = Q\cdot 2^{-n}$$

where $n$ is the number of half-lifes.

Initially, there is quantity $q(0) = Q\cdot 2^0 = Q$ of something.

After 1 half-life, there is $q(1) = Q \cdot 2^{-1} = \frac{Q}{2}$ remaining.

After 2 half-lifes, there is $q(2) = Q \cdot 2^{-2} = \frac{Q}{4}$ remaining.

After 3 half-lifes, there is $q(3) = Q \cdot 2^{-3} = \frac{Q}{8}$ remaining.

After 4 half-lifes...

Now, note that the quantity $\frac{q(n+1)}{q(n)} = \frac{1}{2}$ is constant.

That is to say that, given a quantity at any point in time (not just the "starting" point), one half-life later half of that quantity has decayed. This is the meaning of half-life.

From the Wikipedia article linked:

Half-life (t½) is the amount of time required for a quantity to fall to half its value as measured at the beginning of the time period.

  • $\begingroup$ most recent may be a better phrase than prior.. $\endgroup$
    – user13107
    Commented Mar 6, 2014 at 8:03

Suppose you start with two kilograms of C-14. After 5730 years you have one kilogram left. Call that piece A. Now get another kilogram of C-14, call it piece B, and put it next to piece A.

You now have two identical pieces of C-14, and yet one of them (A) is supposed to half-decay in 2865 years and the other (B) is supposed to half-decay in 5730 years? Do you see how that doesn't make sense?

Hopefully this convinces you that the rate at which a radioactive element decays can only depend on how much of it there is at that moment, not on how much of the original sample is left.

This is something I don't think any of the other answers explicitly brought up, but Nick Stauner alluded to it in a comment.

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    $\begingroup$ The example of splitting some piece in half is a good one, and helps indicate the difference between this type of decay which happens to "all the stuff" at once, as opposed to, say, a river eroding the surface of "some stuff". $\endgroup$ Commented Mar 6, 2014 at 19:50

As a crude analogy to give you some intuition, try the following: put 100 pennies in a shoebox, all with heads up. Shake the shoebox vigorously. Take out all the pennies that have changed to tails up. That's one half-life. Shake the box again, and again take out the pennies that are tails up. Repeat until there are no pennies left in the box.

The idea here is that heads up pennies represent carbon-14 atoms. The tails up pennies represent the atoms that have decayed. For any individual penny, each time you shake the box there is a 50/50 chance it will turn tails up, just as for every individual atom there is a 50-50 chance it will decay during one half-life period.


Half-life is used to describe exponential decay. What you're describing would be linear decay.

In one half-life period, on average, half of the C14 atoms would decay. So one would expect that if you start with four C14 atoms, you would after one half life have two, and after another half life only one would remain.

However, note that this process has a random component. You cannot predict exactly when an individual atom will decay. However once you have a larger number of atoms, you can make accurate predictions of how many will be left after a certain time period.


Basically, nuclear disintegration is probabilistic in its very nature. What it means is that one cannot say with conviction that, say, one atom kept on a table will disintegrate in, say, the next 1 minute. All one can say is that among a given sample of, say, 100 nuclei, 10% of it will disintegrate in the next 1 minute.

Nuclear disintegration follows what is known as first order kinetics which means that rate of reaction is directly proportional to the quantity of reactant present. In other words,

$$ d/dx(C) = -k C $$

where C is the current concentration of reactant and k is proportionality constant.

From this calculation, what one can get is a term called half-life, which means that after this time has elapsed, half of the concentration gets disintegrated (I'm using disintegrated and reacted interchangeably, since the reaction in nuclear disintegration is disintegration).

This means that a sample of 100 atoms after one half-life would remain 50 $=100 * (1/2)^1$, which after 2 half-lives would become 25 $=100 * (1/2)^2 = 50 * (1/2)^1$ and so on...


Suppose that decay worked the way you proposed, half as many atoms take half as long to decay. It sounds sort of plausible at first, but consider this: how does any one atom know when it's allowed to decay? It can't just roll a die and decay if it rolls a 1, it has to know how big the sample it's in is, and adjust its probability of decaying accordingly. If it didn't adjust its probability of decay, one would expect to exponential decay, because:

Thought experiment: roll a 20-sided die. If you roll a 1, it decays. How many rolls does it take to get to a 50% probability of decaying? (hint: it's not 10) How many rolls to get 100%?

Thought experiment 2: roll a hundred 20-sides dice. Any die that lands on 1 decays. How many do you likely have left after the first round? Should it take longer, on average, to make them all decay than if you had only 1?

Thought experiment 3: as many dice as there are atoms in a sizable chunk of material. How does it behave?

It should be clear that on average, you lose 5% of the dice that you had left (not of how many you started out with, which is not something the system can remember) per round - exponential decay. The "half-life" of these dice is about 13.5 rounds, that's how long it takes before approximately half are decayed.


I think you're confused simply by the language. Remember that it's a quarter of the original sample. So it's like compounding interest in the bank. You start with initial principal, once the interest is compounded, you might say that the percentage of that principal is ADDED TO the "principal", and then a percentage of THAT is calculated, and added to that second number. Similarly with nuclear decay, except you're subtracting, and you're subtracting an even half over a year instead of adding something like .05% every month (or whatever number banks use).

Half of that second sample is a quarter of the original. So you could express this fraction of the original as $\frac{1}{2^n}$ where $n = $the unit of time for your constant. In this case, a year. So for every year, $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}$, etc.


The mass of radioactive materials follows the ordinary differential equation: $$ m'(t)=-am(t), $$ where $m$ is the mass and $a$ a positive constant - i.e., constant relative rate of decay.

This implies $$ m(t)=m(0)\mathrm{e}^{-at}. \tag{1} $$ If $T_h$ is half life, then $$m(T_h)=m(0)\mathrm{e}^{-aT_h}=\frac{1}{2}m(0),$$ which implies that $$ T_h=\frac{\log 2}{a}, $$ and hence $(1)$ can be written also as $$ m(t)=2^{-t/T_h}m(0). $$ So the quarter-life is $T_Q$, for which $m(T_Q)=\frac{1}{4}m(0)$ or $$ m(T_Q)=2^{-T_Q/T_h}m(0)=\frac{1}{4}m(0), $$ which holds only if $T_Q=2T_h$!


Imagine a sample of 1000 atoms with a half-life of 1 hour.

That means every hour, the sample is reduced to 50% of its size.

After one hour, you are left with 500 atoms. How much time for that new sample (500 atoms) to be reduced to 50% (250 atoms) ?

In your interpretation :

For the new sample to be reduced to 50%, it needs to lose 250 atoms. Since it lost 500 atoms in 1hour, it should take 30 minutes to lose 250 atoms. And that's where you're wrong. It still needs 1 hour for half of the atoms to decay.

You assume that the number of atoms decaying by time (500/hour) is constant, but it isn't.

What is constant is the probability for each atom decaying in an hour : 50% (In this exemple, that means we can expect 250 atoms to have decayed after 1 hour, and it gets way more precise with "real" atoms number on a longer period)

  • $\begingroup$ I think it is more precise to say each atom has a 50% probability of decaying in an hour, and that probability is what's constant. The decay process has nothing to do with the presence of an ensemble of atoms, so a ratio doesn't really need to be considered. $\endgroup$
    – chase
    Commented Mar 6, 2014 at 10:31
  • $\begingroup$ @chase - Agreed and edited accordingly $\endgroup$
    – Helbrecht
    Commented Mar 6, 2014 at 10:47

putting it simply: activity law states that :

dn/dt is proportional to n. which means

the rate of reaction of any substance depends on the amount of the substance itself. because there is a greater amount of carbon initially, the probability of some of it decaying is higher , than when there is less amount of it left.


It's a single number for compounding decreases

Half lives are basically the same as "doubling time" in investing, but halving.

The Rule of 72: compounding increases

The "rule of 72" lets you estimate how long it takes your money to double, given a certain interest rate. For example, if you get 9% interest per year, 72/9 is 8, so according to the rule, your money would double in about 8 years.

You can see this doubling if you just multiply an amount by 1.09, then multiply the result by 1.09, on and on. (I'm not trying to be precise here, so I'm just using integers for the results.) Every 8 rounds, the amount is about double.

0:    100,000
1:    109,000
2:    118,810
3:    129,503
4:    141,158
5:    153,862
6:    167,710
7:    182,804
8:    199,256 - about double
9:    217,189
10:   236,736
11:   258,,043
12:   281,266
13:   306,580
14:   334,173
15:   364,248
16:   397,031 - about double again
17:   432,763
18:   471,712
19:   514,166
20:   560,441
21:   610,881
22:   665,860
23:   725,787
24:   791,108 - about double again
... and so on

Clearly this isn't super accurate; the rule of 72 is just an approximation that you can do in your head. See the Wikipedia article for the exact formula.

Half lives: compounding decreases

Half lives are the same idea, but with negative interest (imagine fees on a bank account or inflation). If an amount decreases 9% per year, it will halve in about 8 years, halve again in another 8 years, etc.

You can see this if you just multiply an amount by 0.91, then multiply the result by 0.91, on and on. (Again, I'm not trying to be precise here, so I'm just using integers for the results.) Every 8 rounds, the amount is about half.

0:    100,000
1:    92,000
2:    84,640
3:    77,869
4:    71,639
5:    65,908
6:    60,636
7:    55,785
8:    51,322 - about half
9:    47,216
10:   43,439
11:   39,964
12:   36,767
13:   33,825
14:   31,119
15:   28,630
16:   26,339 - about half again
17:   24,232
18:   22,294
19:   20,510
20:   18,869
21:   17,360
22:   15,971
23:   14,693
24:   13,518 - about half again
... and so on

Half lives are easier to compare than percentage decreases

So instead of half lives, we could use percentage decays. But decay rates very extremely widely, so unlike in financial calculations, where "per year" is always a reasonable time scale, we'd have to do one of two awkward things:

  1. Use the same unit for everything, like "percent decay per nanosecond", even though that would be a large percentage for some isotopes and an extremely tiny percentage for others
  2. Use both percentage and time units, like "this isotope decays by 1% per year, and this one by %3 per minute, and this one by 0.001% per millennium", or whatever.

If we instead use half lives, we have a single number that's easy to compare - eg, "half life of one millisecond" vs "half life of 10 billion years".


Half life 1=5,730 years while the ratio is 1:1

Half life 2= 11,460 years while the ratio is 1:3

Half life 3= 17,190 years while the ratio is 1:7

Half life 4= 22,920 years while the ratio is 1:15

To get the 2-4 half lives, you just add the 5,730 each time.

So how I got the second half life you would do 5,730+5,730=11,460

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    $\begingroup$ doesnt answer the question. $\endgroup$
    – jaromrax
    Commented Mar 3, 2017 at 14:57

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