Why do radioactive isotopes have a half-life? I know that they decay in order to become stable but why would it take out enough subatomic particles to be half? Or am I approaching this question the wrong way: the remaining radioactive substance is going to be half in x amount of time and the other half will be stable atoms.

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    $\begingroup$ Half life is a half life because it is defined this way. Wait less than that, and the decay will take out less than half atoms; wait more, and it will take out more than half. $\endgroup$ May 21, 2018 at 5:00
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    $\begingroup$ The half life is used as a convenient measure because a sample of isotopes decays exponentially with time. Each atom has the same chance of decaying in a fixed time interval and this leads to an exponential decay. All atoms behave the same way and continue to decay until all have been changed, whether this takes millennia or seconds. $\endgroup$
    – porphyrin
    May 21, 2018 at 6:45
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    $\begingroup$ You can just as easily define a "one-third" life, a "quarter-life". Radioactive isotopes decay exponentially; half-life is just convenient measure that captures the kinetics of the decay. $\endgroup$
    – getafix
    May 21, 2018 at 9:33
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    $\begingroup$ @Bluedragon01313 We generally discourage crossposting without at least mentioning that you have put the question in a different location. $\endgroup$
    – Tyberius
    May 21, 2018 at 17:53
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    $\begingroup$ Duplicate: How does a half-life work? $\endgroup$
    – Bohemian
    May 21, 2018 at 21:51

2 Answers 2


As commenters have pointed out, there is nothing about half-life that makes it special; we could as easily measure and tabulate one-tenth lives or one-fifth-lives.

Half-life has two marginal advantages: the half that has decomposed and the half that hasn't are equal, so whether we are referring to remaining or already decomposed nuclei is unambiguous, while for a third-life we could be confused as to whether we are talking about the time it takes for a third of the nuclei to decompose or for all but a third of the nuclei to decompose; and experimentally speaking, half is convenient because it's distinctly smaller from the initial amount of nuclei but large enough to be easily measured compared to the initial amount. However, these two advantages are fairly unimportant, and if we had decided to measure and tabulate one-tenth lives for all radioactive isotopes, we would be fine.

What's conceptually more important is that we can use half-lives as a proxy for isotopic disintegration rates, because radioactive decay follows first-order kinetics. Nothing prevents us from using half-life to describe any kinetic process, such as a chemical reaction rate. However, most of the time that's of limited value, because in general, half-lives are not constant, but change depending on the amount of reagents. It is only for first-order kinetics (i.e. when the amount of nuclei decays exponentially with time) that half-lives (or any fractional life) are constant.

For a general kinetic equation with only one reagent $\ce{A}$,

$$ -\frac{\mathrm{d} [\ce{A}]}{\mathrm{d} t} = k [\ce{A}]^n$$

where $n$ is the kinetic order of the process, we can always define the half-life of the process by integrating from an initial amount of $\ce{A}$, $N_\ce{A}$, to half that amount:

$$ \int\limits_{N_\ce{A}}^{N_\ce{A}/2} \frac{\mathrm{d} [\ce{A}]}{[\ce{A}]^n} = \int\limits_{0}^{t_{1/2}} - k\,\mathrm{d} t $$

which means that the half-life can be expressed in terms of the integral on the left side:

$$ t_{1/2} = - \frac{1}{k} \int\limits_{N_\ce{A}}^{N_\ce{A}/2} \frac{\mathrm{d} [\ce{A}]}{[\ce{A}]^n} $$

This integral will in turn depend on the kinetic order $n$. While in general the half-life will take the form

$$ t_{1/2} = \frac{1}{k} \frac{1}{n-1} \left( (N_\ce{A}/2)^{1-n} - N_\ce{A}^{1-n} \right) $$

and therefore will be dependent on the starting conditions $N_\ce{A}$, and will change as $\ce{A}$ is consumed, for first-order kintetics ($n=1$)

$$ t_{1/2} = - \frac{1}{k} \left( \ln{|N_\ce{A}/2|} - \ln{|N_\ce{A}|} \right) = \frac{\ln{2}}{k} $$

so half-life does not change with the amount of $\ce{A}$.

As long as the decay is first-order (or exponential, which is the same), half-life is such a constant that we can measure and tabulate, as it will be the same under all possible circumstances; no matter if we are talking about tonnes of fissile material or a handful of nuclei, the value for half-life will be the same.

If we start with $N_\ce{A}$ radioactive nuclei, after a time $t_{1/2}$ you'll have half that amount, $N_\ce{A}/2$; since now you have $N_\ce{A}/2$, after another $t_{1/2}$ you'll have $N_\ce{A}/4$; after a third interval equal to $t_{1/2}$ you'll have $N_\ce{A}/8$, and so on.

This makes tabulating half-lives equivalent to tabulating kinetic coefficients $k$, as these can be transformed into each other quickly and unambiguously, while for other kintetic orders we'd need additional informations to perform that transformation.

Note as well that the last step also shows why we could arbitrarily decide to use any other fractional lives in the same way: if, for instance, we used one-tenth lives, we'd get

$$ t_{1/10} = \frac{\ln{10}}{k} $$

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    $\begingroup$ I think we should have measured decay based on one-tenth lives, measured as the time it takes for 1/10th of the material. Then we could say we "decimated" the material using the original definition of that word. A chemistry measurement and an English history lesson in one! $\endgroup$
    – Cort Ammon
    May 21, 2018 at 14:44
  • $\begingroup$ @CortAmmon Decimate means to destroy one-temth, and it's not clear if a tenth-life would mean one-tenth survives or has decayed. With half-life either definition works just fine. $\endgroup$
    – aeismail
    May 21, 2018 at 20:58
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    $\begingroup$ @aeismail Which is precisely why it would become an English history lesson! $\endgroup$
    – Cort Ammon
    May 21, 2018 at 21:11
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    $\begingroup$ We should really use 1/e-life, which is the mean life, i.e. mean lifetime of an atom before decay. It's more consistent with other measures like attenuation length or electrical time constant. $\endgroup$
    – user71659
    May 22, 2018 at 0:38
  • $\begingroup$ @user71659 I agree, and "pi is wrong" but we still use it ;-) $\endgroup$
    – uhoh
    May 22, 2018 at 7:16

After the half-life time, half of the atoms you initially had are still there good as new, the others have split off an alpha particle, undergone beta decay, and might even have decayed further.

Sit down 10000 people with three dices each, which they throw every five seconds, and tell them they can leave when they throw a triple of sixes. After a certain time $t_{1/2}$, which I could calculate if I weren't too lazy now, about half of your people will be gone. After another $t_{1/2}$, approximately half of the rest will have left, and so on. With a hundred people, the prediction might be off, but of atoms, you usually have at least billions.

Problem with atoms is I don't know their game. But they always play the same one, and I always measure the same half-life time for the same kind of atom (isotope).

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    $\begingroup$ I don't think your dice example explains what it needs to explain. Any gradual process reaches a point where it's half-complete. The question is why is this a useful measure. The key point with the dice (and radioactive decay) is that, if it takes an hour for half of the people to leave, it will take an hour for half of the remaining people to leave, and so on. This is in contrast to, e.g., the time taken for half of those people to die: if you take 100 people in their 20s, it will probably take 50 years for half of them to die, but it won't take 50 years for half of the remainder to die! $\endgroup$ May 21, 2018 at 16:50
  • $\begingroup$ @DavidRicherby good point $\endgroup$
    – Karl
    May 21, 2018 at 18:06

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