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I wondered this since my teacher told us about half life of radioactive materials back in school. It seems intuitive to me to think this way, but I wonder if there's a deeper explanation which proves me wrong.

When many atoms are involved, half life can statistically hold, but since decaying of an individual atom is completely random and stateless, can't all the atoms in a 1 kg of matter just decide to decay in the next minute, even if the probability of this event occurring is extremely small?

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    $\begingroup$ Hi everybody -- I have deleted several comments that contained partial or complete answers to the question. Please use comments to suggest improvements or request clarifications only. If you have an answer, post it as an answer. Thanks! $\endgroup$ – tpg2114 Jan 11 at 13:03
  • $\begingroup$ Have you ever heard of a nuclear reactor? $\endgroup$ – Hot Licks Jan 11 at 13:57
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    $\begingroup$ @HotLicks: nuclear fission in reactors has to be induced by neutron capture; so not really the same situation. $\endgroup$ – Zorawar Jan 11 at 19:22
  • $\begingroup$ Did you try to do a back-of-an-envelope-calculation? $\endgroup$ – Qmechanic Jan 12 at 16:54
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The short answer is yes. No matter how many atoms there are, there is always a (sometimes vanishingly small) chance that all of them decay in the next minute. The fun answer is actually seeing how small this probability gets for large numbers of atoms.

Let's take iodine-131, which I chose because it has the reasonable half-life of around $8$ days = $\text{691,200}$ seconds. Now $1$ kg of iodine-131 will have around $7.63 \times N_A$ atoms in it, where $N_A$ is Avogadro's constant. Using the formula for probability for the decay of an atom in time $t$:

$$ P(t) = 1-\exp(-\lambda t), $$

and assuming that all decays are statistically independent$^\dagger$, the probability that all the atoms will have decayed in one minute is:

$$ (1-\exp(-\lambda \times 60\,\text{s}))^{7.63\times N_A} $$

where $\lambda$ is the decay constant, equal to $\frac{\ln 2}{\text{half-life}}$, in this case, almost exactly $10^{-6}\,\text{s}^{–1}$. So $$ P = (1-\exp(-6\times10^{-5}))^{7.63\times N_A} \\ \approx(6\times10^{-5})^{7.63\times N_A} \\ \approx (10^{-4.22})^{7.63\times N_A} \\ = 10^{-4.22\times7.63\times N_A} \\ \approx 10^{-1.94\times10^{25}} $$

(I chose iodine-131 as a concrete example, but pretty much any radioactive atom will result in a similar probability, no matter what the mass or the half-life is.) So if you played out this experiment on $10^{1.94\times10^{25}}$ such setups, you would expect all the atoms to decay in one of the setups, on average.

To give you an idea of how incomprehensibly large this number is, there are "only" $10^{78}$ atoms in the universe - that's $1$ followed by $78$ zeroes. $10^{1.94\times10^{25}}$ is $1$ followed by over a million billion billion zeroes. I'd much rather bet on horses.


$^\dagger$ This Poisson distribution model is a simplifying, but perhaps crude approximation in this scenario, since even small deviations from statistical independence can add up to large suppressing factors given the number of atoms, and so $10^{1.94\times10^{25}}$ is certainly an upper bound (of course, the approximation is fully justified if the atoms are separated to infinity at $0 \text{ K}$, or their decay products do not have sufficient energy to make more than a $1/N_A$-order change in the decay probability of other atoms). A more detailed analysis would have to be tailored specifically to the isotope under consideration - or a next-order approximation could be made by making the decay constant $\lambda$ a strictly increasing function of time. Rest assured that the true probability, while much more difficult to calculate than this back-of-the-envelope estimation, will still run into the mind-bogglingly large territory of $1$ in $1$ followed by several trillions of zeroes.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – tpg2114 Jan 13 at 18:58
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TLDR: statistical models are models, and thus by definition not a perfect reflection of reality.

Nihar's answer is good but I'm going to tackle it from a different direction.

First off, if we only look at statistical mechanics you can run through the math and of course you will find an extremely small probability. You might stop there. But statistical mechanics uses statistical models, and all models are wrong. They make assumptions and necessarily simplify reality to solve complicated problems. There could very well be some physical processes unaccounted for in statistical mechanics that negate any possibility of such a rapid decay.

A classic example is having a room and figuring out the probability that all the oxygen all of a sudden is only in one half of the room. From a stat mechanics standpoint, it's basically the probability of flipping a fair coin an unimaginably large number of times and having them all land the same way. But in reality, the unimaginably small number you would calculate wouldn't actually be correct, because the assumptions made by your model wouldn't perfectly reflect reality (particles interact with each other, for one). Much like the ideal gas law, these things are useful but can completely fail if you deviate too far from the assumptions made. This is true of all statistical models, of course.

So if we assume that the stat model of half life is a completely accurate representation of reality, the answer to your question is technically yes. Of course we know it's not, so that leads me to my final point.

There's also a heavy philosophical component to these sorts of questions since we are dealing with probabilities that are so small they are effectively 0. If someone flips a coin a billion times and it lands tails every time no one is going to think it's a fair coin, because it's obviously not*. You could also consider state of the art cryptography. The odds of successfully randomly guessing a key is so low that for all intents and purposes it is 0. Or imagine watching a video of a bunch of shattered glass forming into a vase. Your conclusion wouldn't be 'see ya thermodynamics, wouldn't want to be ya', it would be 'I'm watching a video of a vase shattering in reverse'. Yes, there are technically tiny probabilities associated with these events but it is so small that saying they are technically possible is more of a philosophical statement than anything else.

* The idea of a fair coin is a rabbit hole on its own. How do you determine that a coin is fair? By tossing it a bunch of times and observing a nearly equal number of tails and heads. If it deviates too much from 50/50, we declare it to be biased. But of course no matter what outcome we observe, there's always a chance it was a fair coin, so technically we can never know for sure. In order to make use of statistics then, we must arbitrarily pick a cut off point for random chance. Usually this is 2 sigma, maybe 3. CERN uses 5 sigma for new particle detection but again, this is arbitrary. Applied statistics is very much an art as much as it a branch of math.

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    $\begingroup$ I cannot be more in agreement with @eps. The odds of this happening within the model are so tiny that problems with the modelling assumptions dominate the odds of witnessing such an event. $\endgroup$ – Paul Young Jan 10 at 0:10
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    $\begingroup$ "So if we assume that the stat model of half life is a completely accurate representation of reality, the answer to your question is technically yes. Of course we know it's not, so that leads me to my final point." Could you add to your answer some references to the experiments invalidating the idea of the random decay model that is usually taught (decay being random following a Poisson distribution)? $\endgroup$ – orlp Jan 10 at 0:55
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    $\begingroup$ Yeah I’d really like to know what you think is wrong with the model in this particular case. $\endgroup$ – kaylimekay Jan 10 at 10:31
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    $\begingroup$ Yes, arguing that models can be wrong with some probability larger than the probability they predict does not automatically mean their prediction is unreliable; things are more subtle (I have a paper on it, arxiv.org/abs/0810.5515 ). $\endgroup$ – Anders Sandberg Jan 10 at 10:40
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    $\begingroup$ Mandatory Dilbert: assets.amuniversal.com/321a39e06d6401301d80001dd8b71c47 $\endgroup$ – Peter - Reinstate Monica Jan 10 at 20:38
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One thing to keep in mind is that this is not only a statistics question and the analogy of atoms decaying and flipping coins can be misleading.

For example, uranium 235 has a half life of more than 700 million years, but when brought in the right configuration (close packed) and in the right amount (above critical mass), it decays practically in an instant... Simply because one atom decaying can trigger another to decay and so on in a chain reaction.

So, if you can assume that all decays happen independently of each other, then the answers based purely on statistics are valid. If more physics than statistics is involved, then it depends on the exact material, i.e. what material, is it pure, in what configuration, etc.

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    $\begingroup$ That (decays practically in an instant) sounds like it could be violent, even dangerous. =) $\endgroup$ – Kevin Fegan Jan 10 at 3:14
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    $\begingroup$ @KevinFegan not particularly. We just make bombs out of it sometimes ;) $\endgroup$ – DividedByZero Jan 10 at 13:56
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    $\begingroup$ Mostly, $\mathrm{^{235}U}$ decays by $\alpha$ emission. Those particles have relatively low energy, so are very unlikely to induce further nuclear reactions. Only $2.0×10^{-7}$% of its decays are SF, (spontaneous fission) en.wikipedia.org/wiki/…, but fission releases 2 or 3 neutrons, and those neutrons can induce other $\mathrm{^{235}U}$ nuclei to fission. The SF half-life is around $3.5×10^{17}$ years (compared to $7.04×10^8$ years for the overall half-life), but that figure is rarely mentioned because fission is so dependent on geometry. $\endgroup$ – PM 2Ring Jan 11 at 9:34
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    $\begingroup$ It doesn't really "decay in an instant". That phrase suggests the particles decay. Aloha decay, beta decay, whavmtever. In U-235's case of cjain reaction, almost all the atoms are bombarded to bits from outside (each atom is impacted by neutrons from the wider world beyond that atom). Very very few "decayed". Its a bit different. By very crude analogy you wouldn't describe a human blown to bits by a landmine as "decayng" either. We use that word for very specific kinds of gradual breaking down. Sudden outside impact is not usually considered that. $\endgroup$ – Stilez Jan 11 at 23:57
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The answer is 'no'. This 'no' is on the same level like:

  • Can it happen that you float for 15 minutes in the middle of your room. (Statistical mechanics tells technically yes, but again with a for all practical purposes zero probability)
  • Can you put a monkey in front of a typewriter and get Shakespeare novels out of it?
  • Can you walk through a solid wall (non zero tunnel probability due to quantum mechanics)
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    $\begingroup$ The last one is actually not true. Quantum tunneling is not possible on macroscopic levels due to the quantum decoherence. $\endgroup$ – Petr Fiedler Jan 11 at 17:38
  • $\begingroup$ Re "...get Shakespeare novels out of it": You can do a lot of things with the infinite improbability generator. $\endgroup$ – Peter Mortensen Jan 11 at 19:58
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    $\begingroup$ Shakespeare was pretty good at writing plays and poems. Pity he never wrote any novels... $\endgroup$ – PM 2Ring Jan 13 at 1:41
  • $\begingroup$ @PetrFiedler Welcome on StackExchange $\endgroup$ – HolgerFiedler Jan 14 at 5:57
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I see that people on this site mostly seem to think you can just multiply numbers together to get probabilities, and thus the answer is that the probability is something of order $10^{-10^{25}}$.

The trouble with this is that the decay events are not entirely independent events, so this method of calculation is wrong. It is ok as a first very VERY rough approximation, and the answer will certainly be a tiny number, but the answer will not be this particular tiny number. You will see by reading on why I put the second "very" in capitals.

There are cooperative effects throughout physics. For example, in the decaying solid the particles emitted by any one nucleus will disturb the others. This is a tiny effect, but when we are considering events of tiny probability we have to think about such tiny effects. Another factor is the surrounding electromagnetic field, which may be in a thermal state, but even in its vacuum state it produces correlated effects across the sample. Electromagnetic fields have almost no effect on radioactive decay, but anything that can affect all the nuclei at once will have a non-negligible influence compared to the tiny numbers that emerge from any assumption that all the nuclei behave independently.

Let's get some rough feeling for the influence of these cooperative effects. For $n$ independent events, each of probability $p_0$, the overall probability is $p_0^n$. But suppose that if one event happens, then the probability for the others is increased a tiny bit, from $p_0$ to $p_1 = p_0(1 + \epsilon)$ for some very small $\epsilon$. If those further events were independent then now the overall probability is of order $p_0 p_1^{n-1}$. This is larger than $p_0^n$ by the ratio $$ \frac{p_0 (p_0 + \epsilon p_0)^{n-1}}{p_0^n} = (1 + \epsilon)^{n-1} $$ With $n$ of the order of Avogadro's number, you can see that values of $\epsilon$ of the order of $1/N_A$ would suffice to introduce a non-negligible increase in the overall probability, where by "non-negligible" I mean "by a factor of order $1$". But the overall probability remains tiny.

That was just one atom influencing the others. If they each have that kind of effect then one gets the $(1 + \epsilon)$ factor raised to a power of order $N_A^2$. So by this sort of argument the number $10^{-10^{25}}$ which I started with is wrong by a factor which could easily be as big as $2^{N_A}$. I'm not trying to state the imprecision with any care. I'm just saying that the calculation based on $N_A$ independent processes gives a final answer which is wrong by an enormous factor.

Let's consider next some sort of cooperative effect such as a fluctuation in the electromagnetic field sufficient to stimulate all the nuclei, enough to get them over the energy barrier so the electron or alpha particle or whatever can escape. To disturb nuclei one needs energies of order mega-electron volt, whereas at room temperature the thermal radiation has photons of energies of order $k_B T \simeq 0.026$ eV. But if we trust the Boltzmann factor then we might roughly estimate a chance of $\exp(-E/k_B T)$ to get an excitation of a mode of energy $E$. With $E = 1$ MeV that gives $\exp(-4 \times 10^7)$ at room temperature. With "all these" gamma ray photons around, the radioactive decay process is going to happen slightly differently. Of course this probability is again tiny, but it is vastly larger than $10^{-10^{25}}$, so it has to be taken into consideration prior to announcing that that latter number is even close to right. This is because even the tiniest amount of any sort of correlation or cooperative effect will be sufficient to overwhelm the probability of multiple independent events.

One could estimate the effect of these thermal gamma rays by finding out the cross-section for gamma-stimulated decay and doing a scattering calculation. I don't know the answer but it will be huge compared to $10^{-10^{25}}$.

In summary, the short answer to the originally posed question is "no, that can't happen". The longer answer then admits that physics suggests there is a non-zero very very small probability that it could happen, just as there is for a number of other bizarre occurrences. For the value of the probability, no quick calculation can get even close to the right order of magnitude. To estimate it, first one does the independent-decay calculation to satisfy oneself that that is not the most likely route by which it could happen. Then one is left with the much more difficult problem of thinking what sort of physical effects can cause several nuclei to decay at once, and estimating those. I think the answer must be small compared to that number $\exp(-4 \times 10^7)$ which I mentioned above, but I have little notion of what the probability really is. Maybe as low as $10^{-10^{10}}$?

Perhaps it might be valuable to re-emphasize the point I am making. When we calculate more ordinary physical scenarios, such as a body sliding down a slope or a pendulum or an atom etc., we correctly ignore any negligible effects such as the gravitational attraction to planets lightyears away or other such things, and focus on the main contribution. In a similar way, in the present case a correct approach will simply recognize as negligible the contribution to the probability owing to all the nuclei just happening to decay in the same minute, and focus on the much larger probabilities associated with other ways in which the outcome can happen. A calculation which does not do this is, simply, wrong. It is like stating that a time is of the order of 1 femtosecond when in fact it is of the order of 1 petasecond. That would not be regarded as a reasonable estimate, but simply wrong, and by an embarrassingly large factor.

If we want to understand what goes on in real-world processes, as opposed to idealized models, then real-world processes are what we have to think about.

Finally, I want to re-emphasize that the effects I have mentioned are indeed vanishingly small. But in comparison to $10^{-10^{25}}$ they are enormous.

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    $\begingroup$ I'd be interested in reading more if you can add some sources. From what I've seen the consensus over the last century has been that there isn't an effect on the decay rate. There are some papers claiming double-digit percent differences based on other factors, but I haven't seen anything like you say. I would think enough study has been done on chain reactions that we would have a better idea if that could be a factor. Little Boy released about 10x the energy in gamma rays in 10 nanoseconds as this would produce in 60,000,000,000 nanoseconds, and that doesn't seem to have affected it. $\endgroup$ – Jason Goemaat Jan 13 at 17:19
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    $\begingroup$ The question was asking 'can't all the atoms in a 1 kg of matter just decide to decay in the next minute,' so I think treating them as independent events is warranted. Even if not, then the energy release would blow the mass apart, lowering the energy density. I think the Little Boy bomb's chain reaction only lasted 10 nanoseconds and only involved the decay of 1kg of the 64kg of enriched uranium because of this. Fat Man did better because the implosion kept the chain reaction going for hundreds of nanoseconds. $\endgroup$ – Jason Goemaat Jan 13 at 18:35
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    $\begingroup$ @JasonGoemaat My response is driven by (1) the desire to think it through and ask what is genuinely the case in the physical world, not in some idealized model; and (2) to alert people to the fact that correlated effects, even very small, will dominate over an accumulation of uncorrelated ones in the limit of large numbers (Avogadro's number being sufficiently large in the present context). So much so that the idealized model becomes simply irrelevant, the same way that we ignore negligible contributions when we do any other calculation in science. I added a remark at the end of my answer. $\endgroup$ – Andrew Steane Jan 13 at 21:36
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In order for that to happen in the real world you need to start with about 3.8 million kilograms of that material.

Here is how you come up with that number. You start from the the formula connecting the half-life to the number of particles over time

$$ N(t) = N_0 \left(\frac{1}{2}\right)^\frac{t}{t_{1/2}} $$

Now you replace $N(t)$ with what you would like to have $$ N_0 - 1~\text{kg} = N_0 \left(\frac{1}{2}\right)^\frac{t}{t_{1/2}} $$ And you solve for $N_0$ $$ N_0 = \frac{1~\text{kg}}{1-\left(\frac{1}{2}\right)^\frac{t}{t_{1/2}}}$$ At this point is just a matter of plugging in $t=60~\text{s}$ and $t_{1/2}=5~\text{y}$.

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  • $\begingroup$ I think the question was about a "piece" of 1kg decaying in one Minute (so it is not part of a bigger compound), but this answer seems interesting $\endgroup$ – Jonas Jan 11 at 16:51
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    $\begingroup$ @Jonas I thought that too, but it is not clearly stated in the question so I just offered a different point of view to the matter. $\endgroup$ – DarioP Jan 11 at 16:54
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@Nihar has an excellent answer: It's possible but with a chance of 1 in $10^{1.94\times10^{25}}$

That is a truly large number. When you use exponents that need to be represented with their own exponents, it can sometimes be difficult to think about what they actually mean. for some perspective:

  • There are about $5\times10^{19}$ atoms in a grain of sand
  • There are about $8\times10^{18}$ grains of sand in the world
  • That's about $4\times10^{38}$ atoms in all the sand in the world
  • There are about $1.33\times10^{50}$ atoms of all kinds in the world
  • There are about $10^{56}$ atoms in the solar system
  • There are between $10^{78}$ and $10^{82}$ atoms in the universe

Using the largest estimate of $1\times10^{82}$ atoms in the universe, we've gone only from an exponent of 19 to 82 comparing a grain of sand and the entire universe. This exponent is 1,940,000,000,000,000,000,000,000,000.

How many trials would we have to do to get a reasonable chance of this happening? The formula for figuring out the odds of a random event happening at least once is $1-(1-P)^y$ where P is the probability $1/{10^{1.94\times10^{25}}}$. I couldn't find any app that would give sensical results given large values for y, but if y = P then the odds approach ${-(1-e)}/e$ as P gets large. That's about 63.2%. So if we do $10^{1.94\times10^{25}}$ trials, there's about a 63.2% chance of it happening at least once and about a 37.8% chance of it not happening at all.

So how can we imagine doing $10^{1.94\times10^{25}}$ trials?

If we take all the atoms in the universe and change them all into separate 1kg bundles of iodine-131, we would have about $2.2\times10^{57}$ of them. Spread out over the volume of the visible universe ($3.57\times10^{80} m^3$), that's one bundle every $1.6\times10^{23}$ cubic meters, that's a cube that is 57,000 kilometers per side with a 1kg bundle of iodine-133 in the center. The age of the universe is estimated at 13.772 billion years, that's about $7.24\times10^{15}$ minutes. If we took all those bundles of iodine-133 and re-ran our experiment every minute (converting the decayed atoms back to iodine-131 for each trial) from the big bang until now, that's about $1.6\times10^{73}$ individual trials.

That exponent of 73 is no-where near the exponent we need to achieve a 63.2% chance of it happening. There would need to be about $2.66\times10^{23}$ universes of atoms converted to iodine-131 re-running the experiment each minute for 13.777 billion years to have a 63.2% chance of it happening at least once.

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  • $\begingroup$ I agree with your instinct here, to bring home to us quite what is meant by a number such as $10^{10^{25}}$. However, you should keep in mind that that number is itself incorrect as an estimate of the probability here. It is wrong by a factor of order $10^{10^{25}}$. $\endgroup$ – Andrew Steane Jan 13 at 13:17
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To understand this, you need to see what triggers a nuclear decay. The answer is a beautiful example of quantum mechanical behavior. Nothing triggers it. It is just that the world is fundamentally quantum mechanical, and probabilistic.

All the other answers that "no, there is no triggering event, it just happens, quantum mechanics is like that" are perfectly right.

What happens before a radioactive element decays?

All you can do is calculate the probabilities.

So the answer to your question is, that yes, there is a non-zero probability for the material to decay in the next minute.

But your question is more about if there is a chance that all the atoms in the material decay simultaneously in the next minute. And the answer is again yes, there is a non-zero probability for that to happen, but it just happens so that the probability is so little, that even on giant timescales like the age of our universe, there is very little probability for us to observe that to happen.

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