Why is radioactive half-life constant?

Question

Say you have just four radioactive atoms with a half-life of one hour. (I am using a small number of atoms to keep it simple and illustrate my confusion more clearly). So that means one hour from now, two of the atoms will have decayed (on average) and two will remain undecayed (on average). Now, I am struggling to understand why the last two undecayed atoms won't, on average, both decay in the following one hour. After all, if it took one hour for the first two atoms to decay, then surely it should take one more hour for two more atoms to decay...

In general, if it takes x years for half of a sample to decay, shouldn't it then logically take another x years for the ENTIRE other half of the sample to decay? Obviously, this isn't the case, but I am struggling to understand why it isn't the case... It's almost as if the atoms in a sample somehow 'know' how many other atoms are in the sample...

Answer 1

It might help you think of it in terms of tossing coins or rolling dice. Say you toss a bunch of coins every minute and get rid of all the ones that turn up tails. The coin collection would have a half-life of a minute. The point is that the coins (atoms) don't know anything about how many others there are or how long they've been waiting around to decay.

Answer 2

It doesn't work on something like 4 atoms, or even a hundred atoms. It's statistical so you need many many atoms.

What you are doing is like calculating the derivative to find the slope at one point on a curve, and then saying you do not understand why the curve does not have a constant slope. You have mistakenly assumed that the value you have calculated is fundamental to the behaviour of the thing you are observing.

Fundamentally, it is NOT that are are "n atoms decaying every minute". That is your error. Rather, it is that each specific individual atom has a probability of decaying over some time interval.

But when you have a block of material you don't care about a particular individual atom decaying. You just care that an atom has decayed since all the atoms are interchangeable.

Imagine the probability of one person getting struck by lightning; It's very low. But if your population is enormous there is nearly a 100% chance that some number of people, somewhere will be struck by lightning over the next year. This results in a rate where some number of people get struck by lightning every year.

Even though the probability of a particular person getting struck by lightning remains the same, the rate at which people get struck by lightning depends on how many people are remaining. As you are left with fewer and fewer people, the people getting struck by lightning every year decreases since that same probability is applied over fewer and fewer people.

If your objective is to predict the population, it's inconvenient to use probability but you also can't use a rate because it's always changing because it's dependent on the population.

That's what half-life resolves. It's rather easy to measure, intuitive to use, and remains the same even as the population changes and remains valid so long as the population is statistically large.

Suppose the probability of someone getting killed by lightning is extremely deadly 5% every year. At the end of the first year there are only 95% of the original people left. But in the second year the probability is has not changed and is still 5%. So now 5% is applied to the remaining people. It doesn't matter how many people you started with. All that matters is how many people you have right now. Instead of using 5% over a year, you could instead choose to accumulate that probability over some number of years where the cumulative probability for any one person is 50%. In this case that would be 13.5 years. That's what half-life is. Every 13.5 years, half of your population at the beginning of that disappears.

However, if you reduce the population enough, at some point there is no longer a near 100% chance that a non-zero number of people will be struck by lightning over the next year. That's the point at which the statistical nature of half-life can no longer be applied. In practice it will stop being valid at some point.

Another consequence of the population being very large, (as it is like the number of atoms) is that there will be some very, very lucky atoms amongst them that are able to survive for a very long time. It might be one in a trillion trillion trillion chance that a particular atom can survive all those half-life intervals, but you have many magnitudes of atoms more than that. After many half-life intervals, when you have these long-lived atoms in front of you, you never know how much longer any particular one will last; the probability of each one surviving the next half-life interval has never changed. But what you do know is that they survived all the previous half-life intervals because they are still in front of you. But if there are still enough of them in front of you to treat statistically that means there are a great number of even luckier atoms hidden in the bunch that will survive many more half-life intervals. That's why the number of atoms remaining is mathematically asymptotic, but as I mentioned, when the population gets small enough then it cannot be treated statistically anymore.

Hopefully this illustrates how decay is not an "n amount of atoms decay every minute" but is instead a "probablity of one specific atom decaying" translated to a macro-scale metric when the population is large enough to be treated statistically.

Answer 3

It's almost as if the atoms in a sample somehow 'know' how many other atoms are in the sample.

This is the key. It is actually exactly the opposite, it is precisely because they don’t know how many other atoms are in the sample that the half life is constant.

Survival curves don’t just happen with atoms, they happen with all sorts of objects. You can think of the number of rabbits in a population that die, or the number of some type of car parts that fail, and so forth. These things generally do not have constant failure rates, or half lives.

The failure rates of car parts tends to be high for new parts, some parts break in the first hundred hours because of manufacturing defects. Then the failure rate levels off, and becomes stable, almost like a constant half life. In this range it is just a random lottery, with every part playing the game every hour it is in use, with equal odds of each part failing in any given hour it is used. Then, the failure rate begins to increase again as the parts become mechanically worn. They failure as higher rate towards the end of their usable life. So the failure rate is not constant as the part "ages".

Similarly, healthy adult rabbits can live with a low failure rate if there is a modest number rabbits in the population. Each rabbit has essentially an independent random chance of dying any day. But if there are too many they die more frequently by starvation or if there are too few rabbits in the population then they die at a much higher rate by predation. So their individual death rate depends on the number of other rabbits.

Atoms are not like cars. There aren’t defective atoms and they don’t wear out, they only have that constant middle part. Atoms are also not like rabbits, they don’t care how many they are. There isn’t an age or population based adjustment to the constant random decay rate.

So why is the half life constant, it’s because atoms of a particular type are all identical.

Answer 4

Say you have just 4 radioactive atoms with a half-life of 1 hour. . . . . . So that means 1 hour from now, 2 of the atoms will have decayed (on average) and 2 will remain undecayed (on average).
That word average is the word that you have misunderstood.

There is a finite probability that all four will decay in 1 hour and it is $1/2 \times 1/2 \times 1/2 \times 1/2=1/16$.
The probability that all four will not decay in 1 hour and it is $1/2 \times 1/2 \times 1/2 \times 1/2=1/16$.

The probability that only one will decay in a half life is a little more difficult to compute.
Labelling the atoms $\mathbf {a,\,b,\,c,\,d}$ then the probability that atom $\mathbf a$ will decay and the rest will not is $1/2 \times 1/2 \times 1/2 \times 1/2=1/16$.
The probability is the same if only one of the other atoms decays and so the probability only one atom will decay within one half life is $4\times 1/16 = 1/4$.

Now we come to an interesting result for the probability of two atoms decaying in one half life which you might expect to be $1/2$?
The probability of only $\mathbf a$ and $\mathbf b$ decaying in one half life is $1/16$ but what is needed is the probability of any two atoms decaying within one half life.
So what is needed is the number of ways which two items can be chosen from four items and there a six ways: $\{\mathbf {a,b}\}, \, \{\mathbf {a,c}\}, \,\{\mathbf {a,d}\}, \,\{\mathbf {b,c}\}, \,\{\mathbf {b,d}\}, \,\{\mathbf {c,d}\}$.
Thus the probability of only 2 atoms decaying is $6\times 1/16 = 3/8$.
Thus only two atoms decaying in one half-life is the most probable event but it does not occur half the time.

Only three atoms decaying in one half-life requires the number of ways of choosing three items from four $(4)$ and so the probability of three atoms decaying in one half-life is $4\times 1/16 = 1/4$.

So the probability distribution for a number $x$ atoms decaying from an original number of four atoms looks like this,

To find the average number of decays in one half life you need to compute a weighted average which in this case is $(1/16 \times 0) + (1/4 \times 1) +(3/8 \times 2) +(1/4 \times 3) +(1/16 \times 4) = \mathbf 2$ and there is the expected number of atoms which decay in one half life, $\mathbf 2$.

Moving on.
. . . I am struggling to understand why the last 2 undecayed atoms won't, on average, both decay in the following 1 hour.
Using the ideas used before you find that the probability of no atoms decaying in one half-life is $1/4$, one atom decaying is $1/2$ and both atoms decaying is $1/4$ with a weighted average of one atom decaying in a half-life.
So why is the average not two? It is because there are two other decay modes, neither atom decays and one atom decays, which have to be taken into account.

Another important idea it that the probability of getting exactly half the atoms to decay in one half life is very small. Indeed for $10,000$ unstable nuclei the probability of exactly $5,000$ decaying during one half life is $0.007979$.

The probability distribution which I have used indirectly is called the binomial distribution and there are many binomial distribution calculators available of which I give you link one and link two but note that there are many others.

Answer 5

Atoms only have a few distinguishing properties like the number of protons, neutrons, electrons and their energy levels. Atoms which have these same few properties are identical and indistinguishable from one another. An atom's "age" is not a distinguishing property, an atom newly formed in a nuclear reaction may be entirely indistinguishable from one that's been around for millennia. Atoms have no "memory" or any property that can capture something like "time until decay".

This property indicates that radioactive decay must follow an exponential distribution, which has a fixed percentage of events every time period. In one half life, each atom has a 50% chance of decay. Of the ones that remain, they have a 50% chance to decay over the next half life, and 50% after that, and after that. If you saw 50% of atoms decay in the first half life and then 100% of the remaining ones decay in the second half life, it would imply that those remaining atoms "knew" they had survived longer, since they would exhibit a time-dependent decay rate. But our basic premise is that an atom's properties aren't time-dependent - a new atom and and old atom can't be told apart.

The exponential distribution may be a bit counterintuitive since so many things do exhibit age-dependent survival, like living beings, electronics, buildings, etc. - as they get older, they generally become more likely to fail. All of these things have some mechanism of "wear" which physically encodes the object's history and makes them more likely to fail the older they get. This is not the case for atoms, an atom cannot "wear out" or encode its own history in some way. With age-dependent survival, it's often true that every day brings you closer to death, as your probability of death increases over time. But an atom that survives one half life is no closer to "death" than when it started - an atom that has survived a million half lives has exactly the same expected lifetime going forward as a newly minted atom. An intuitive example of the exponential distribution is your odds of winning the lottery assuming you play each day. If you play enough, you will eventually win, but playing each day gets you no closer to winning - having played and lost every day for 10 years does not make you any more likely to win than someone buying their first lottery ticket.

If one particular atom has a 50% chance of decay in a time period, so too do all other atoms of that type. If this wasn't the case, it would contradict the notion that atoms are indistinguishable from one another.

Answer 6

Let us suppose you have a kilo of some radioactive element. After time equal to the half life you now have half a kilo of said radioactive element left (plus the decay products of the other half kilo). After another half life, you know have a quarter of a kilo, etc.

Now suppose you take that kilo and cut it in half. You now have two half kilos of the radioactive element, the same amount as was left after a half life with the one kilo sample. Wait a half life, and you have two lots of a quarter kilo of radioactive element. Or a combined half kilo, the same thing as before.

The half life doesn't result from the atoms knowing how much of the atom they are together with, it results from them not knowing. If the half life wasn't constant then you could change the result of waiting a half life simply by cutting your sample in half.

Answer 7

Afterall, if it took 1 hour for the first 2 atoms to decay, then surely it should take 1 more hour for 2 more atoms to decay.

Believe it or not, I started the same thought experiment at the exact same moment, also with 4 atoms, so I have 2 atoms now – just like you.

You expect your 2 atoms to decay in 1 hour. My situation is exactly like yours, so I bet you expect my 2 atoms to decay in 1 hour. This means you expect our 4 atoms to decay in 1 hour.

How are our 4 atoms different than your 4 atoms from an hour ago? Experiments and observations show they are not different. Radioactive decay is in fact one of such observations. This is what happens: if 2 of your 4 atoms decayed in an hour then 2 of our 4 should decay in the next hour. On average. On average these 2 will be 1 of mine and 1 of yours.

Oh, wait, I think there are 78 more fellow users of Physics SE who started the same thought experiment. We are 80 users and we started with 320 atoms and now we have 160. Our current situation is like it was for 40 of us an hour ago. So in an hour the total number of atoms should be what it is for these 40 now: 80 atoms. And then it will be like the starting situation of some 20 of us, so in yet another hour the possession of all 80 users should be like of 20 users now: 40 atoms. And this will be like the starting situation of 10 of us. So in yet another hour…

Hopefully this will help you build better intuition.

---

What about the intuition you had so far? Your reasoning was flawed because by saying

if it took 1 hour for the first 2 atoms to decay, then surely it should take 1 more hour for 2 more atoms to decay

you assume the two atoms that have decayed are the right sample to predict what will happen to your two atoms in the next hour. What makes you think this sample is better than any other sample? Your sample is biased because you chose exactly the atoms that had decayed.

There exists a set of two atoms that haven't decayed. Deliberately choosing this sample would be equally biased. You would say:

If these 2 atoms survived 1 hour, then surely they will survive 1 more hour.

And what if you consider yet another possible 2-atom set? one atom that has decayed and one atom that hasn't? Then:

If only 1 of these 2 atoms survived 1 hour, then surely exactly 1 of my 2 remaining atoms will survive 1 more hour.

To predict the behavior of your remaining two atoms, basing on your experience with your four atoms in the past hour, you need to consider all available samples. E.g. you can say something like this:

From my original 4 atoms, I could choose a 2-atom sample from which 0 atoms survived the last hour. It was equally possible to choose a 2-atom sample from which 2 atoms survived. These possibilities average to 1 surviving atom. Any other sample results in 1 surviving atom, so overall they must average to 1 surviving atom. Therefore, basing on what happened to my 4 atoms, I conclude that (on average) in an hour there will be 1 atom left from my remaining 2 atoms.

Now we could ask if "1 survivor" is the most probable outcome. It is, but I won't elaborate or do the math here, this is not the point. My point is you based your conclusion on a single 2-atom sample, ignored one totally independent 2-atom sample and few other possible 2-atom samples you could choose from your original set of 4 atoms. You cannot arbitrarily pick a sample if you want to know something "on average".

In many cases a random sample gives you a good chance of getting a result close enough to the average. Your scenario is such case, especially if the number of atoms is huge; for 4 atoms to a lesser degree, but still. E.g. to predict what will happen to our (80 user's) 160 atoms in an hour, you can randomly pick 160 atoms out of the original 320 and count how many survived. 80 is the most probable count and "about 80" is "almost sure" (for wide enough "about" and "almost").

Unfortunately, by choosing a sample that is biased for sure and totally ignoring the complementary sample, you gave yourself no chance of getting a result that is "close enough" to the average.

Answer 8

This overlaps other answers but maybe the different phrasing will suit you.

Suppose that it worked as you expected then odd things would happen. Suppose you have $4$kg of radioactive stuff with a half life of $1$ hour. You watch it for $1$ hour and, as expected, $2$kg remains. You expect the rest to go in one more hour. Now, I walk in but don't talk to you. I see $2$kg of stuff so I expect to see $1$kg after the same hour. So, after that hour is there $0$kg or $1$kg? The stuff will need to know who is looking at it and for how long.

Now, we will use the coin analogy. Start with $4$ coins and toss them every hour. Push aside the tails as decayed and throw the remainder after another hour. The most likely case is $4, 2, 1$ but other results are quite likely. Now throw $4000$, you probably won't get exactly $2000$ remaining after the first hour but the relative error will be smaller. By the time that you get to realistic quantities, e.g. $1$ mole which is $6 \times 10^{23}$ atoms, the difference from $3\times 10^{23}$ after one hour will be too small to measure. This time, the stuff does not need to know who is looking or for how long. Also, each atom does not need to know about the others.

Answer 9

Radioactive decay is a random process at the level of single atoms. A nucleus of a radionuclide has no “memory”. A nucleus does not “age” with the passage of time. Thus, the probability of its breaking down does not increase (or decrease) with time, but stays constant no matter how long the nucleus has existed.

Answer 10

Labelled and Unlabelled Atoms

Afterall, if it took 1 hour for the first 2 atoms to decay, then surely it should take 1 more hour for 2 more atoms to decay...

You propose two situations

1. There are 4 atoms. We ask how long on average until 2 of them decay.

2. There are 2 atoms. We ask how long on average until 2 of them decay.

Situation 2 has only one way the event "two atoms decay" can happen $-$ both atoms decay. This takes one hour on average.

Situation 1 is different. Call the four atoms Arnold, Bernice, Charlie and Danielle.

The average time until Arnold and Bernice decay is 1 hour. The average time until Bernice and Charlie decay is 1 hour. Each of these counts as the event "two atoms decay".

There is a total of ${4 \choose 2} = 6$ ways the event "two atoms decay" can happen. Since there are more ways the event can happen, the average time until it happens is smaller.

Answer 11

I feel that everyone is getting a little off topic and too elaborate with their answers.

They also aren't answering the question which is why did the two items not decay in that 1 hour time. The answer is, because half-life refers to the chance that a specific atom will decay, usually 50%, it doesn't mean that an atom must decay within that period.

Answer 12

As other answers have pointed out, the half-life of a substance is based on probability. A particle has a certain probability that it'll decay at any given moment.

If you have 200 particles with a 1/100 chance per second to decay, after the first second, on average two will have decayed. If you only have two particles with a 1/100 chance per second to decay, it's much less likely that either of them decay.

To illustrate this, I wrote a simulation using p5.js. There are 1000 particles with a 1/1000 chance per frame (1/30 of a second) to decay. Once half the particles have decayed the "Measured HL" is updated. The measured half-life of these particles is about 25 seconds.

Note, once you get to just a few particles the half-life is more likely to deviate from the norm since the sample size is so low.

https://editor.p5js.org/d5c4b3/sketches/WbBfFnynj

Answer 13

It is useful to recall what is actually going on in an atomic nucleus.

The below is highly oversimplified, but serves to give an intuitive feel for roughly what is going on inside a nucleus.

You have a whole bunch of nucleons (protons and neutrons) which are constantly jiggling about in a volume of space, under the influence of two forces:

a) the electromagnetic force, which tends to push protons apart (and gets stronger the closer they get to each other), but has no effect on neutrons

b) the strong force, which tends to glue the nucleons (protons and neutrons) together, and is several orders of magnitude stronger than the electromagnetic force -- but only works over short distances.

Now nucleons take up volume, and the more nucleons you have in that small volume, the bigger that cluster of nucleons becomes. There comes a stage (when you've got approximately 200 nucleons in the same nucleus) that the range of action of the Strong Force is comparable to the diameter of that cluster. So if you had a bunch of nucleons which form a cluster whose diameter is much larger than the range of the Strong Force, the Electromagnetic Force is going to cause such a cluster to break up.

Hence we can see that the size of the nucleus (that cluster of nucleons) is limited to the range over which the Strong Force operates.

Now consider what happens when the nucleus is at that limiting size. Imagine a chaotic barn dance in which everybody does their own thing but has to stay within a circle. People holding hands in little bunches (4 is a popular grouping), stomping around enthusiastically, bumping into each other at random according to whatever they happen to be doing, and sooner or later a group of 4 (it's mostly always a group of 4) will get bounced out of the ring. Who knows when? Nobody does, until it happens. But if you have many millions of these chaotic barn dances going on, all with the same properties (number and size of people, size of ring, etc. etc.) then there will be a more-or-less constant probability that over the course of some period of time that a bunch of 4 will get bounced out of that ring.

(Of course, after some people have got bounced out, there will be more room for the rest of the people in the ring, and so the probability that another bunch of people get bounced out will be different.)

Answer 14

The kinetics of irreversible processes (chemical reactions, deformation, heat flow, etc) are not always linear, i.e. the second half of the process won't complete as quickly as the first. In general there is no way to predict what the kinetics of a process will be ahead of actually observing the process occur and measuring either the rate of creation of "products" (in this case, radioactive particles or radiation) or else the rate of disappearance of "reactants" (here, radioactive nuclei). So let's look at the observed facts on radioactivity first.

Radioactive decay is not observed to be a process in which (on average) a set proportion of the original mass at some initial time $t = 0$ is lost over equal time periods until it loses all radioactivity.

By continually measuring the radiation intensity from a given mass of radioactive material with some suitable Geiger counter, you won't observe a linear drop-off to an eventual zero (or ambient background intensity level) radiation. Instead you will see a fall-off in intensity whereby the loss in radiation intensity per unit time is itself continually decreasing with time from the start of measurement.

This observed curve signifies a proportional loss mechanism whereby a proportion of the current radioactive mass is being lost continually throughout its full radioactive life. The reason for this mechanism operating here is, as said by other answers, due to the fact that radioactive emission of $\alpha$-particles, $\beta$-particles or $\gamma$-rays is equally likely for all unstable nuclei in the radioactive mass. If 10% of the mass has unstable nuclei at some point in time, then the radiation intensity then will be proportional to that unstable number of nuclei. And the resulting loss in the number of unstable nuclei will also be proportional to the current number of unstable nuclei since every radioactivity event results in loss of an unstable nucleus.

Schematically we might write a sort of nuclear chemical equation like:

$ N$ unstable nuclei $\Rightarrow (N - \lambda N)$ unstable nuclei + $\lambda N$ counts of radiation

where $N$ is the current number of unstable nuclei and $\lambda$ is the proportion of unstable nuclei emitting radiation.

So:

$$ - \frac{dN}{dt} = \lambda N $$

Rearranging:

$$ \frac{dN}{N} = - \lambda dt $$

This can be integrated and leads us to:

$$ N(t) = N_0 e^{-\lambda t} $$

where $N_0$ is the original number of radioactive atoms at some initial time. This function has the shape of the graph above.

Regardless of what size $N_0$ may be, when the left-hand side of this equation is some known fraction of $N_0$ then the value of $t$ depends only on the decay-proportion rate ($\lambda$) and that fraction of $N_0$ that we have on the LHS - both of which are constant for a given substance.

For example, if $N(t) = N_0 / 2 $ we can get the time for half the radioactive mass present at $t = 0$ to decay (i.e. the half-life) via:

$$ t_\frac{1}{2} = \frac{ln2}{\lambda} $$

For a given radioactive substance, \lambda is constant, so its half-life is also constant.

The assertion that the second half of a radioactive substance's mass should need no longer than its half-life to fully expire its radioactivity is clearly not applicable here. As explained, the second half of the radioactive mass will be lost at an ever reducing rate and thus take much longer. For example a half of the remaining half of the mass (= 1/4 of the initial mass) will expire in another half-life. After that, half of the remaining mass (= 1/8 of the initial mass) will decay over the next half-life period and so on. In fact, to be exact about it, the entire remaining mass will take forever to lose full radioactivity as its decay rate slows down to near nothing towards the end.

The situation is analogous to unfilling a unit square by continually taking half of the square's remaining area. In this way after taking $n$ withdrawals we have a remaining area of:

$$ A = 1 - (\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{2^n}) $$

As $n$ becomes large, the removed area tends to unity and the remaining area to zero. But this process never fully removes all the remaining area - it just takes half of whatever is left.

So it takes an infinite number of withdrawals to completely unfill the unit square - although of course we can reduce it to a very small size by a relatively small number of extractions.

The notion of half-life to measure the persistence of a radioactive source is useful insofar as it gives us a relatable means of describing how long a radioactive substance is significantly present and emitting potentially dangerous radiation to the atmosphere. If we are happy with 0.1 % of the radioactive mass being there and know its half-life to be 3 months, we can find the number of half-lives $N$ required to achieve this reduction via:

$$ (\frac{1}{2})^n = \frac{0.125}{100} $$

$$ \Rightarrow n \approx 10 $$

So after 10 x 3 months $\approx 2.5$ years there will be less than 0.1 % of the radioactive mass still there.

Answer 15

The universe is ultimately quantum mechanical, and does not care about our classical way of thinking. Every single atom in your sample has the same exact probability of decaying, and this is because the same exact QM processes govern all indistinguishable atoms (' decay) in your sample.

A half-life often describes the decay of discrete entities, such as radioactive atoms. In that case, it does not work to use the definition that states "half-life is the time required for exactly half of the entities to decay". For example, if there is just one radioactive atom, and its half-life is one second, there will not be "half of an atom" left after one second. Instead, the half-life is defined in terms of probability: "Half-life is the time required for exactly half of the entities to decay on average". Note that after one half-life there are not exactly one-half of the atoms remaining, only approximately, because of the random variation in the process. Nevertheless, when there are many identical atoms decaying (right boxes), the law of large numbers suggests that it is a very good approximation to say that half of the atoms remain after one half-life.

https://en.wikipedia.org/wiki/Half-life

When we say the phrase "the half life is constant" we simply admire this fact about the universe, and acknowledge that these processes are ultimately governed by QM, and that the very same processes are governing each and every indistinguishable atom (and their decay) in your sample. Just as you can see from the nice answer by @dale, this can be expressed in a phrase "the atoms don't know how many other atoms are in the sample". The whole point is it does not matter if the atoms are in a bunch together or you put every atom on a different planet (disregard different gravitational fields) in the universe. QM governs all processes the same way in the known universe and each atom is indistinguishable and the decay process is governed by the same QM rules, throughout the decay of the whole ensemble.

Your question and the answer to it is a very good example of how QM is determining the half life and governs the decay process and the rules are the same regardless of if there is only a single atom or a whole sample, the underlying rules are always based on QM.

Answer 16

Here is some more information. Radioactive decay is a random process that can be accurately modeled using the Poisson probability distribution. Let $X$ be a random variable representing the number of decays in a specific time $t$ for a specific quantity (mass) of radioactive material that contains a very large number of decaying particles. Let $q$ be the probability that exactly one decay happens within time $dt$. The probability that exactly $x$ decays occur in time $t$ is ${e^{-qt}(qt)^{-x} \over x!}$. See, for example, Evans, the Atomic Nucleus, for more details.