Why does a collection of radioactive atoms show predictable behaviour while a single one is highly random?

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Solution 1

Law of large numbers

This law simply states that if you repeat a trial many times, the result tends to be the expected value. For example if you roll a 6-sided die, you could get any of the six results 1, 2, 3, 4, 5, 6. But the average of the six results is 3.5, and if you roll the 6-sided die a million times and take the average of all of them, you are extremely likely to get an average of about 3.5.

But you 1) might not get a number close to 3.5, in fact there's a non-zero chance you get an average of, for example, 2 or 1, and 2) still can't predict which result you will get when you roll a single die.

In the same way, you might not be able to predict when a single atom will decay (i.e. when you roll a single die), but you can make very good predictions when you have lots of atoms (i.e. equivalent to rolling the die millions of times).

Solution 2

As an illustration, we can simulate the radioactive decay, using various starting number of atoms. We get something like this:

simulated decay

The two plots show proportion of remaining atoms as a function of time. The bottom panel uses a logarithmic scale to better see what is happening. Each curve shows a simulation with a given starting population (from 1 to 1000 atoms). As you can see, as you increase the number of atoms the curves converge rapidly to the limit curve (in blue). As the numbers of atoms in a lot of problems in much larger than 1000, it makes sense to use the limit curve to model the atom population.

Solution 3

Radioactive decay is entirely random and it is impossible to predict when a specific atom will decay. However, at any point in time, each radioactive atom in a sample has the same probability of decaying. Therefore, the number of decay events (or reduction in the number of atoms) $-dN$ in a small time interval $dt$ is proportional to the number of atoms $N$.

So $-\frac{dN}{dt} = kN$. The solution to this differential equation is $N(t)=N(0)e^{-kt}$.

So when there is a sufficiently large number of atoms in a sample their number can be treated as continuous and a differential equation can be used to solve for the amount of sample.

In other words, after one half-life there is not always exactly half of the atoms remaining because of the randomness in the process. But when there are many identical atoms decaying it is a pretty good approximation to say that half of the atoms remain after one half-life (for large enough numbers of atoms large fluctuations are unlikely to occur).

Solution 4

Loosely speaking a random number is always Poisson distributed, if we have a "large" number of possible events, each of which is "rare" and independent of each another. This can be shown mathematically (look up Poisson process). Since this applies to the number of spam mails received per hour, and to the decay of a radioactive isotope, both are distributed as $$ Pr(X=k) = \frac{\lambda^k e^{-\lambda}}{k!} $$ where $\lambda$ is the (dimensionless) rate constant of the Poisson process, which is equal to the average value, $E[X]=\lambda$ as well as to the variance, $Var[X]=\lambda$. In Physics we usually replace $\lambda \to \tilde\lambda \cdot t$, where $\tilde\lambda$ has dimension $s^{-1}$.

To simplify the above argument, one could say that the $e^{- \tilde\lambda t}$ law of radioactive isotopes is due to an average effect.

Solution 5

The underlying reason is down to the probabilistic nature of quantum events. At the quantum level, after a given length of time every event has a particular probability of occurring. Just like rolling a die, you never know when you will roll a six but you know that one will presently turn up. If you roll hundreds or thousands of times, the maths of probability will give you a good idea of what the distribution of sixes will be.

So it is with radioactivity. You never know when a given atom will "roll a six" and decay. But you know what the distribution of decay events in a lump of atoms will be.

You may still want to know, why are quantum events probabilistic? Augh! It is one of life's deepest mysteries. The maths works, that is all we can say for sure.

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Sabbir Ahmed
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Sabbir Ahmed

Completed B.Sc. in EEE from Bangladesh University of Engineering and Technology (BUET) in year 2014. Worked as telecommunication engineer in the state owned telecom company named BTCL for two years. Currently working as Instrumentation and Control (I&C) engineer in Rooppur Nuclear Power Plant. I have keen interest in many things. Like to learn everyday.

Updated on July 23, 2020

Comments

  • Sabbir Ahmed
    Sabbir Ahmed over 3 years

    Well, we know that it is impossible to say exactly when a radioactive atom will go on decay. It is a random process. My question is why then a collection of them decays in a predictable nature (exponential decay)? Does the randomness disappear when they get together? What is the cause of this drastic change of their behaviour?

    • David Z
      David Z over 3 years
      Comments are not for extended discussion; this conversation has been moved to chat.
  • Sabbir Ahmed
    Sabbir Ahmed over 3 years
    So, does this mean that when the number of atoms in the sample gets lower over the time, they don't strictly follow the law? Like their half life will be varied?
  • MSalters
    MSalters over 3 years
    @SabbirAhmed: Well, once you're down to 3 atoms, the problem with a half-life should be obvious.
  • Caleth
    Caleth over 3 years
    Also, you most likely don't get exactly 3.5, instead a number like 3.4985 or 3.50103, but then you round those to 3.5
  • Kai
    Kai over 3 years
    Note that when physicists talk about such systems we usually are considering them in the thermodynamic limit, i.e. we assume the number of particles to be effectively infinite. The standard deviation of the mean decreases as $1/\sqrt{N}$.
  • Oscar Bravo
    Oscar Bravo over 3 years
    @SabbirAhmed Correct. If you have very small numbers of atoms, the half-lives of different samples will vary. For example, if you had a little group of 10 atoms and measured the time until you got 5 decays, you would get a number that was different to the published half-life. If you did this again, you'd get another number. If you did it a thousand times and then averaged all the numbers, the average would approach the known half-life again.
  • Yakk
    Yakk over 3 years
    @OscarBravo Also, people doing the experiment would make mistakes, and get really really bored. :)
  • cmaster - reinstate monica
    cmaster - reinstate monica over 3 years
    And remember, there are $N_A = 6.02214076\cdot 10^{23}$ atoms in a mol, so when you repeat an experiment $N_A$ times, you get the average with a precision of about $\frac{1}{\sqrt{N_A}} = 7.8\cdot 10^{-12}$, i.e. 11 digits of precision!
  • JesseM
    JesseM over 3 years
    And, even though the average is near 3.5, in none of those millions upon millions of dice rolls will you ever get a single "3.5" result. Averages (like "expected outcomes") and statistics in general all apply to large sets, not individual outcomes.
  • CJ Dennis
    CJ Dennis over 3 years
    1 die, 2 dice. You roll a die, and you roll multiple dice.
  • Peter
    Peter over 3 years
    @CJDennis In primary school I was also taught 1 dice, 2 dice.
  • Chris
    Chris over 3 years
    @CJDennis As with many things, singular "die" versus "dice" depends on your dialect. In particular, while it is often considered incorrect in American English, the latter is standard in British English.
  • I'm with Monica
    I'm with Monica over 3 years
    @Chris And I thought the singular must be douse. English is such a peculiar language... ;)
  • J.G.
    J.G. over 3 years
    In particular, if the mean number of decays expected in a period is a large value, say $n$, the actual number is approximately $\operatorname{Poisson}(n)$-distributed, with relative error $1/\sqrt{n}$.
  • JimmyJames
    JimmyJames over 3 years
    @Chris Would you say "the dice is cast"?
  • David Hammen
    David Hammen over 3 years
    @JimmyJames - No, that would be grammatically incorrect, but it would also be grammatically incorrect to say "the dice are cast". The correct usage is "the dies are cast" if more than one die has been cast.That use of die refers to a tool or device used to impart a particular shape, form, or finish to a material. The plural of that meaning of die is dies rather than dice. English is a funny language.
  • Michael Ferguson
    Michael Ferguson over 3 years
    @DavidHammen no, the phrase "the die is cast" definitely refers to the small, numbered cubes used in games of chance, not the tools used for shaping material. While I guess it makes sense that once you cast (in metal) a manufacturing die you can't change it it's also true that once you cast (throw) a die you can't take it back, and the origin of the phrase (attributed to Caesar) definitely refers to those small chance cubes.
  • JimmyJames
    JimmyJames over 3 years
    @DavidHammen I also used to think that 'the die is cast' referred to a 'tool die' and was surprised to learn the correct origin.
  • Michael
    Michael over 1 year
    This explanation seems to imply that radioactive elements are like constantly rolling n-sided dice where less stable element have smaller n, and rolling a "1" cause the element to decay.