I read somewhere in some textbook that

"The concept of half life and mean half life becom meaningless ehen you have a sample containing only a few nuclei. For example, if you are given say, 100 active nuclei on a sample holder, and the half life of the material is one hour, then it does not mean that you will be left with 50 nuclei after 1 hour! As the number of nuclei in the sample is increased-though we can not say which nuclei will decay at a given instant- we can say that about half of them will decay after one half life".

I didn't get the thing why one can't discuss half life concept for few number of nuclei. 2nd is the example taken here I didn't understand it what does it trying to convey to readers that after 1 half life 50 wont decay.

  • 2
    $\begingroup$ Toss 100 coins. Is it guaranteed that you will get exactly 50 heads? $\endgroup$
    – PM 2Ring
    Mar 20, 2022 at 7:57

1 Answer 1


Consider tossing a coin $2$ times which is equivalent to observing 2 unstable nuclei in a time interval equal to one half life.
What is the probability that there will be $0,\,1,\,2$ heads (nucleus decays).
The answer is $0.25,0.5,0.25$ consisting of $\rm TT,\,HT,\, TH,\,HH$ so the chance of just one head (decay) is $0.5$.
So far so good!

Now what about $10$ coin tosses (unstable nuclei)?
The probability of four or less heads is $\frac{193}{512}$ which is the same as the probability of six or more heads whilst the probability of exactly five heads is $\frac {63}{256}$.
So the probability of exactly $5$ of the $10$ unstable nuclei decaying in a time of one half life is $\frac {63}{256}$.

For your example of $100$ unstable nuclei the probability of exactly $50$ decaying in one half life is $0.07959$

As the number of nuclei increases the probability of exactly half of them decaying gets smaller so for $10,000$ unstable nuclei the probability of exactly $5,000$ decaying is $0.007979$.

The distribution of such events is called a binomial and I used the WolframAlpha binomial distribution calculator to crunch the numbers.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.