If radioactive decay is measured by halflife does it mean that few atoms of a radioactive material can be considered practically stable? If we try to measure the time when 0,1% of the atoms of the radiactive material are still stable we will get a huge dismeasure between the time when the first bunch of atoms decayed and these last 0,1%? Let say we collect all atoms of a radioactive material that would decay first and make a body of 1kg from them will they decay rapidly or their time of decay will be just what statistic tells fot that element?
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$\begingroup$ A related question from a few months ago: physics.stackexchange.com/q/606390/123208 $\endgroup$– PM 2RingCommented May 2, 2021 at 15:34
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3 Answers
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does it mean that [a] few atoms of a radioactive material can be considered practically stable?
No, it doesn't mean that. If you have a chunk of some radioactive isotope, then all of the atoms in that chunk are equally unstable. That is, they all have the same probability of decaying during a given time period.
Also, there's no measurement you can make on any of those atoms that will tell you if it's going to decay within the next half-life, or if it will still be undecayed ten half-lives from now. (There may be observable changes to the shape of a large nucleus just before it fissions, but the timescale for those changes is very short).
A common classroom activity to demonstrate how half-life works uses coin tossing, with each student representing a radioactive atom. Each student has a coin and a watch.
At the end of each minute (as determined by their watch), the student tosses their coin. If the coin comes up tails, then their atom has decayed, so they leave the room. If the coin is heads, they stay, and toss the coin again at the end of the next minute.
If there are ~30 people in the class, then room is likely to be empty after 5 minutes or so, because $2^5=32$.
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Yeah, statistics can be counter-intuitive.
Imagine the following scenario:
You have very large population, and every year every member of that population does a round of russian roulette, with the terribly bad odds of having a 1 in 2 chance of dying.
Each year half the population does not survive the round of russian roulette.
But if the population is sufficiently large then a small portion will survive multiple decades, even though they go through a round of russian roulette every year.
Yet, it is wrong to suggest that the small, small population that is left standing is very good at playing russian roulette.
That's not how it works. If the odds of dying are 1 in 2, and the starting population is large enough, then some members of the initial population survive for multiple decades.
For each member of the population the odds are just as bad (1 in 2) for every round.
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The decay of a radioactive atoms is independent of the existence of other radioactive atoms. So let's suppose that we start at time $t=0s$ with $1\, 000$ atoms of a radioactive isotope with half-life $\tau$, and we give each atom a "name", $\{A_1, A_2, \ldots, A_{1000}\}$. After time $t=\tau$ the probability that the first atom $A_1$ has decayed is $p=1/2$. Hence, imagine that nature is flipping a coin and if the coin lands on heads the first atom has decayed. The same is true for all the other atoms. Thus, after time $t=\tau$ nature flips $1\,000$ coins and all atoms decay, if their coins lands heads. Therefore, on average $500$ radioactive atoms remain after time $t=\tau$.
The same process is done for the next time step, $\tau \to 2\tau$. However, since we have only (approx.) 500 radioactive atoms left, nature flips only 500 coins. Again, on average 250 coins land heads and the corresponding atoms decay.
Having the described process in mind, we can look at the expected time for the first 0.1% of atoms to decay. Even without going into the calculation we immediately see that it must be smaller than the half-life $\tau$. However, the last 0.1% will last much longer. To see this, we do the following estimate: After 10 half-life we have $2^{-10} = 1/1024 \approx 0.1\%$ of the initial atoms left. However, in order to reach 0.05% it takes a full half-life $\tau$.
