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Can we refer the half life of radioactive materials as another form of Zeno's paradox. Half and half and half forever. Does that mean that if we keep a piece of uranium in a laboratory then there will always exist an atom of uranium that has not been broken down.

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    $\begingroup$ No because there is a discrete amount of such atoms. The last disintegration will happen at an unknown time but will happen. $\endgroup$ – Slereah Dec 5 '17 at 12:00
  • $\begingroup$ In the continuous approximation, yes: in an exponential decay, zero is reached only as $t\to 0$. $\endgroup$ – stafusa Dec 5 '17 at 12:23
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No. There is a discrete number of uranium atoms, and eventually they will all decay. A kg of uranium contains 4.2 mol $= 2.5\cdot 10^{24}$ atoms. If we say the half-life $\tau$ is 4.5 billion years (oversimplifying a bit since different isotopes are different), then after that time there will be $1.25\cdot 10^{24}$ atoms, and so on - but eventually the halvings will reach halving a single atom.

On average, there will be $N(t)=N_0 2^{-t/\tau}$ atoms, and if we calculate the $t$ when $N(t)=1$ we get $t=\tau \log(1/N_0)/\log(2)$. In this case, after 364 billion years (81 half lives, since $2^{81}\approx N_0$) we reach one atom. Which will eventually, sooner or later, decay.

As long as the number of atoms is large, one can treat them as a continuous magnitude. But once they become a mere handful decays are best analysed as a discrete random process.

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Let us suppose that you are left with one Uranium nucleus.

The probability that it will decay within the period of a half life is $\dfrac 12$.

The probability that it will not decay in that period of a half life is $\dfrac 12$.

That Uranium nucleus which did not decay within the period of one half life has a probability of $\dfrac 12$ of decaying in the next period of a half life and so the probability of it surviving a period of one life and then decaying during the next period of a half life is $\dfrac 12 \times \dfrac 12 = \dfrac {1}{2^2}$.

Surviving a period of two half lives and then decaying during the third half life period has a probability of $\dfrac {1}{2^2} \times \dfrac 12 = \dfrac {1}{2^8}$.

So what about the chances of surviving $N$ half lives?

Well that will be $\dfrac {1}{2^N}$

So you will see that the chances of survival get smaller and smaller but as you quite rightly pointed out still finite.

The probability of decaying within $N$ half lives is $\dfrac 12 + \dfrac {1}{2^2} +\dfrac {1}{2^3}+ \,.\,.\,.\,+ \dfrac {1}{2^N} = 1- \dfrac {1}{2^N}$ and note that this sum never becomes exactly equal to one but if you wait enough half lives that chances are that the Uranium atom will have decayed.

Of course the other way of doing the sum is to say that the probability of surviving $N$ half lives is $\dfrac {1}{2^N}$ and the probability of not surviving $N$ half lives is one (certainty) minus the probability of surviving $N$ half lives which is equal to $1- \dfrac {1}{2^N}$.

If you picked out one Uranium nucleus then in a time approximately equal to age of the Universe (13.8 gigayears) approximately three half lives (3 x 4.5 gigayears) of U-238 would have elapsed so the probability of observing the U-238 nucleus under observation decaying is only $\dfrac 78$.


Update as the result of a comment.

Wait a minute, are you saying that two equal pieces of Uranium, by mass, can decay after unequal intervals of time<

The short answer to your question is "Yes".

Inherent in the nucleus is a parameter which is called the decay or transformation constant or the half life. That parameter is unaffected by temperature, chemical composition or the age of the nucleus. The key word in the definition of such a parameter is average or mean. So when you say that the half life of $\rm U-238$ is $4.5 \, \rm gigayears$ it means that on average after one half life, half of the $\rm U-238$ nuclei would have decayed.

If you start with four $\rm U-238$ nuclei then after one half life the probability of one nucleus decaying is $\frac 12$ whereas the probabilities of none or both nuclei decaying are $\frac 14$ and $\frac 14$.

Starting with $100$ $\rm U-238$ nuclei the probability of exactly 500 nuclei decaying is $0.025$ which is roughly the same as the probability of $499$ decaying or $501$ decaying but note that after one half life you cannot be certain that exactly half of the nuclei would have decayed.
At the other end of the spectrum the probability that none of the thousand $\rm U-238$ nuclei decayed is approximately $10^{-301}$ and that only one decayed is approximately $10^{-298}$.

So when you start off with your equal numbers of $\rm U-238$ nuclei it is highly unlikely that the last nucleus in each of the specimens will decay at the same time.

I hope that I have shown you that two equal pieces of Uranium, by mass, can decay after unequal intervals of time.

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  • $\begingroup$ Wait a minute. ARE YOU SAYING THAT TWO EQUAL PIECES OF URANIUM, by mass can decay after unequal intervals of time. $\endgroup$ – user174052 Dec 6 '17 at 2:35
  • $\begingroup$ @user497826 I have added an update to my answer. $\endgroup$ – Farcher Dec 6 '17 at 10:43

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