# Does the half-life of an element mean it will never decay completely?

Example: Half life of Polonium-194 is 0.7 seconds. If we supposedly take 50g of Polonium, there will surely be a time when no more of this Polonium will be left because if we consider the decay discretely, in the form of individual atoms, won't there be a time when the last atom decays completely? Does this mean an element can decay completely? If so, why don't we actually 'run out' of natural radioactive elements? Is it so because the elements they decay into combine to form the parent element again?

There will certainly come a time at which we can say "it is more likely than not that not even one atom of the original Polonium sample is left". So, yes, the sample can decay completely.

The fact is, the earth is running out of natural radioactive elements. Most of what is left are Uranium, Thorium and Potassium because they have half-lives which are not tiny compared to the age of the solar system.

The reason why we had any radioactive elements to start out with is that the solar system formed from a cloud of dilute gas which contained debris from an exploded supernova. In the violence of a supernova explosion smaller nuclei can be slammed together so hard that they fuse into the heavier radioactive elements.

In reactors we can make samples of heavy radioactive elements - but usually at the cost of many uranium atoms. Other than that the number of radioactive nuclei is winding down here on earth.

• Until we get fusion power going ... Commented Feb 19, 2020 at 0:24

No, not really. For example, suppose you have a sample of $$2^{1000}$$ atoms with half-life $$t$$. (Note: there are only about $$10^{80}$$ protons in the Universe.)

• If you wait for a time $$t$$, then half of them have decayed, and you have on average $$2^{999}$$ remaining.
• If you wait $$1000t$$, then you will have on average a single radioactive atom remaining.
• And if you wait $$2000t$$, then the average number of remaining radioactive atoms is $$2^{-1000}$$. That means it is overwhelmingly likely that actual number of remaining atoms is zero.

Your half-second polonium isotope was among the isotopes produced in the supernova explosion whose detritus recombined to form our solar system, five-ish billion years ago. That "primordial" polonium is entirely gone. The radioactive elements that remain are the ones with billion-year half-lives.

A famous consequence of this is the natural fission reactor at Oklo, Gabon, where several tons of uranium ore underwent rainwater-moderated fission about two billion years ago. This was possible because, on a younger Earth, there was more of the highly-fissionable, shorter-lived isotope U-235 in uranium ores than there is today. A similar structure formed by geological processes today wouldn't fission because natural uranium is no longer sufficiently enriched.

• If you removed "No, not really" from the beginning of your answer then with light editing we could probably delete mine ... Commented Dec 30, 2018 at 19:15
– rob
Commented Dec 30, 2018 at 19:18
• While it's quite likely that a supernova triggered the formation of our solar system and contributed some radioactives, the ingredients for the solar system came from a very large number of stars (many of them not supernovae), as explained here by Rob Jeffries. Commented Dec 30, 2018 at 23:19

When discussing radioactivity, it is important remember that all $$^{194}$$Po atoms are independent. No atom "cares" if other atoms have or have not decayed. If the decay probability per unit time is constant ($$\lambda$$), then the probability that a given atom has decayed by time $$t$$ is given by the CDF of the decay probabilty:

$$P_i(t) = 1-e^{-\lambda t}$$

This applies to each atom in your $$\frac 1 4$$-mole of polonium independently. Discussing collective phenomenon like the time it takes half the sample to disappear obscures this fact: each atom operates independently.

If you have $$N$$ atoms, the probability that they all have decayed is the product of the probabilities the each has decayed:

$$P(t) = (1-e^{-\lambda t})^N \approx 1-Ne^{-\lambda t}$$

which gets close to 1 when

$$e^{\lambda t} \gg N$$

or

$$t \gg \frac{1}{\lambda}\ln{N} = \tau\ln N$$

where $$\tau$$ is the mean-life.

As pointed out in the other answers, half-life is usually defined for a very big collection of particles, where we can take thermodynamic limit, i.e., assume that the number of particles is infinite for all the practical purposes.

However, an exact solution is available in terms of pure death process (see this answer for the mathematical details and references), which gives the probability that at time $$t$$ we still have $$n$$ non-decayed atoms left as $$P(n, t|N_0) = \begin{cases} {N_0\choose n}e^{-n\lambda t}\left(1-e^{-\lambda t}\right)^{N_0-n}, \text{ for } n\leq N_0,\\ 0, \text{ otherwise}, \end{cases}$$ where $$N_0$$ is the initial number of atoms. Thus, the probability that all the atoms have decayed by time $$t$$ is $$P(0, t|N_0)=\left(1-e^{-\lambda t}\right)^{N_0}.$$ Note that in this case all the atoms might have decayed before time $$t$$, which is just a parameter. If we want to know the time at which all the atoms decay (which is obviously also random), we can obtain the distribution function for this time as (see here for some basic survival analysis math) $$f(t|N_0)= \partial_t P(0, t|N_0) = \lambda N_0 e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{N_0-1}$$ One can now calculate its moments $$\langle t\rangle$$, $$\langle t^2\rangle$$ and other statistical characteristics.

We are running out of some radioactive elements. It is those which are not replaced by human activity and natural processes.

Nuclear reactors are a source of radioactive elements which replace those which have decayed with the Manual for reactor produced radioisotopes showing what can be done.

A example of replenishment by natural processes is the production of the radioisotope $$\rm ^{14}C$$ in the upper atmosphere.

Examples of depletion without natural replacement are the four decay chains, Thorium series, Uranium series, Actinium series and the Neptunium series with only the last two isotopes of the Neptunium series occurring naturally because the parent and its other offspring have a comparatively short half life.

Going back to your example if you read the relevant part of this paper in which the production of $$\rm ^{194}Po$$ is described it should be apparent that starting with $$50\, \rm g$$ of $$\rm ^{194}Po$$ is nigh impossible.
So you can say that this radioisotope does not occur naturally.

Take an ordinary banana for example. It contains approximately 60 micrograms of Potassium-40 ($$\,^{40}K$$), a naturally occurring radioactive isotope. This is about $$9\times10^{17}$$ atoms. It takes 60 half-lives for this to decay into less than a single atom. Since the half-life of $$\,^{40}K$$ is 1.25 billion years, it will take 75 billion years before the last atom decays.

• The half life of a banana is very much shorter than that. Commented Dec 30, 2018 at 19:25
• I am sure it is a violation of the rules for me to say so, but that made me LOL my2cts Commented Dec 30, 2018 at 19:38

The probability that all nuclei have decayed in a 50g sample of $$\rm^{194}Po$$ is a function rapidly approaching unity after a few periods of $$0.392* ~^2 \! \log(1.5 \cdot 10^{23}) \approx 31$$ seconds.