How are the half lives (or whole lives) calculated for slowly decaying radioactive substances such as tellurium? I must be having a fundamental misunderstanding of how we are able to predict the amount of time it will take for an isotope or radioactive material to decay fully including for ones that can remain unstable for practically eons.

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    $\begingroup$ Same way as any other half-life: by counting the decay events. Did you think that the half-life of 1600 years for radium was obtained by counting atoms in a flask left by Marie Curie that long ago? Well, no. $\endgroup$ Commented Jun 27, 2020 at 8:05
  • $\begingroup$ @IvanNeretin How in the world does somebody practically measure these decay events? $\endgroup$
    – Ark Lomas
    Commented Jun 27, 2020 at 8:46
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    $\begingroup$ "How in the world does somebody practically measure these decay events?"...Well by using instruments. $\endgroup$ Commented Jun 27, 2020 at 9:39
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    $\begingroup$ Well, 226Ra is fun, but 128Te with T1/2=2.2e24 years less so. In 1 mol of 128Te, the decay rate is 1 kernel in several years. It is worse than catching neutrinos. $\endgroup$
    – Poutnik
    Commented Jun 27, 2020 at 11:44
  • $\begingroup$ Even 209Bi is fun, compared to 128Te. $\endgroup$
    – Poutnik
    Commented Jun 28, 2020 at 6:32

1 Answer 1


Suppose you have a sample of $N$ atoms of a radioactive material. As it is radioactive, you may state that a small fraction $k$ is decomposed by second. This fraction does not depend on the total number of atoms. If this sample is placed in a Geiger counter, each decomposition of its atoms produces a ray that crosses the inner volume of the Geiger counter, and it will be detected and counted. In a given time, the number of rays is equal to the number of disintegrations. If $\mathrm dN/\mathrm dt$ is the number of atoms disintegrated per second (or per minute), the factor $k$ can be calculated by dividing the number of rays detected every second in the counter, by the total number of atoms $N$ of the radioactive material. $$k=\frac{\mathrm dN}{\mathrm dt}\cdot\frac1N$$

Once $k$ is known, the half-life $T$ can be derived from the integrated law, if $N_0$ is the original number to atoms at $t=0$ $$N/N_0=\mathrm e^{-kt}$$ The half-life is the time $T$ after which $N=N_0/2$, which gives $$1/2=\mathrm e^{-kT}$$ with the following result: $$T=\frac{\ln2}k=0.693/k$$

So the period may be obtained from the knowledge of $k$. It is not necessary to wait up to the radioactive material has decreased to one half.

For example, if $0.0238\ \mathrm g$, or $6.02·10^{19}$ atoms $\ce{U-238}$ is inserted in a Geiger counter, $292$ rays will be detected per second. From this measurements, the constant $k$ can be calculated: $$k=\frac{292\ \mathrm{s^{-1}}}{6.02\times10^{19}}=4.87\times10^{-18}\ \mathrm{s^{-1}}$$ and the half-life is $$T=\frac{0.693}{4.87\times10^{-18}\ \mathrm{s^{-1}}}=1.423\times10^{17}\ \mathrm s=4.51\times10^{9}\ \mathrm a$$


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