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# 4.3.2: The Decay Constants and Half-Lives


Radioactive decay is a stochastic, a.k.a. a random process. Suppose that we have just one radioactive nucleus: there is no theory that would allow to predict the moment at which the decay would occur. Or, suppose that we have two such nuclei. Is there a way to say which one will decay first? No, there is no such way.

Trustworthy prediction can be made, yes, but only concerning a large number of nuclei of a given isotope. Suppose that the number of nuclei in a specimen at an initial moment of time $$t_{0}=0$$ is $$N$$, and that we monitor the number of decays $$d N$$ occurring in a short period of time, $$d t$$. Alternatively, $$d N$$ can also be thought of as the "decrease in the population of radioactive nuclei in the specimen occurring over a time period $$d t$$ " - and then it should be a negative number. Naturally, one can expect that the number of decays is proportional to length of the time period considered, and to the number of nuclei in the specimen - if there twice as many nuclei in the specimen, there should be twice as many of decays, right? - so we can write:

$dN \propto-N d t, \quad \text { or } \quad d N=-\lambda N dt$

where $$\lambda$$ is just the coefficient of proportionality.

When the decay process continues, the nuclei population gradually decreases. In order to find out how many nuclei will remain in the specimen after a longer elapse of time - say, from $$t_{0}$$ to $$t_{1}-$$ one can treat the $$N$$ in Eq. 4.2 as a function of time: $$N=N(t)$$. Then, after rewriting the equation as:

$\frac{d N}{N}=-\lambda d t$

one can perform an integration:

$\int_{N\left(t_{0}\right)}^{N\left(t_{1}\right)} \frac{d N}{N}=\int_{t_{0}}^{t_{1}}-\lambda d t$

Here one can use the known general integration formulas, in which $$x$$ is a variable, and $$a, b$$ and $$c$$ are constant numbers:

$\int_{a}^{b} \frac{d x}{x}=[\ln x]_{a}^{b}=\ln b-\ln a=\ln \frac{b}{a}$

and

$\int_{a}^{b} c d x=c \int_{a}^{b} d x=c[x]_{a}^{b}=c(b-a)$

By applying them to the Eq. $$4.3$$, we obtain:

$\ln \frac{N\left(t_{1}\right)}{N\left(t_{0}\right)}=-\lambda\left(t_{1}-t_{0}\right)$

Recall the definition of the natural logarithm function: if $$y=\ln x$$, it means that $$x=e^{y}$$ (where $$e=2.718281 \ldots$$ is the base of the natural logarithm). One can write then: T

$\frac{N\left(t_{1}\right)}{N\left(t_{0}\right)}=e^{-\lambda\left(t_{1}-t_{0}\right)} \Rightarrow N\left(t_{1}\right)=N\left(t_{0}\right) \times e^{-\lambda\left(t_{1}-t_{0}\right)}$

This formula can be simplified: $$t_{1}$$, the moment "the stopwatch is started", can be taken as zero; the symbol for the initial population of radioactive nuclei, $$N\left(t_{0}\right)$$, can be replaced a simpler one, $$N_{0}$$, and the subscript at $$t_{1}$$ can be dropped, yielding a simple elegant final formula:

$N(t)=N_{0} e^{-\lambda t}$

This is an important equation, expressing what is usually called "the exponential law of radioactive decay". The $$\lambda$$ parameter in the equation is usually referred to as "the decay constant". It's an individual characteristic of the isotope described by the equation. There is no theory making it possible to calculate the $$\lambda$$ value for any radioisotope with given $$A$$ and $$Z$$ numbers: it has to be determined by experimental measurements.

There is, however, a more "user friendly" and conceptually clearer parameter to be used in the Eq. 4.4 instead of $$\lambda$$. Let's ask the following question: if the initial population of radioactive nuclei in a given specimen is $$N_{0}$$, how much time will elapse until the population drops to one-half of its initial vale, i.e., to $$N(t)=\frac{1}{2} N_{0}$$ ? Let's call it "the half-life time", and introduce for it the $$\tau_{1 / 2}$$ symbol. The relation between $$\lambda$$ and $$\tau_{1 / 2}$$ can be readily found from the Eq. $$4.4$$ if we do the following:

$\frac{1}{2} N_{0}=N_{0} e^{-\lambda \tau_{1 / 2}}$

Here $$N_{0}$$ cancels out, and if we then take the logarithms of both sides, we obtain:

$\ln \frac{1}{2}=\ln \left(e^{-\lambda \tau_{1 / 2}}\right) \Rightarrow-\ln 2=-\lambda \tau_{1 / 2} \Rightarrow \lambda=\frac{\ln 2}{\tau_{1 / 2}}$

By inserting this result to Eq. 4.4, we obtain a new form of this equation:

$N(t)=N_{0} e^{-\frac{\ln 2}{\tau_{1 / 2}} t} \text { or } \quad N(t)=N_{0} e^{-\frac{0.693}{\tau_{1 / 2}} t}$

The physical meaning of the half-lifetime parameter $$\tau_{1 / 2}$$ is "intuitionally clear", and therefore it is commonly used as a radioisotope characteristic. It should be stressed that radioactive decay is a phenomenon completely insensitive to any extraneous influence: one may apply ultra-low or ultra-high temperature, ultra-high pressure, high magnetic or high electric field to an isotope sample - and measurements of $$\tau_{1 / 2}$$ in such drastic conditions will always yield the same result. And no matter whether the investigated sample has a solid, a liquid, or a gas form. It has been confirmed by "zillions" of experiments carried out on a wide variety of radioactive materials.

The half-lifetimes of known radioisotopes vary from milliseconds to billions of years. On Earth, the Naturally Occurring Radioactive Materials (NORM) are:

• Either "primordial" radioisotopes with their $$\tau_{1 / 2}$$-s longer, or at least not considerably shorter than the age of Earth (about 5 billion years): Uranium-238 ( $$\tau_{1 / 2}=4.5 \times 10^{9}$$ years), Thorium-232 $$\left(\tau_{1 / 2}=1.4^{*} 10^{10}\right.$$ years), Potassium-40 $$\tau_{1 / 2}=1.25 \times 10^{9}$$ years), Uranium-235 $$\left(t_{1 / 2}=\right.$$ $$7^{*} 10^{8}$$ years). For U-235 about seven half-life periods have elapsed since the Earth was formed; it means that there had been over 100 times more U-235 on Earth than there is now. At present, in natural uranium mix, there is only $$0.7 \%$$ of $$\mathrm{U}-235$$, and the rest is $$\mathrm{U}-238$$.
• Any radioisotope with a $$\tau_{1 / 2}$$ considerably shorter than 5 billion years, if present on Earth at the time of its formation, would long be gone. However, such elements may emerge as products of the decay of Uranium or Thorium which do not transform at once to a stable daughter isotope, but their decay initiate a chain of successive decay events, consisting of morethan 10 different transformations untilit finally reaches a stable nucleus. Some of the elements in such chains may have a relatively long lifetimes. For instance, Polonium $$-210\left(\tau_{1 / 2}=138\right.$$ days $$)$$ and Radium-226 ( $$\tau_{1 / 2}=1601$$ years), the first radioactive elements other than Uranium and Thorium - discovered by Maria and PierreCurie, for which they were awarded two Nobel Prizes.
• Any NORM which relatively short $$\tau_{1 / 2}$$ period - if it is not an element of the decay chain - must have emerged as the result of the bombardment of Earth by high energy radiation from outer space. This is the case, for instance, with the famous $${ }_{6}^{14} \mathrm{C}$$ isotope used for "radiocarbon dating" (196o Nobel Prize in Chemistry). The C-14 isotope is produced in the atmosphere, when a high-energy neutron penetrates a nitrogen nucleus and knocks a proton out of it:
${ }_{7}^{14} \mathrm{~N}+{ }_{0}^{1} n \rightarrow{ }_{6}^{14} \mathrm{~N}+{ }_{1}^{1} p$

There is one more comment worth making, about a misconception that is not so uncommon. Namely, after learning that there is such a thing as the half-life time, some students may have a tendency of thinking: $$O K$$, if after $$\tau_{1 / 2}$$ period one-half of the initial number of radioactive nuclei remains, then after another $$\tau_{1 / 2}$$ period all will be gone. This is not true, of course. After another $$\tau_{1 / 2}$$ period $$1 / 4$$ of the initial population will remain, and after yet another one, $$1 / 8$$ of it $$-$$ and so on.

4.3.2: The Decay Constants and Half-Lives is shared under a CC BY 1.3 license and was authored, remixed, and/or curated by Tom Giebultowicz.

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