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Radioactive Half-Lives

Radioactive Half-Lives

Radioactive decay follows first-order kinetics. Since first-order reactions have already been covered in detail in the kinetics tutorial, we will now apply those concepts to nuclear decay reactions. Each radioactive nuclide has a characteristic, constant half-life (t1/2), the time required for half of the atoms in a sample to decay. An isotope’s half-life allows us to determine how long a sample of a useful isotope will be available, and how long a sample of an undesirable or dangerous isotope must be stored before it decays to a low-enough radiation level that is no longer a problem.

For example, cobalt-60, an isotope that emits gamma rays used to treat cancer, has a half-life of 5.27 years (see the figure below). In a given cobalt-60 source, since half of the \({}_{27}^{60}\text{Co}\) nuclei decay every 5.27 years, both the amount of material and the intensity of the radiation emitted is cut in half every 5.27 years. (Note that for a given substance, the intensity of radiation that it produces is directly proportional to the rate of decay of the substance and the amount of the substance.) This is as expected for a process following first-order kinetics. Thus, a cobalt-60 source that is used for cancer treatment must be replaced regularly to continue to be effective.

A graph, titled “C o dash 60 Decay,” is shown where the x-axis is labeled “C o dash 60 remaining, open parenthesis, percent sign, close parenthesis” and has values of 0 to 100 in increments of 25. The y-axis is labeled “Number of half dash lives” and has values of 0 to 5 in increments of 1. The first point, at “0, 100” has a circle filled with tiny dots drawn near it labeled “10 g.” The second point, at “1, 50” has a smaller circle filled with tiny dots drawn near it labeled “5 g.” The third point, at “2, 25” has a small circle filled with tiny dots drawn near it labeled “2.5 g.” The fourth point, at “3, 12.5” has a very small circle filled with tiny dots drawn near it labeled “1.25 g.” The last point, at “4, 6.35” has a tiny circle filled with tiny dots drawn near it labeled.”625 g.”

For cobalt-60, which has a half-life of 5.27 years, 50% remains after 5.27 years (one half-life), 25% remains after 10.54 years (two half-lives), 12.5% remains after 15.81 years (three half-lives), and so on.

Since nuclear decay follows first-order kinetics, we can adapt the mathematical relationships used for first-order chemical reactions. We generally substitute the number of nuclei, N, for the concentration. If the rate is stated in nuclear decays per second, we refer to it as the activity of the radioactive sample. The rate for radioactive decay is:

decay rate = λN with λ = the decay constant for the particular radioisotope

The decay constant, λ, which is the same as a rate constant discussed in the kinetics tutorial. It is possible to express the decay constant in terms of the half-life, t1/2:

\(\lambda =\;\cfrac{\text{ln 2}}{{t}_{1\text{/}2}}\;=\;\cfrac{0.693}{{t}_{1\text{/}2}}\phantom{\rule{3em}{0ex}}\text{or}\phantom{\rule{3em}{0ex}}{t}_{1\text{/}2}=\;\cfrac{\text{ln 2}}{\lambda }\;=\;\cfrac{0.693}{\lambda }\;\)

The first-order equations relating amount, N, and time are:

\({N}_{t}={N}_{0}{e}^{-kt}\phantom{\rule{3em}{0ex}}\text{or}\phantom{\rule{3em}{0ex}}t=-\cfrac{1}{\lambda }\;\text{ln}\;(\cfrac{{N}_{t}}{{N}_{0}}\;)\)

where N0 is the initial number of nuclei or moles of the isotope, and Nt is the number of nuclei/moles remaining at time t. The example below applies these calculations to find the rates of radioactive decay for specific nuclides.

Example

Rates of Radioactive Decay

\({}_{27}^{60}\text{Co}\) decays with a half-life of 5.27 years to produce \({}_{28}^{60}\text{Ni}.\)

(a) What is the decay constant for the radioactive disintegration of cobalt-60?

(b) Calculate the fraction of a sample of the \({}_{27}^{60}\text{Co}\) isotope that will remain after 15 years.

(c) How long does it take for a sample of \({}_{27}^{60}\text{Co}\) to disintegrate to the extent that only 2.0% of the original amount remains?

Solution

(a) The value of the rate constant is given by:

\(\lambda =\;\cfrac{\text{ln 2}}{{t}_{1\text{/}2}}\;=\;\cfrac{0.693}{5.27\;\text{y}}\;=0.132\;{\text{y}}^{-1}\)

(b) The fraction of \({}_{27}^{60}\text{Co}\) that is left after time t is given by \(\;\cfrac{{N}_{t}}{{N}_{0}}.\) Rearranging the first-order relationship Nt = N0eλt to solve for this ratio yields:

\(\;\cfrac{{N}_{t}}{{N}_{0}}\;={e}^{-\lambda t}={e}^{-(0.132\text{/y})(15.0\text{/y})}=0.138\)

The fraction of \({}_{27}^{60}\text{Co}\) that will remain after 15.0 years is 0.138. Or put another way, 13.8% of the \({}_{27}^{60}\text{Co}\) originally present will remain after 15 years.

(c) 2.00% of the original amount of \({}_{27}^{60}\text{Co}\) is equal to 0.0200 \(×\)N0. Substituting this into the equation for time for first-order kinetics, we have:

\(t=-\cfrac{1}{\lambda }\;\text{ln}\;(\cfrac{{N}_{t}}{{N}_{0}}\;)\;=\;-\cfrac{1}{0.132\;{\text{y}}^{-1}}\;\text{ln}\;(\cfrac{0.0200\;×\;{N}_{0}}{{N}_{0}}\;)\;=29.6\;\text{y}\)

Because each nuclide has a specific number of nucleons, a particular balance of repulsion and attraction, and its own degree of stability, the half-lives of radioactive nuclides vary widely. For example: the half-life of \({}_{\phantom{\rule{0.5em}{0ex}}83}^{209}\text{Bi}\) is 1.9 \(×\) 1019 years; \({}_{\phantom{\rule{0.5em}{0ex}}94}^{239}\text{Ra}\) is 24,000 years; \({}_{\phantom{\rule{0.5em}{0ex}}86}^{222}\text{Rn}\) is 3.82 days; and element-111 (Rg for roentgenium) is 1.5 \(×\) 10–3 seconds. The half-lives of a number of radioactive isotopes important to medicine are shown in the table below, and others are listed in this appendix.

Half-lives of Radioactive Isotopes Important to Medicine
TypeDecay ModeHalf-LifeUses
F-18β+ decay110. minutesPET scans
Co-60β decay, γ decay5.27 yearscancer treatment
Tc-99mγ decay8.01 hoursscans of brain, lung, heart, bone
I-131β decay8.02 daysthyroid scans and treatment
Tl-201electron capture73 hoursheart and arteries scans; cardiac stress tests

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