Physics » Radioactivity and Nuclear Physics » Half-Life and Activity

# Human and Medical Applications

Human-made (or artificial) radioactivity has been produced for decades and has many uses. Some of these include medical therapy for cancer, medical imaging and diagnostics, and food preservation by irradiation. Many applications as well as the biological effects of radiation are explored in Medical Applications of Nuclear Physics, but it is clear that radiation is hazardous. A number of tragic examples of this exist, one of the most disastrous being the meltdown and fire at the Chernobyl reactor complex in the Ukraine (see this figure).

Several radioactive isotopes were released in huge quantities, contaminating many thousands of square kilometers and directly affecting hundreds of thousands of people. The most significant releases were of $${}^{\text{131}}\text{I}$$, $${}^{\text{90}}\text{Sr}$$, $${}^{\text{137}}\text{Cs}$$, $${}^{\text{239}}\text{Pu}$$, $${}^{\text{238}}\text{U}$$, and $${}^{\text{235}}\text{U}$$. Estimates are that the total amount of radiation released was about 100 million curies.

## Human and Medical Applications

### Example: What Mass of $${}^{\text{137}}\text{Cs}$$ Escaped Chernobyl?

It is estimated that the Chernobyl disaster released 6.0 MCi of $${}^{\text{137}}\text{Cs}$$ into the environment. Calculate the mass of $${}^{\text{137}}\text{Cs}$$ released.

Strategy

We can calculate the mass released using Avogadro’s number and the concept of a mole if we can first find the number of nuclei $$N$$ released. Since the activity $$R$$ is given, and the half-life of $${}^{\text{137}}\text{Cs}$$ is found in this appendix to be 30.2 y, we can use the equation $$R=\cfrac{0\text{.}\text{693}N}{{t}_{1/2}}$$ to find $$N$$.

Solution

Solving the equation $$R=\cfrac{0\text{.}\text{693}N}{{t}_{1/2}}$$ for $$N$$ gives

$$N=\cfrac{{\text{Rt}}_{1/2}}{\text{0.693}}\text{.}$$

Entering the given values yields

$$N=\cfrac{(6.0 MCi)(\text{30}\text{.}2 y)}{0\text{.}\text{693}}\text{.}$$

Converting curies to becquerels and years to seconds, we get

$$\begin{array}{lll}N& =& \cfrac{(6\text{.}0×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{Ci})(3\text{.}7×{\text{10}}^{\text{10}}\phantom{\rule{0.25em}{0ex}}\text{Bq/Ci})(\text{30.2 y})(3\text{.}\text{16}×{\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{s/y})}{\text{0.693}}\\ & =& \text{}3\text{.}1×{\text{10}}^{\text{26}}\text{.}\end{array}$$

One mole of a nuclide $${}^{A}X$$ has a mass of $$A$$ grams, so that one mole of $${}^{\text{137}}\text{Cs}$$ has a mass of 137 g. A mole has $$6\text{.}\text{02}×{\text{10}}^{\text{23}}$$ nuclei. Thus the mass of $${}^{\text{137}}\text{Cs}$$ released was

$$\begin{array}{lll}m& =& \left(\cfrac{\text{137 g}}{\text{6.02}×{\text{10}}^{\text{23}}}\right)(3\text{.}1×{\text{10}}^{\text{26}})=\text{70}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{g}\\ & =& \text{}\text{70 kg}\text{.}\end{array}$$

Discussion

While 70 kg of material may not be a very large mass compared to the amount of fuel in a power plant, it is extremely radioactive, since it only has a 30-year half-life. Six megacuries (6.0 MCi) is an extraordinary amount of activity but is only a fraction of what is produced in nuclear reactors. Similar amounts of the other isotopes were also released at Chernobyl. Although the chances of such a disaster may have seemed small, the consequences were extremely severe, requiring greater caution than was used. More will be said about safe reactor design in the next tutorial, but it should be noted that Western reactors have a fundamentally safer design.

Activity $$R$$ decreases in time, going to half its original value in one half-life, then to one-fourth its original value in the next half-life, and so on. Since $$R=\cfrac{0\text{.}\text{693}N}{{t}_{1/2}}$$, the activity decreases as the number of radioactive nuclei decreases. The equation for $$R$$ as a function of time is found by combining the equations $$N={N}_{0}{e}^{-\mathrm{\lambda t}}$$ and $$R=\cfrac{0\text{.}\text{693}N}{{t}_{1/2}}$$, yielding

$$R={R}_{0}{e}^{-\mathrm{\lambda t}}\text{,}$$

where $${R}_{0}$$ is the activity at $$t=0$$. This equation shows exponential decay of radioactive nuclei. For example, if a source originally has a 1.00-mCi activity, it declines to 0.500 mCi in one half-life, to 0.250 mCi in two half-lives, to 0.125 mCi in three half-lives, and so on. For times other than whole half-lives, the equation $$R={R}_{0}{e}^{-\mathrm{\lambda t}}$$ must be used to find $$R$$.

### PhET Explorations: Alpha Decay

Watch alpha particles escape from a polonium nucleus, causing radioactive alpha decay. See how random decay times relate to the half life.

Alpha Decay