# 31.5 Half-life and activity  (Page 5/16)

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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 [link] ). 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 The Chernobyl reactor. More than 100 people died soon after its meltdown, and there will be thousands of deaths from radiation-induced cancer in the future. While the accident was due to a series of human errors, the cleanup efforts were heroic. Most of the immediate fatalities were firefighters and reactor personnel. (credit: Elena Filatova)

## 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 Appendix B to be 30.2 y, we can use the equation $R=\frac{0\text{.}\text{693}N}{{t}_{1/2}}$ to find $N$ .

Solution

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

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

Entering the given values yields

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

Converting curies to becquerels and years to seconds, we get

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

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

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

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 chapter, 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 $R=\frac{0\text{.}\text{693}N}{{t}_{1/2}}$ , 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 $R=\frac{0\text{.}\text{693}N}{{t}_{1/2}}$ , yielding

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