## Activity, the Rate of Decay

What do we mean when we say a source is highly radioactive? Generally, this means the number of decays per unit time is very high. We define **activity** \(R\) to be the **rate of decay** expressed in decays per unit time. In equation form, this is

\(R=\cfrac{\Delta N}{\Delta t}\)

where \(\text{Δ}N\) is the number of decays that occur in time \(\text{Δ}t\). The SI unit for activity is one decay per second and is given the name **becquerel** (Bq) in honor of the discoverer of radioactivity. That is,

\(1\phantom{\rule{0.25em}{0ex}}\text{Bq}=\text{1 decay/s.}\)

Activity \(R\) is often expressed in other units, such as decays per minute or decays per year. One of the most common units for activity is the **curie** (Ci), defined to be the activity of 1 g of \({}^{\text{226}}\text{Ra}\), in honor of Marie Curie’s work with radium. The definition of curie is

\(\text{1 Ci}=3\text{.}\text{70}×{\text{10}}^{\text{10}}\phantom{\rule{0.25em}{0ex}}\text{Bq,}\)

or \(3\text{.}\text{70}×{\text{10}}^{\text{10}}\) decays per second. A curie is a large unit of activity, while a becquerel is a relatively small unit. \(\text{1 MBq}=\text{100 microcuries}\phantom{\rule{0.25em}{0ex}}(\mu \text{Ci})\). In countries like Australia and New Zealand that adhere more to SI units, most radioactive sources, such as those used in medical diagnostics or in physics laboratories, are labeled in Bq or megabecquerel (MBq).

Intuitively, you would expect the activity of a source to depend on two things: the amount of the radioactive substance present, and its half-life. The greater the number of radioactive nuclei present in the sample, the more will decay per unit of time. The shorter the half-life, the more decays per unit time, for a given number of nuclei. So activity \(R\) should be proportional to the number of radioactive nuclei, \(N\), and inversely proportional to their half-life, \({t}_{1/2}\). In fact, your intuition is correct. It can be shown that the activity of a source is

\(R=\cfrac{0\text{.}\text{693}N}{{t}_{1/2}}\)

where \(N\) is the number of radioactive nuclei present, having half-life \({t}_{1/2}\). This relationship is useful in a variety of calculations, as the next two examples illustrate.

### Example: How Great Is the \({}^{\text{14}}\text{C}\) Activity in Living Tissue?

Calculate the activity due to \({}^{\text{14}}\text{C}\) in 1.00 kg of carbon found in a living organism. Express the activity in units of Bq and Ci.

**Strategy**

To find the activity \(R\) using the equation \(R=\cfrac{0\text{.}\text{693}N}{{t}_{1/2}}\), we must know \(N\) and \({t}_{1/2}\). The half-life of \({}^{\text{14}}\text{C}\) can be found in this appendix, and was stated above as 5730 y. To find \(N\), we first find the number of \({}^{\text{12}}\text{C}\) nuclei in 1.00 kg of carbon using the concept of a mole. As indicated, we then multiply by \(1\text{.}3×{\text{10}}^{-\text{12}}\) (the abundance of \({}^{\text{14}}\text{C}\) in a carbon sample from a living organism) to get the number of \({}^{\text{14}}\text{C}\) nuclei in a living organism.

**Solution**

One mole of carbon has a mass of 12.0 g, since it is nearly pure \({}^{\text{12}}\text{C}\). (A mole has a mass in grams equal in magnitude to \(A\) found in the periodic table.) Thus the number of carbon nuclei in a kilogram is

\(N({}^{12}\text{C})=\cfrac{6.02×{\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}{\text{mol}}^{–1}}{12.0\text{ g/mol}}×\text{(1000 g)}=\text{5.02}×{\text{10}}^{\text{25}}\text{.}\)

So the number of \({}^{\text{14}}\text{C}\) nuclei in 1 kg of carbon is

\(N({}^{\text{14}}\text{C})=(5.02×{\text{10}}^{\text{25}})(1.3×{\text{10}}^{\text{−12}})=6.52×{\text{10}}^{\text{13}}\text{.}\)

Now the activity \(R\) is found using the equation \(R=\cfrac{0\text{.}\text{693}N}{{t}_{1/2}}\).

Entering known values gives

\(R=\cfrac{0\text{.}\text{693}(6\text{.}\text{52}×{\text{10}}^{\text{13}})}{\text{5730 y}}=7\text{.}\text{89}×{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}{\text{y}}^{–1},\)

or \(7\text{.}\text{89}×{\text{10}}^{9}\) decays per year. To convert this to the unit Bq, we simply convert years to seconds. Thus,

\(R=(\text{7.89}×{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}{\text{y}}^{–1})\cfrac{1.00 y}{3\text{.}\text{16}×{\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{s}}=\text{250 Bq,}\)

or 250 decays per second. To express \(R\) in curies, we use the definition of a curie,

\(R=\cfrac{\text{250 Bq}}{3.7×{\text{10}}^{\text{10}}\phantom{\rule{0.25em}{0ex}}\text{Bq/Ci}}=6.76×{\text{10}}^{-9}\phantom{\rule{0.25em}{0ex}}\text{Ci.}\)

Thus,

\(R=6.76\phantom{\rule{0.25em}{0ex}}\text{nCi.}\)

**Discussion**

Our own bodies contain kilograms of carbon, and it is intriguing to think there are hundreds of \({}^{\text{14}}\text{C}\) decays per second taking place in us. Carbon-14 and other naturally occurring radioactive substances in our bodies contribute to the background radiation we receive. The small number of decays per second found for a kilogram of carbon in this example gives you some idea of how difficult it is to detect \({}^{\text{14}}\text{C}\) in a small sample of material. If there are 250 decays per second in a kilogram, then there are 0.25 decays per second in a gram of carbon in living tissue. To observe this, you must be able to distinguish decays from other forms of radiation, in order to reduce background noise. This becomes more difficult with an old tissue sample, since it contains less \({}^{\text{14}}\text{C}\), and for samples more than 50 thousand years old, it is impossible.