42.5.3 Activity

The activity $R$ of a radioactive sample is the number of decays per second. This is simply the absolute value of $d N / d t$ , or

R = | \frac{d N}{d t} | = \frac{N}{τ} = \frac{N_{0}}{τ} e^{- t / τ} = R_{0} e^{- t / τ} = R_{0} {(\frac{1}{2})}^{t / t_{1 / 2}}

(42.13)

where $R_{0} = N_{0} / τ$ is the activity at $t = 0$ . The activity of a sample decreases exponentially along with the number of remaining nuclei.

The SI unit of activity is the becquerel, defined as

1 becquerel = 1 Bq \equiv 1 decay/s or 1 s^{- 1}

An older unit of activity, but one that continues in widespread use, is the curie. The curie was originally defined as the activity of 1 g of radium. Today, the conversion factor is

1 curie = 1 Ci \equiv 3.7 \times 10^{10} Bq

One curie is a substantial activity. The radioactive samples used in laboratory experiments are typically $\approx 1 μ Ci$ or, equivalently, $\approx 40, 000 Bq$ . These samples can be handled with only minor precautions. Larger sources of radioactivity require lead shielding and special precautions to prevent exposure to high levels of radiation.

EXAMPLE 42.3 A laboratory source

The isotope $^{137} Cs$ is a standard laboratory source of gamma rays. The half-life of $^{137} Cs$ is 30 years.

How many $^{137} Cs$ atoms are in a $5.0 μ Ci$ source?
What is the activity of the source 10 years later?

MODEL The number of $^{137} Cs$ atoms decays exponentially.

SOLVE

The number of atoms can be found from $N_{0} = τ R_{0}$ . The initial activity in SI units is

$R_{0} = 5.0 \times 10^{- 6} Ci \times \frac{3.7 \times 10^{10} Bq}{1 Ci} = 1.85 \times 10^{5} Bq$

To find the lifetime, first convert the half-life to seconds:

$t_{1 / 2} = 30 years \times \frac{3.15 \times 10^{7} s}{1 year} = 9.45 \times 10^{8} s$

Then

$τ = \frac{t_{1 / 2}}{ln 2} = 1.36 \times 10^{9} s$

Thus the number of $^{137} Cs$ atoms is

$\begin{array}{l} N_{0} & = & τ R_{0} = (1.36 \times 10^{9} s) (1.85 \times 10^{5} Bq) \\ = & 2.5 \times 10^{14} atoms \end{array}$
The activity decreases exponentially, just like the number of nuclei. After 10 years,

$R = R_{0} {(\frac{1}{2})}^{t / t_{1 / 2}} = (5.0 μ Ci) {(\frac{1}{2})}^{(10 yr)/(30 yr)} = 4.0 μ Ci$

REVIEW Although $N_{0}$ is a very large number, it is a very small fraction $(\approx 10^{- 10})$ of a mole. The sample is about 60 ng of $^{137} Cs$ .