Solve problems involving the exponential decay of radioactive nuclides, half-life and decay constant, and apply to radiometric dating and medical applications

What this dot point is asking

QCAA wants you to apply the exponential decay law to find the number of nuclei remaining (or activity) at a given time, and to convert between half-life and decay constant. The dot point also covers two big applications: radiometric dating and nuclear medicine.

Exponential decay

Radioactive decay is a first-order process: each nucleus has a constant probability per unit time of decaying, independent of how old it is. This means the population follows an exponential curve:

where $N_0$ is the number at $t = 0$ and $\lambda$ is the decay constant (s$^{-1}$). The activity (decays per second) is

SI unit of activity: becquerel (Bq, $1$ Bq = $1$ decay s$^{-1}$).

Half-life

The half-life $T_{1/2}$ is the time for half the nuclei in a sample to decay. From the exponential law:

For integer numbers of half-lives, use the convenient form:

After $1$ half-life: $1/2$ remains. After $2$: $1/4$. After $n$: $(1/2)^n$.

Radiometric dating

Carbon-14 ($T_{1/2} = 5730$ years) is continuously produced in the upper atmosphere and absorbed by living things in equilibrium with atmospheric levels. When an organism dies, intake stops and the C-14 decays. Measuring the remaining C-14 fraction gives the time since death.

Uranium-238 ($T_{1/2} = 4.5 \times 10^9$ years) and other long-lived isotopes are used to date rocks.

Nuclear medicine

Technetium-99m ($T_{1/2} = 6$ hours) is the most-used medical radionuclide. It emits a gamma photon that an imaging camera detects, and its short half-life means most of the dose has decayed away by the next day. Iodine-131 ($T_{1/2} = 8$ days) is used to treat thyroid disorders because it concentrates in the thyroid gland.

Worked example

A sample of iodine-131 has an initial activity of $4.0 \times 10^9$ Bq. What is the activity after $24$ days? ($T_{1/2} = 8.0$ days.)

Number of half-lives: $n = 24 / 8 = 3$.

$A = A_0 (1/2)^n = (4.0 \times 10^9) \times (1/8) = 5.0 \times 10^8$ Bq.

Common traps

Treating decay as linear. Half of the remaining sample decays each half-life, not half of the original. After $2$ half-lives, $25\%$ remains, not $0\%$.

Mixing units of time. If $T_{1/2}$ is in days and $t$ is in seconds, the ratio is nonsense. Convert before substituting.

Using natural log in the wrong direction. $\lambda = \ln 2 / T_{1/2}$, not $T_{1/2} / \ln 2$.

Treating activity as constant. Activity drops exponentially along with $N$. The activity now is much smaller than the activity at $t = 0$.

How this appears in IA1 and EA

IA1 data test. Often a decay curve to read (activity vs time), with a half-life to extract and a future activity to predict.

EA Paper 1. Multiple choice on $(1/2)^n$ for integer $n$.

EA Paper 2. A two-part calculation: convert $T_{1/2}$ to $\lambda$, then find activity or number remaining at a non-integer number of half-lives using the exponential formula.

In one sentence

Radioactive decay follows the exponential law $N = N_0 e^{-\lambda t}$ with decay constant $\lambda = \ln 2 / T_{1/2}$, so after each half-life the number of remaining nuclei (and the activity) halves; integer-half-life problems use $N = N_0 (1/2)^n$, and applications include carbon-14 dating and medical isotopes like technetium-99m.