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.

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.

Examples in context

Example 1. The Royal Brisbane and Women's Hospital uses technetium-99m for bone scans. Its half-life is $6.0 \text{ h}$. A patient is injected at $0800$ with $800 \text{ MBq}$; by the scanner at $1100$ ($3.0 \text{ h}$ later, $t/T_{1/2} = 0.5$), activity is $A = A_0 (1/2)^{0.5} = 800 \times 0.707 = 566 \text{ MBq}$. By $2000$ that evening ($t/T_{1/2} = 2$) the activity has dropped to $A_0/4 = 200 \text{ MBq}$, well below detection threshold. Decay-constant accounting $\lambda = \ln 2 / T_{1/2}$ underpins the dosing-and-schedule pharmacy software used statewide.

Example 2. Carbon-14 dating of charcoal from a Mt Cotton archaeological site yields a $^{14}\text{C}/^{12}\text{C}$ ratio that is $0.30$ of modern. With $T_{1/2} = 5730 \text{ years}$, $N/N_0 = 0.30$ gives $t = -T_{1/2} \ln(0.30) / \ln 2 = 5730 \times 1.737 = 9.95 \times 10^3 \text{ years}$. QCAA EA Unit 1 thematic items often pair such datasets with a brief on the assumptions (constant atmospheric ratio, closed system since burial), allowing a top-band candidate to flag where uncertainty in $\lambda$ propagates into the inferred date.

Try this

Q1. State the half-life relationship for the number of nuclei, and find $N/N_0$ after three half-lives. [2 marks]

• Cue. $N = N_0 (1/2)^{t/T_{1/2}}$; after three half-lives, $N/N_0 = 1/8$.

Q2. A sample of iodine-131 ($T_{1/2} = 8.0 \text{ days}$) has an initial activity of $480 \text{ MBq}$. Calculate the activity after $24 \text{ days}$ and the decay constant $\lambda$. [3 marks]

• Cue. $t/T_{1/2} = 3$, so $A = 60 \text{ MBq}$; $\lambda = \ln 2 / (8 \times 86400) = 1.00 \times 10^{-6} \text{ s}^{-1}$.

Q3. A bone-scan dose contains $30 \text{ MBq}$ of technetium-99m ($T_{1/2} = 6.0 \text{ h}$). (a) Calculate the activity after $18 \text{ h}$. (b) Determine the time for the activity to fall to $1.0 \text{ MBq}$. (c) Justify why short-half-life isotopes are preferred for diagnostic imaging. [2+3+2 marks; ISMG: Analysis and interpretation, Evaluation]

• Cue. (a) $3.75 \text{ MBq}$; (b) $t = T_{1/2} \log_2(30) = 29.4 \text{ h}$; (c) lower lifetime patient dose plus rapid clearance.