Half-life and exponential radioactive decay (QCE Physics Unit 1)

Syllabus dot point

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

8 min readUpdated 22 May 2026

Quick answer

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.

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What this dot point is asking
Exponential decay
Half-life
Radiometric dating
Nuclear medicine
How this appears in IA1 and EA
Examples in context
Try this
Exam-style questions

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:

N(t) = N_0 e^{-\lambda t}

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

A(t) = \lambda N(t) = A_0 e^{-\lambda t}

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:

T_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda}

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

N = N_0 \left(\tfrac{1}{2}\right)^{t / T_{1/2}}

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 and a half-life of $T_{1/2} = 8.0$ days. We find the activity after $24$ days.

Step 1: Determine how many half-lives have elapsed

The key to using the integer form of the decay law is finding how many complete half-lives fit into the elapsed time. Dividing the total time by the half-life gives this count.

n = \frac{t}{T_{1/2}} = \frac{24}{8.0} = 3

Step 2: Apply the half-life formula to find the remaining activity

After each half-life the activity halves, so after $n$ half-lives the activity is multiplied by $\left(\tfrac{1}{2}\right)^n$ . With $n = 3$ , the factor is $\tfrac{1}{8}$ .

A = A_0 \left(\frac{1}{2}\right)^n = (4.0 \times 10^9) \times \frac{1}{8} = 5.0 \times 10^8\ \text{Bq}

Final answer: the activity after $24$ days is $5.0 \times 10^8$ Bq, one-eighth of the original value.

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.

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.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Year 11 SAC-style4 marks

A sample contains

8.0 \times 10^{12}

atoms of cobalt-60, with half-life

5.27

years. (a) How many atoms remain after

21.08

years? (b) Calculate the decay constant in s

^{-1}

Show worked answer →

(a) Number of atoms after $21.08$ years.

Number of half-lives: $n = 21.08 / 5.27 = 4.0$ .

$N = N_0 (1/2)^n = (8.0 \times 10^{12}) (1/2)^4 = (8.0 \times 10^{12})/16 = 5.0 \times 10^{11}$ atoms.

(b) Decay constant.

$T_{1/2} = 5.27$ years $= 5.27 \times 365.25 \times 24 \times 3600 = 1.66 \times 10^8$ s.

$\lambda = \ln 2 / T_{1/2} = 0.693 / (1.66 \times 10^8) = 4.17 \times 10^{-9}$ s $^{-1}$ .

Markers reward the use of $(1/2)^n$ with integer $n$ , conversion of years to seconds, and the formula $\lambda = \ln 2 / T_{1/2}$ .

Practise this

Sources & how we know this

Physics General senior syllabus 2025 — QCAA