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VICPhysicsSyllabus dot point

How is the timing of nuclear decay used in science and medicine?

Solve problems involving exponential decay and half-life ($N = N_0 (\tfrac{1}{2})^{t/T_{1/2}}$), and apply to dating techniques (carbon-14, uranium-lead) and nuclear medicine (technetium-99m, iodine-131)

Q: Year 11 SAC HSC: A wooden artefact's carbon-14 activity is $25$% of that found in living wood. Use $T_{1/2} = 5730$ years to estimate the artefact's age.

Activity ratio = number ratio (because $A \propto N$). $0.25 = (1/2)^n$ where $n$ is the number of half-lives. $(1/2)^n = 1/4 = (1/2)^2$, so $n = 2$. Age = $2 \times 5730 = 11\,460$ years. Markers reward equating activity ratio to number ratio, identifying $n = 2$, and the multiplication.

A focused answer to the VCE Physics Unit 1 dot point on half-life and applications. Applies the integer half-life formula $N = N_0 (1/2)^n$ and the continuous form $N = N_0 e^{-\lambda t}$ with $\lambda = \ln 2 / T_{1/2}$, and works the VCAA SAC-style carbon-14 dating and Tc-99m medical-isotope problems.

Generated by Claude OpusReviewed by Better Tuition Academy5 min answerUpdated 2026-05-19

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What this dot point is asking

VCAA wants you to apply the exponential decay law to find the number of radioactive nuclei (or activity) at any later time, to convert fluently between half-life and decay constant, and to use this in named 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.

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

Activity $A(t) = \lambda N(t)$ . SI unit of activity: becquerel (Bq, $1$ Bq = $1$ decay s $^{-1}$ ).

Half-life

The time for half the nuclei to decay:

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

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

N = N_0 \left(\frac{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). Atmospheric CO $_2$ contains C-14 at a known ratio that living organisms incorporate. Once the organism dies, intake stops and C-14 decays. Measuring the C-14 fraction in a sample dates the death. Useful up to about $50\,000$ years.

Uranium-238 ( $T_{1/2} = 4.5 \times 10^9$ years), via the U-238 to Pb-206 chain. Used for dating rocks billions of years old.

Nuclear medicine

Technetium-99m ( $T_{1/2} = 6$ hours). Most-used medical radionuclide. Emits a gamma photon at $140$ keV that imaging cameras detect. Short half-life: most of the dose has decayed by the next day.

Iodine-131 ( $T_{1/2} = 8.0$ days). Concentrates in the thyroid; used to treat hyperthyroidism and thyroid cancer.

Fluorine-18 ( $T_{1/2} = 110$ minutes). Beta-plus emitter used in PET imaging.

VCAA exam style

Year 11 SAC tasks include:

Compute number remaining after $n$ integer half-lives.
Use the continuous formula for non-integer times.
Compute the age of a sample given the remaining fraction.
Choose between candidate isotopes for an application based on half-life.

Common traps

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

Mixing units of time. $T_{1/2}$ and $t$ must be in the same units.

Using $\lambda = T_{1/2} / \ln 2$ . It is the other way: $\lambda = \ln 2 / T_{1/2}$ .

Treating activity as constant. Activity drops exponentially along with the number of nuclei.

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 both halve; integer-half-life problems use $N = N_0 (1/2)^n$ , and applications include carbon-14 dating ( $T_{1/2} = 5730$ years), uranium-lead dating, and medical isotopes such as Tc-99m and I-131.

Past exam questions, worked

Real questions from past VCAA papers on this dot point, with our answer explainer.

Year 11 SAC4 marksA wooden artefact's carbon-14 activity is $25$% of that found in living wood. Use $T_{1/2} = 5730$ years to estimate the artefact's age.

Show worked answer →

Activity ratio = number ratio (because $A \propto N$ ).

$0.25 = (1/2)^n$ where $n$ is the number of half-lives.

$(1/2)^n = 1/4 = (1/2)^2$ , so $n = 2$ .

Age = $2 \times 5730 = 11\,460$ years.

Markers reward equating activity ratio to number ratio, identifying $n = 2$ , and the multiplication.