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)

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.

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:

For integer numbers of half-lives, use the simplified 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). 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.

Examples in context

Example 1. ANSTO Lucas Heights technetium-99m production for Melbourne hospitals. ANSTO's OPAL reactor at Lucas Heights produces molybdenum-99 (half-life $66$ hours), which decays to technetium-99m (half-life $6.0$ hours) used in approximately $500$ scans per day across Australian hospitals including the Royal Melbourne. From the moment a generator leaves Sydney, the $99$Mo activity falls by $N/N_0 = (1/2)^{t/66}$. After $24$ hours of road transport plus storage, $N/N_0 = (1/2)^{24/66} = 0.78$, so $78\%$ of activity remains. The $99$mTc daughter is then eluted on the morning of imaging, providing fresh $140$ keV gamma emitters whose $6$ hour half-life clears from the patient overnight.

Example 2. Carbon-14 dating of charcoal from Mungo Lake hearths. Carbon-14 (half-life $5730$ years) decays steadily after an organism dies. A charcoal sample from a Lake Mungo hearth in NSW gives $N/N_0 = 0.025$. Solving $0.025 = (1/2)^{t/5730}$ yields $t = 5730 \log_2(40) = 5730 \times 5.32 \approx 30\,500$ years before present. The result aligns with luminescence dating of surrounding sediments and supports the antiquity of human occupation at Mungo. Carbon-14 is only useful out to about $50\,000$ years because after that $N/N_0 < 0.002$ and the small remaining signal cannot be reliably separated from background.

Try this

Q1. Define half-life and state the SI unit. [2 marks]

• Cue. Time for half of the original radioactive nuclei in a sample to decay; measured in seconds (or any time unit).

Q2. A medical $^{131}$I sample (half-life $8.0$ days) has an initial activity of $200$ MBq. Calculate (a) the activity after $24$ days, and (b) the time at which activity falls to $12.5$ MBq. [2+2 marks]

• Cue. (a) $200 \times (1/2)^3 = 25$ MBq. (b) $12.5/200 = 1/16 = (1/2)^4$ so $t = 32$ days.

Q3. Refer to ANSTO production. (a) Calculate the fraction of $^{99}$Mo remaining after $24$ hours given a $66$ hour half-life. (b) Determine the activity ratio of $^{99}$mTc daughter to $^{99}$Mo parent at secular equilibrium. (c) Outline two reasons why $^{99}$mTc is preferred for diagnostic imaging. [2+3+2 marks]

• Cue. (a) $(1/2)^{24/66} = 0.78$. (b) Activities approximately equal at secular equilibrium. (c) Short half-life clears quickly; $140$ keV gamma is well matched to detectors.