Every radioactive nucleus has the same chance of decaying in the next second, whatever its age. That is exactly a first-order process, so the equations you know apply directly, and they explain the half-lives you met before.
Key idea
$N = N_0e^{-kt}$, $\ln\dfrac{N_0}{N} = kt$, and $t_{1/2} = \dfrac{0.693}{k}$, constant for each isotope. Activity (decays per second, in becquerels) is $A = kN$, so it falls by the same law. These work at any time, not just at whole numbers of half-lives.
Why it works
A nucleus has no memory and no timer. In any short time interval $dt$, each one has the same small probability $k\,dt$ of decaying. So the number that decay is proportional to the number present: $-dN/dt = kN$. That is the first-order rate law, and its solution is the exponential. Temperature, pressure and chemical bonding have essentially no effect on $k$, because the nucleus is shielded by its electrons.
Worked examples
Worked example: Iodine-131
Iodine-131 has a half-life of $8.02\ \text{days}$. What percent of a sample is left after $30.0\ \text{days}$?
- $k = \dfrac{0.693}{8.02\ \text{d}} = 0.0864\ \text{d}^{-1}$; $kt = (0.0864)(30.0) = 2.59$.
- $\dfrac{N}{N_0} = e^{-2.59} = 0.0748$: 7.48% is left. (Same answer: $(\tfrac{1}{2})^{30.0/8.02} = (\tfrac{1}{2})^{3.74}$.)
Answer: 7.48%.
Worked example: Radiocarbon dating
Wood from an old site has 40.0% of the carbon-14 per gram of living wood. Using $t_{1/2} = 5700$ years, how old is it?
- $k = \dfrac{0.693}{5700\ \text{yr}} = 1.216 \times 10^{-4}\ \text{yr}^{-1}$.
- $$t = \dfrac{1}{k}\ln\dfrac{N_0}{N} = \dfrac{\ln(1/0.400)}{1.216 \times 10^{-4}} = \dfrac{0.916}{1.216 \times 10^{-4}} = 7.53 \times 10^3\ \text{yr}$$
- Check: 40% is between 50% (5700 yr) and 25% (11 400 yr), as 7500 years is.
Answer: About 7500 years.
Worked example: Activity from a count of atoms
A sample has $1.0 \times 10^{15}$ atoms of iodine-131 ($t_{1/2} = 8.02$ days). Find its activity.
- $k$ in $\text{s}^{-1}$: $t_{1/2} = 8.02 \times 86\,400 = 6.93 \times 10^5\ \text{s}$, so $k = \dfrac{0.693}{6.93 \times 10^5} = 1.00 \times 10^{-6}\ \text{s}^{-1}$.
- $$A = kN = (1.00 \times 10^{-6})(1.0 \times 10^{15}) = 1.0 \times 10^9\ \text{Bq}$$
Answer: $1.0 \times 10^9$ decays per second.
Common mistake
Thinking the half-life gets shorter as the sample shrinks, as if each nucleus carried a timer. Each nucleus's chance of decaying stays the same; only the number of nuclei changes. Check: in a first-order process every half-life is the same length.
Living things keep a steady carbon-14 level by exchanging carbon with the air; after death the C-14 decays without being replaced. The measured half-life is about 5700 years (many older books use 5730), and real radiocarbon ages are corrected with calibration curves because the level in the air has varied a little over time.
Watch half-lives date ancient wood and bone in the Science Lab's Build an Atom.