A sample has 25% parent atoms remaining. How many half-lives have passed?

After 1 half-life: 50%. After 2 half-lives: 25%. Answer: 2 half-lives.

A bone has 12.5% of its original C-14. How old is it? (t₁/₂ = 5,730 yr).

12.5% = (1/2)³ → 3 half-lives. Age = 3 × 5,730 = 17,190 years.

Why can't C-14 dating be used for a 100-million-year-old dinosaur bone?

After ~10 half-lives (~57,300 years), virtually all C-14 has decayed to undetectable levels. For very old samples, use K-40 or U-238 with billion-year half-lives.

Pro 🔒~20 min

Radiometric Dating

Exponential decay and half-life in geological dating

How it works

Radiometric Dating demonstrates a key principle: Radioactive isotopes decay at a constant rate characterized by their half-life — the time for half the parent atoms to convert to daughter atoms. Radioactive isotopes decay at a constant rate characterized by their half-life — the time for half the parent atoms to convert to daughter atoms. After 1 half-life, 50% remain; after 2, 25%; after 3, 12.5%. This exponential decay follows N(t) = N₀ × (1/2)^(t/t₁/₂). Different isotopes have vastly different half-lives: C-14 (5,730 years, good for organic remains up to ~50,000 years), K-40 (1.25 billion years, for ancient rocks), U-238 (4.47 billion years, for the oldest rocks and meteorites). By measuring the ratio of parent to daughter isotopes in a sample, geologists calculate its age.

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Step-by-step

Select an isotope to see its half-life.
Press 'Start Decay' to watch parent atoms (orange) transform into daughter atoms (blue).
The decay curve plots the fraction remaining over time.
Vertical dashed lines mark each half-life.
Drag the time slider to jump to any point.
The data panel calculates the age from the current parent/daughter ratio.

Key formulas

$N(t) = N_0 \left(\frac{1}{2}\right)^{t/t_{1/2}}$ Number of parent atoms remaining: N₀ = initial count, t = elapsed time, t₁/₂ = half-life
$t = t_{1/2} \cdot \frac{\ln(N_0/N)}{\ln 2}$ Age calculation from parent/daughter ratio: rearranged decay equation

Frequently asked questions

A sample has 25% parent atoms remaining. How many half-lives have passed?: After 1 half-life: 50%. After 2 half-lives: 25%. Answer: 2 half-lives.
A bone has 12.5% of its original C-14. How old is it? (t₁/₂ = 5,730 yr).: 12.5% = (1/2)³ → 3 half-lives. Age = 3 × 5,730 = 17,190 years.
Why can't C-14 dating be used for a 100-million-year-old dinosaur bone?: After ~10 half-lives (~57,300 years), virtually all C-14 has decayed to undetectable levels. For very old samples, use K-40 or U-238 with billion-year half-lives.