Isotope dating — the half-life

What just happened?

You set 2 half-lives — $\tfrac14$ of the C-14 nuclei are left. Every half-life $T_{1/2}$ removes half of what is still left, so the population falls like a power of two, not linearly.

Decay law:

— the number of nuclei that have not yet decayed falls exponentially. $N_0$ — the initial number, $T_{1/2}$ — the half-life (for C-14 ≈ 5730 years), $t$ — time.

What does it mean in practice?

After each successive $T_{1/2}$ one half drops away: in turn $\tfrac12,\ \tfrac14,\ \tfrac18,\ \tfrac{1}{16}\ldots$ Inverting the formula, the remaining fraction gives you the age of the sample:

Why does this date artefacts?

A living organism keeps a constant share of C-14. After death the intake stops and the isotope decays. By measuring how much C-14 is left relative to $N_0$, we work out how many $T_{1/2}$ have passed — that is the age of the remains.

Why you need this for the exam:

Decay problems come down to counting half-lives: divide the time by $T_{1/2}$, and compute the fraction as $(\tfrac12)^n$. Trick: count how many times “half of a half”, instead of integrating.

Remember: after each half-life $T_{1/2}$ half of the previous amount remains — N = N₀(½)^(t/T½). This is the essence of radioactive decay and isotope dating.