N = N₀(½)^(t/t½) — Radioactive Half-Life Law

Radioactive half-life decay — how much remains after a given time — with a worked carbon-14 example.

N = N_0\left(\frac{1}{2}\right)^{t/t_{1/2}}

Remaining nuclei equal the initial number times one-half raised to the number of elapsed half-lives.

What it means

Radioactive decay is exponential: every half-life, half the remaining nuclei decay — regardless of how many you started with. After one half-life 1/2 remains, after two 1/4, after three 1/8. The decay never truly reaches zero, but it becomes negligible.

Variables

Symbol	Name	Unit	Meaning
$N$	Remaining nuclei	count	Nuclei surviving after time t
$N₀$	Initial nuclei	count	Nuclei at time zero
$t$	Time	s (or years)	Elapsed time
$t₁/₂$	Half-life	s (or years)	Time for half the nuclei to decay

Assumptions

Decay probability per nucleus is constant
Large number of nuclei (statistical law)
No new nuclei are produced during the interval

Derivation

Each half-life multiplies the survivors by 1/2, so after $n = t/t_{1/2}$ half-lives the fraction is $(1/2)^n$ . This is the discrete form of the exponential decay law $N = N_0 e^{-\lambda t}$ .

Worked example

Problem

A sample's carbon-14 has decayed to 1/8 of its original amount. How old is the sample? (t₁/₂ = 5,730 years.)

Given:

Fraction remaining = 1/8
t₁/₂ = 5,730 years (carbon-14)

Write the law: $N = N_0(1/2)^{t/t_{1/2}}$
Set $N/N_0 = 1/8 = (1/2)^3$ , so $t/t_{1/2} = 3$
Calculate: $t = 3 \times 5730 = 17{,}190\ \mathrm{years}$

17,190 years

1/8 = (1/2)³, so three half-lives have passed: 3 × 5,730 = 17,190 years. Archaeologists use exactly this logic to date ancient organic remains.