Half-Life Calculator

Radioactive decay (and several other natural processes) shrink by half over a fixed time interval, repeatedly — never fully disappearing in a finite time, only ever getting closer to zero.

Inputs

Result

25

25% remaining

How the half-life calculator works

Remaining amount = initial amount × 0.5^(elapsed time ÷ half-life).

Worked example: Carbon-14, half-life 5,730 years, 11,460 years elapsed

  1. Elapsed time ÷ half-life = 11,460 ÷ 5,730 = 2 half-lives exactly.
  2. Remaining = 100 × 0.5² = 100 × 0.25 = 25.
  3. So 25% of the original Carbon-14 remains after exactly two half-lives — this is the basis of radiocarbon dating.

Common mistakes to avoid

Assuming the substance is completely gone after one half-life

One half-life means half remains, not none — the amount keeps halving indefinitely (25% after two half-lives, 12.5% after three), approaching but never mathematically reaching exactly zero.

Treating half-life as a linear decay rate

The decay is exponential, not linear — the substance doesn't lose the same fixed amount every year; it loses a fixed proportion of whatever remains, so the actual amount lost keeps shrinking over time even as the percentage rate stays constant.

Frequently asked questions

How is half-life used in radiocarbon dating?

By measuring how much Carbon-14 remains in an organic sample relative to the amount expected when it was alive, and knowing Carbon-14's half-life (5,730 years), the elapsed time since the organism died can be calculated using this same formula rearranged for time.

Does half-life apply to anything besides radioactivity?

Yes — the same exponential-decay mathematics describes drug elimination from the body (pharmacokinetics), certain chemical reaction rates, and even some population decline models, wherever a quantity decreases by a fixed proportion per fixed time interval.

Why can't a substance ever reach exactly zero using this model?

Because each half-life only ever halves the remaining amount — mathematically, repeatedly multiplying by 0.5 approaches zero asymptotically but never reaches it exactly, though in practice a physically negligible amount remains after enough half-lives.

How would I find the half-life if I know the remaining fraction and elapsed time?

Rearrange the formula: half-life = elapsed time × ln(0.5) ÷ ln(remaining fraction ÷ initial amount) — using natural logarithms to solve for the exponent.

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