How do you calculate half-life?

Use N = N₀ × (½)^(t/T½). Rearranged, the time is t = T½ × log₂(N₀/N) and the half-life is T½ = t × ln2 / ln(N₀/N). If 25% of a sample remains after 10 days, two half-lives have passed and T½ = 5 days.

Why are there two carbon-14 half-lives, 5730 and 5568 years?

5568 years is Libby's original measurement. By international convention, published conventional radiocarbon ages (BP, relative to AD 1950) still use it and are then calibrated to calendar years with the IntCal curve. The Cambridge value of 5730 ± 40 years is closer to the true half-life and is what most textbook problems use; ages differ by about 3%.

What is the difference between becquerels and curies?

One becquerel is one decay per second. The curie was originally the activity of 1 g of radium-226 and is now defined as 3.7 × 10¹⁰ Bq. Since A = λN, shorter half-lives and lighter isotopes give higher activity per gram: 1 mg of cobalt-60 is about 41.8 GBq (1.13 Ci).

How is radioactive decay related to first-order kinetics?

Both have a rate proportional to the amount present, so the maths is identical: ln[A] = ln[A]₀ − kt and t½ = ln2 / k, independent of the starting concentration. The decomposition of N₂O₅, sucrose hydrolysis and much drug elimination are first order. Tick first-order kinetics to work with the rate constant k directly.

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☢️Half-Life Calculator - Radioactive Decay, Carbon-14 Dating, Activity

After every half-life T½ a radioactive sample has half as many atoms left: N = N₀ × (½)^(t/T½) = N₀e^(−λt), with decay constant λ = ln2 / T½ and mean life τ = 1/λ ≈ 1.443 T½. Ten grams of iodine-131 (T½ ≈ 8.02 days) shrink to 10 × ⅛ = 1.25 g after 24.06 days, three half-lives; a sample with 25% of its carbon-14 left has been through two half-lives, about 2 × 5730 = 11,460 years. This calculator solves for any one of N₀, N, t and T½, with times in seconds, minutes, hours, days or years, and includes more than 30 common isotopes — carbon-14, uranium-238, iodine-131, cobalt-60, caesium-137, radium-226, radon-222, technetium-99m, fluorine-18 and more — with evaluated NNDC/NUBASE half-lives. A decay curve marks each half-life and lets you drag to read the amount left at any time. The activity tab turns a sample mass into the number of atoms, the activity A = λN in becquerels and curies, the specific activity and the activity after a given time. The carbon-14 tab dates a sample from the percent remaining or a measured specific activity, giving both the 5730-year result and the conventional radiocarbon age with the Libby half-life of 5568 years. Tick first-order kinetics to use the same maths for [A] = [A]₀e^(−kt) and t½ = ln2 / k, entering a rate constant directly.

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How to use Half-Life Calculator

1Pick an isotope (or type a custom half-life) and choose the quantity to solve for
2Fill in the other three values and units; read the result, λ, τ and the decay curve
3Use the Activity or Carbon-14 dating tabs as needed, or tick first-order kinetics for chemistry problems

FAQ

How do you calculate half-life?: Use N = N₀ × (½)^(t/T½). Rearranged, the time is t = T½ × log₂(N₀/N) and the half-life is T½ = t × ln2 / ln(N₀/N). If 25% of a sample remains after 10 days, two half-lives have passed and T½ = 5 days.
Why are there two carbon-14 half-lives, 5730 and 5568 years?: 5568 years is Libby's original measurement. By international convention, published conventional radiocarbon ages (BP, relative to AD 1950) still use it and are then calibrated to calendar years with the IntCal curve. The Cambridge value of 5730 ± 40 years is closer to the true half-life and is what most textbook problems use; ages differ by about 3%.
What is the difference between becquerels and curies?: One becquerel is one decay per second. The curie was originally the activity of 1 g of radium-226 and is now defined as 3.7 × 10¹⁰ Bq. Since A = λN, shorter half-lives and lighter isotopes give higher activity per gram: 1 mg of cobalt-60 is about 41.8 GBq (1.13 Ci).
How is radioactive decay related to first-order kinetics?: Both have a rate proportional to the amount present, so the maths is identical: ln[A] = ln[A]₀ − kt and t½ = ln2 / k, independent of the starting concentration. The decomposition of N₂O₅, sucrose hydrolysis and much drug elimination are first order. Tick first-order kinetics to work with the rate constant k directly.