Physics, units and measurement
Half-life and radioactive decay calculator
Solve N = N0 (1/2)^(t/T½) for the remaining amount, initial amount, elapsed time or half-life, with the decay constant and mean lifetime.
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Result
- Elapsed time t
- 11460 years
- Fraction remaining
- 25.00 %
- Half-lives elapsed
- 2.000
- Decay constant λ = ln 2 ÷ T½
- 1.210 × 10⁻⁴ per year
- λ in SI
- 3.833 × 10⁻¹² s⁻¹
- Mean lifetime τ = T½ ÷ ln 2
- 8267 years
Working
- N = N₀ e^(−λt) = N₀ (1/2)^(t/T½)
- t = T½ × log₂(N₀/N) = 5730 years × log₂(100.0 / 25.00) = 11460 years
| Half-lives | Time | Amount | Remaining |
|---|---|---|---|
| 0 | 0 years | 100.0 % | 100.0 % |
| 1 | 5730 years | 50.00 % | 50.00 % |
| 2 | 11460 years | 25.00 % | 25.00 % |
| 3 | 17190 years | 12.50 % | 12.50 % |
| 4 | 22920 years | 6.250 % | 6.250 % |
| 5 | 28650 years | 3.125 % | 3.125 % |
| 6 | 34380 years | 1.563 % | 1.563 % |
| 7 | 40110 years | 0.7813 % | 0.7813 % |
| 8 | 45840 years | 0.3906 % | 0.3906 % |
| 10 | 57300 years | 0.09766 % | 0.09766 % |
- The amount can be a mass, a number of atoms or an activity (Bq, Ci, counts per minute): all decay with the same half-life. Keep both amounts in the same unit.
- One year here is the Julian year (365.25 days), as used for half-lives in nuclear data tables.
- Rounded to 4 significant figures.
Method and limits
Pick the unknown and enter the other values. The amount can be a mass, a number of atoms or an activity: they all fall by half in each half-life, so only the ratio N ÷ N₀ matters.
N = N₀ (1/2)^(t/T½) = N₀ e^(−λt) λ = ln 2 ÷ T½ τ = 1 ÷ λ = T½ ÷ ln 2
t = T½ × log₂(N₀ ÷ N) T½ = t × ln 2 ÷ ln(N₀ ÷ N)
Activity A = λN, so activities decay with the same half-life. The table lists what remains after whole numbers of half-lives. A year here is the Julian year of 365.25 days.
Limits
Radioactive decay is random: the law describes large numbers of atoms, and a count of a few hundred events has a statistical uncertainty of several percent. Daughter products that are themselves radioactive (decay chains) need the Bateman equations. Half-lives should come from an evaluated source such as the NNDC NuDat database. Answers keep the significant figures of the least precise number you enter (at least 3, at most 6).
Worked example
Radiocarbon: a sample keeps 25.00 % of its carbon-14. With the conventional Libby–Cambridge half-life of 5730 years, t = 5730 × log₂(100.0 ÷ 25.00) = 2 half-lives = 11 460 years. (The evaluated half-life is 5700 ± 30 years; radiocarbon ages are still reported with Libby’s 5568 years and then calibrated.)
References
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