How much is left after its half-life?
Fill any three of the initial amount, the amount left, the time passed, and the half-life. The half life calculator finds the fourth, with the decay constant, the mean lifetime, and a decay curve.
- Amount left (N)
- 29.8292
After 10,000 yr, 29.8292 of 100 is left, with a half-life of 5,730 yr.
- Percent left
- 29.8292%
- Half-lives passed
- 1.7452
- Decay constant (λ, per year)per year
- 0.000120968
- Mean lifetime (τ)
- 8,266.64 yr
Amount left (N): 29.8292. After 10,000 yr, 29.8292 of 100 is left, with a half-life of 5,730 yr.
How much is left after each half-life?
How to calculate
Works out radioactive decay with the half-life formula N = N₀ × (1/2)^(t ÷ T½): the amount left, the initial amount, the time passed, or the half-life, with the decay constant and mean lifetime.
Example with the default inputs (Initial amount (N₀) 100, Time passed (t) 10,000 yr, Half-life (T½) 5,730 yr): After 10,000 yr, 29.8292 of 100 is left, with a half-life of 5,730 yr.
Formula: N = N₀ × (1/2)^(t ÷ T½); N₀ = N × 2^(t ÷ T½); t = T½ × log₂(N₀ ÷ N); T½ = t ÷ log₂(N₀ ÷ N); λ = ln 2 ÷ T½; τ = T½ ÷ ln 2.
- Decay is first order: the same fraction decays in each equal time, whatever the amount (OpenStax Chemistry 2e, 21.3).
- The amount left and the initial amount are in the same unit; any unit works.
- A year is 365 days.
Worked examples
Each example is checked against the calculator on every build.
- Initial amount (N₀) 1, Time passed (t) 15 yr, Half-life (T½) 5.27 yr gives Amount left (N) 0.139052, Percent left 13.905237%, Decay constant (λ, per year) 0.131527.Source: OpenStax, Chemistry 2e, section 21.3 Radioactive Decay (λ = ln 2 ÷ t½, Nₜ = N₀e^(−λt)), https://openstax.org/books/chemistry-2e/pages/21-3-radioactive-decay, retrieved 2026-10-02, example 21.6 (cobalt-60; the book rounds λ to 0.132 per year and gets 13.8%)
- Initial amount (N₀) 13.6, Amount left (N) 10.8, Half-life (T½) 5,730 yr gives Time passed (t) 1,906 yr, Half-lives passed 0.332575.Source: OpenStax, Chemistry 2e, section 21.3 Radioactive Decay (λ = ln 2 ÷ t½, Nₜ = N₀e^(−λt)), https://openstax.org/books/chemistry-2e/pages/21-3-radioactive-decay, retrieved 2026-10-02, example 21.7 (Dead Sea Scrolls, carbon-14 half-life 5,730 years, about 1,900 years old)
- Initial amount (N₀) 100, Amount left (N) 25, Time passed (t) 0.0274 yr gives Half-life (T½) 0.0137 yr.
- Amount left (N) 10, Time passed (t) 3 yr, Half-life (T½) 1 yr gives Initial amount (N₀) 80.
- Initial amount (N₀) 100, Time passed (t) 10,000 yr, Half-life (T½) 5,730 yr gives Amount left (N) 29.829244.
How it works
Fill any three of the initial amount N₀, the amount left N, the time passed t, and the half-life T½, and leave one empty:
- N = N₀ × (1/2)^(t ÷ T½)
- N₀ = N × 2^(t ÷ T½)
- t = T½ × log₂(N₀ ÷ N)
- T½ = t ÷ log₂(N₀ ÷ N)
This is the first-order decay law of OpenStax Chemistry 2e, section 21.3, N = N₀e^(−λt), written with λ = ln 2 ÷ T½. The calculator also shows:
- Percent left: N ÷ N₀ × 100.
- Half-lives passed: t ÷ T½.
- Decay constant (λ, per year): ln 2 ÷ T½, with T½ in years of 365 days.
- Mean lifetime (τ): T½ ÷ ln 2, shown in years of 365 days.
Limits: N₀ and N are more than 0 and at most 10^15, in the same unit (any unit); t is from 0 to 10^20 seconds; T½ is from 10^-12 to 10^20 seconds. A solved value outside these limits gives no answer: solving for t with N larger than N₀ gives a negative time, which has no answer, and N = N₀ gives t = 0. When solving for T½ with N at least N₀, with t = 0, or with a result under 10^-12 seconds, the page says: "No half-life of 10⁻¹² seconds or more fits: the amount left must be less than the initial amount, and the time more than 0." Times can be typed in seconds, minutes, hours, days, or years; a year is 365 days (31,536,000 s).
The chart draws N₀ × (1/2)^h for h from 0 to the larger of 5 and t ÷ T½ rounded up, with a mark at h = t ÷ T½.
Worked examples by hand
Cobalt-60, T½ = 5.27 years, after 15 years. (1/2)^(15 ÷ 5.27) = 2^(−2.8463) = 0.13905, so 13.905% is left; λ = 0.693147 ÷ 5.27 = 0.131527 per year. OpenStax rounds λ to 0.132 and gets 13.8%.
Carbon dating, 10.8 against 13.6 counts per minute per gram, T½ = 5,730 years. log₂(13.6 ÷ 10.8) = 0.332575 half-lives, so t = 5,730 × 0.332575 = 1,905.66 years (60,096,789,469 s). OpenStax gives about 1,900 years.
100 g to 25 g in 10 days. log₂(100 ÷ 25) = 2, so T½ = 10 ÷ 2 = 5 days (432,000 s).
10 g left after 3 years, T½ = 1 year. N₀ = 10 × 2³ = 80 g.
100 g, T½ = 5,730 years, after 10,000 years. 100 × 2^(−10,000 ÷ 5,730) = 29.83 g.
Other questions people ask
What is a half-life?
The time it takes for half of a radioactive substance to decay. After one half-life 1/2 is left, after two 1/4, after three 1/8. Carbon-14 has a half-life of 5,730 years; cobalt-60, 5.27 years.
How do I calculate the amount left?
Use N = N₀ × (1/2)^(t ÷ T½). For cobalt-60 (T½ = 5.27 years) after 15 years: (1/2)^(15 ÷ 5.27) = 0.139, so 13.9% is left.
How do I find the half-life from two amounts?
Use T½ = t ÷ log₂(N₀ ÷ N). If 100 g drops to 25 g in 10 days, log₂(100 ÷ 25) = 2 half-lives passed, so T½ = 10 ÷ 2 = 5 days.
How does carbon dating use the half-life?
Living things keep the same share of carbon-14; after death it decays. Comparing the activity now with that of living matter gives the age: t = T½ × log₂(N₀ ÷ N). For 10.8 counts per minute per gram against 13.6, t = 5,730 × log₂(13.6 ÷ 10.8) = about 1,906 years.
What is the decay constant?
λ = ln 2 ÷ T½, the fraction of atoms that decays per unit of time, used in N = N₀e^(−λt). For cobalt-60, λ = 0.693 ÷ 5.27 = 0.1315 per year. The calculator gives λ per year.
What is the mean lifetime?
The average time an atom lasts before it decays: τ = T½ ÷ ln 2, about 1.44 half-lives. After one mean lifetime, 1/e (about 36.8%) of the substance is left.
Does the half-life depend on the amount?
No. Radioactive decay is first order, so the same fraction decays in each half-life whether you start with 1 g or 1 kg.