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Formula
N = N₀ × (1/2)^(t ÷ T½)
- N
- amount remaining after time t (same unit as N₀)
- N₀
- initial amount, in any unit (grams, atoms, becquerels, curies, percent)
- t
- time elapsed
- T½
- half-life, the time for the amount to fall by half
- λ
- decay constant, λ = ln 2 ÷ T½; the mean lifetime is τ = 1 ÷ λ
What half-life means
A half-life is the time it takes for half of something to disappear through decay. Radioactive nuclei, first-order chemical reactions and many drugs in the body all lose a fixed fraction of what is left in each interval of time. That gives the exponential decay law:
N = N₀ × (1/2)^(t ÷ T½)
After one half-life, 50% remains. After two, 25%. After three, 12.5%. The half-life stays the same no matter how much you start with, which is why it is a useful fingerprint for an isotope or a reaction.
How to use the half-life calculator
Pick the quantity you don’t know:
- Amount remaining: enter the initial amount, the time elapsed and the half-life.
- Initial amount: enter what is left now, the time elapsed and the half-life to work backward.
- Time elapsed: enter the starting and remaining amounts and the half-life. This is the radiocarbon dating case.
- Half-life: enter two measurements and the time between them.
Amounts can be in any unit, from grams and atoms to becquerels, millicuries or percent, as long as both use the same one. Time and half-life each have their own unit selector, so a half-life in days and an elapsed time in years is fine. Every answer comes with the steps, the number of half-lives, the decay constant λ, the mean lifetime and a table of the amount left after each half-life.
How to calculate half-life problems by hand
Amount remaining
Cobalt-60 has a half-life of 5.2711 years. How much of a 100 mCi source is left after 20 years?
- Number of half-lives: 20 ÷ 5.2711 = 3.794
- Fraction left: (1/2)^3.794 = 0.0721
- Amount left: 100 × 0.0721 = 7.21 mCi
Time elapsed
How long does it take cesium-137 (half-life 30.018 years) to fall to 10% of its starting activity?
t = T½ × log₂(N₀ ÷ N) = 30.018 × log₂(10) = 30.018 × 3.3219 = 99.7 years
On a basic calculator, log₂(x) = ln(x) ÷ ln(2).
Half-life from data
A sample’s count rate falls from 800 to 50 counts per minute over 12 days. N₀ ÷ N = 16, and log₂(16) = 4 half-lives, so T½ = 12 ÷ 4 = 3 days.
Decay constant and mean lifetime
The decay constant λ = ln 2 ÷ T½ is the fraction decaying per unit time when the fraction is small. For cobalt-60, λ = 0.693 ÷ 5.2711 = 0.1315 per year. The mean lifetime, τ = 1 ÷ λ = 7.60 years, is the average time a nucleus survives, which is always about 1.44 times the half-life.
Fraction left after each half-life
| Half-lives elapsed | Fraction left | Percent left |
|---|---|---|
| 1 | 1/2 | 50% |
| 2 | 1/4 | 25% |
| 3 | 1/8 | 12.5% |
| 4 | 1/16 | 6.25% |
| 5 | 1/32 | 3.125% |
| 7 | 1/128 | 0.78125% |
| 10 | 1/1,024 | 0.0977% |
Half-lives of common radioactive isotopes
Recommended values from the Decay Data Evaluation Project (DDEP), published by LNHB.
| Isotope | Use | Half-life |
|---|---|---|
| Fluorine-18 | PET imaging | 1.82890 hours |
| Technetium-99m | Medical imaging | 6.0067 hours |
| Iodine-131 | Thyroid treatment | 8.0233 days |
| Cobalt-60 | Radiotherapy, sterilization | 5.2711 years |
| Hydrogen-3 (tritium) | Glow signs, tracers | 12.312 years |
| Cesium-137 | Gauges, fallout studies | 30.018 years |
| Carbon-14 | Radiocarbon dating | 5,700 years |
Many textbooks, including OpenStax Chemistry 2e, still give carbon-14 as 5,730 years. The newer evaluated value is 5,700 ± 30 years. Both are within the uncertainty for most homework problems, but say which one you used.
Where half-life calculations are used
- Nuclear medicine: working out how much activity a tracer has at the time of a scan, or when a patient’s dose has decayed.
- Radiocarbon dating: estimating the age of wood, bone or cloth from the carbon-14 left.
- Chemical kinetics: predicting how long a first-order reaction takes to reach a target conversion.
- Pharmacology: estimating how much of a single dose remains. A common rule of thumb is that a drug is largely eliminated after about 5 half-lives, when about 3% remains.
- Radiation safety: deciding how long to store waste before its activity is low enough to handle.
The halving pattern is exponential decay, the mirror image of exponential growth. The population growth calculator handles the growing case, and the exponent calculator evaluates powers such as (1/2)^3.794.
Common mistakes
- Dividing the time by the half-life and stopping there. t ÷ T½ is the number of half-lives, not the fraction left. You still need (1/2) to that power.
- Assuming straight-line decay. Half of the sample is gone after one half-life, but the rest does not vanish after a second one; a quarter remains.
- Mixing time units. A half-life in hours and an elapsed time in days must be converted to the same unit. The calculator does this for you.
- Using half-life for reactions that aren’t first order. For zero- and second-order reactions, the half-life changes as the concentration falls.
- Applying it to repeated doses. Medicines taken regularly build up toward a steady state, which a single-decay formula doesn’t describe.
Glossary
- Half-life (T½): time for the amount to fall to half its value.
- Decay constant (λ): ln 2 ÷ T½, the probability per unit time that a nucleus decays.
- Mean lifetime (τ): 1 ÷ λ, the average survival time.
- Activity: decays per second, measured in becquerels (Bq) or curies (Ci). It is proportional to the number of nuclei, so it follows the same half-life.
- First-order reaction: a reaction whose rate is proportional to the concentration of one reactant.
Worked examples
Iodine-131 after four weeks
An 80 mCi dose of iodine-131 (half-life 8.0233 days). How much activity is left after 28 days?
Inputs (Amount remaining): Initial amount (N₀) = 80; Time elapsed (t) = 28; Time unit = Days; Half-life (T½) = 8.0233; Half-life unit = Days
N = N₀ × (1/2)^(t ÷ T½) = 80 × (1/2)^(28 days ÷ 8.0233 days) = 80 × (1/2)^3.48984 = 80 × 0.0890133 = 7.12106
Answer: After 28 days, 7.12106 of 80 remains (half-life 8.0233 days).
Radiocarbon age of a sample
Wood with 35% of the carbon-14 of living wood, using the 5,700-year half-life.
Inputs (Time elapsed): Initial amount (N₀) = 100; Amount remaining (N) = 35; Half-life (T½) = 5700; Half-life unit = Years; Show time in = Years
t = T½ × log₂(N₀ ÷ N) = 5,700 years × log₂(100 ÷ 35) = 5,700 years × 1.51457 = 8,633.07 years
Answer: Decaying from 100 to 35 with a half-life of 5,700 years takes 8,633.07 years (1.51457 half-lives).
Half-life from a lab measurement
A count rate falls from 1,200 to 300 counts per minute in 90 minutes.
Inputs (Half-life): Initial amount (N₀) = 1200; Amount remaining (N) = 300; Time elapsed (t) = 90; Time unit = Minutes; Show half-life in = Minutes
T½ = t ÷ log₂(N₀ ÷ N) = 90 min ÷ log₂(1,200 ÷ 300) = 90 min ÷ 2 = 45 min
Answer: Going from 1,200 to 300 in 90 min means a half-life of 45 min.
Starting activity of a medical isotope
A fluorine-18 PET tracer (half-life 1.8289 hours) reads 15 mCi after 5.5 hours. What was the starting activity?
Inputs (Initial amount): Amount remaining (N) = 15; Time elapsed (t) = 5.5; Time unit = Hours; Half-life (T½) = 1.8289; Half-life unit = Hours
N₀ = N ÷ (1/2)^(t ÷ T½) = 15 ÷ (1/2)^(5.5 h ÷ 1.8289 h) = 15 × 2^3.00727 = 120.606
Answer: 15 remaining after 5.5 h (half-life 1.8289 h) means 120.606 at the start.
Edge case: zero time elapsed
A technetium-99m sample at the moment it is measured: nothing has decayed yet.
Inputs (Amount remaining): Initial amount (N₀) = 50; Time elapsed (t) = 0; Time unit = Hours; Half-life (T½) = 6.0067; Half-life unit = Hours
N = N₀ × (1/2)^(t ÷ T½) = 50 × (1/2)^(0 h ÷ 6.0067 h) = 50 × (1/2)^0 = 50 × 1 = 50
Answer: After 0 h, 50 of 50 remains (half-life 6.0067 h).
Assumptions and limitations
- Decay is exponential (first order), so the same fraction is lost in every equal time interval and the half-life does not depend on how much is present. This holds for radioactive decay and first-order reactions, not for zero- or second-order reactions.
- N₀ and N can be in any unit, as long as both use the same one. The answer for an amount is in that unit.
- Time and half-life can use different units; the calculator converts them. A year is the Julian year of 365.25 days, the convention used for half-lives.
- The calculator follows one substance. It does not add the decay of daughter products, ongoing production or repeated doses.
- Applied to medicines, it describes a single dose that is eliminated by first-order kinetics. It is an estimate, not medical advice.
- Results are shown to 6 significant figures.
Frequently asked questions
What is the half-life formula?
The amount left is N = N₀ × (1/2)^(t ÷ T½). Rearranged, the time elapsed is t = T½ × log₂(N₀ ÷ N) and the half-life is T½ = t ÷ log₂(N₀ ÷ N). With the decay constant, the same law is N = N₀ × e^(−λt), where λ = ln 2 ÷ T½.
How do you calculate half-life from two measurements?
Divide the time between them by the number of halvings, log₂(N₀ ÷ N). If activity drops from 800 to 50 counts per minute in 12 days, that is log₂(16) = 4 halvings, so the half-life is 12 ÷ 4 = 3 days.
How much is left after 3 half-lives?
One eighth, or 12.5%. Each half-life halves what remains, so after n half-lives the fraction left is (1/2)ⁿ. After 5 half-lives about 3.1% is left, and after 10 half-lives about 0.1%.
How do you find the half-life of a first-order reaction?
Use T½ = ln 2 ÷ k ≈ 0.693 ÷ k, where k is the rate constant. A reaction with k = 0.0231 per minute has a half-life of about 30 minutes, whatever the starting concentration.
Does a substance ever completely decay?
Mathematically the amount never reaches exactly zero, but for real samples it becomes negligible. After 20 half-lives less than one millionth remains, so a sample of a few thousand atoms is effectively gone.
Can I use this for a drug's half-life?
For a single dose of a drug that is eliminated by first-order kinetics, yes. Enter the dose as N₀ and the drug's half-life to estimate how much remains. It does not model repeated doses or absorption, and it is not medical advice.
Sources
- Chemistry 2e, 21.3 Radioactive Decay — OpenStax (Rice University). Checked October 10, 2026.
- Chemistry 2e, 12.4 Integrated Rate Laws — OpenStax (Rice University). Checked October 10, 2026.
- Atomic and Nuclear Data: DDEP recommended decay data — Laboratoire National Henri Becquerel (LNHB). Checked October 10, 2026.