Radioactive Decay Law
The decay law describes the exponential decrease of the undecayed atomic nuclei of a radioactive substance.
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Formula
N(t) = N_0 \cdot e^{-\lambda t}Variables & units – Radioactive Decay Law
| Symbol | Meaning | Unit |
|---|---|---|
| N(t) | Number of undecayed nuclei at time t | dimensionless |
| N₀ | Initial number of nuclei (t = 0) | dimensionless |
| λ | Decay constant (substance-specific) | 1/s |
| t | Time | s |
Derivation & background – Radioactive Decay Law
Around 1902 Rutherford and Soddy recognised that per unit of time a fixed fraction of the existing nuclei decays, which necessarily leads to an exponential function. The half-life is related to the decay constant via T½ = ln(2)/λ. After n half-lives the fraction (1/2)ⁿ remains. The activity A = λ·N is measured in becquerels.
Exam blueprint
Validity range
Holds statistically for large numbers of nuclei; for a single nucleus only a decay probability can be given. λ is substance-specific and independent of temperature, pressure or chemical bonding.
Derivation steps
Per unit of time a fixed fraction of the remaining nuclei decays, which defines an exponential function.
- 1Ansatz: dN/dt = −λ·N (decay rate proportional to the stock).
- 2Integration yields N(t) = N₀·e^(−λt).
Rearrangements
Half-life from the decay constant
Follows from N = N₀/2; ln 2 ≈ 0.693.
Time from the remaining fraction
The basic equation of every radiometric dating method.
Form with half-life
Often quicker: count the number of half-lives.
Task variant
After how many half-lives is 6.25% of a sample left?
6.25% = 1/16 = (1/2)⁴, so after 4 half-lives.
A C-14 sample shows 25% activity (T½ = 5,730 a). How old is it?
25% = (1/2)², i.e. 2 half-lives: t = 2 × 5,730 = 11,460 years.
Common mistakes
Assuming everything has decayed after two half-lives.
Each half-life only halves the remainder: after two, 25% is left.
Confusing λ and T½ or substituting one for the other.
They are linked via T½ = ln(2)/λ, not simple reciprocals.
Mixing units of λ and t (e.g. λ per day, t in years).
λ·t must be dimensionless; use the same unit of time.
Assuming linear decay.
The decay is exponential; equal time spans mean equal factors, not equal differences.
Exam context
- Standard: dating (C-14), activity decrease in nuclear medicine, semi-log evaluation of measurement series.
These mistakes cost points in real exams. The set drills them until they stick.
Formula cluster
Nuclear physics
Decay, mass defect and radiation belong together in final exams.
Worked example
Iodine-131 has T½ = 8.02 d, so λ = ln(2)/8.02 ≈ 0.0864 d⁻¹. After t = 24.1 d (3 half-lives): N = N₀ × e^(−0.0864 × 24.1) ≈ 0.125·N₀, one eighth left.
Applications
Radiocarbon dating (C-14), nuclear medicine (dose planning), final repository planning, geochronology
Quanta exam set
Curated exam set for "Radioactive Decay Law":
Question (front)
Which formula describes Radioactive Decay Law?
Answer in your set
Question (front)
How do you rearrange N(t) = N₀·e^(−λt) for Half-life from the decay constant?
Answer in your set
Question (front)
Which common mistake happens with Radioactive Decay Law?
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Scientific sources
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Frequently asked questions about Radioactive Decay Law
How do you calculate with the decay law N(t) = N₀·e^(−λt)?+
First find the decay constant from the half-life: λ = ln(2)/T½. Then insert λ and t into the exponential; both must carry the same unit of time. Example iodine-131 (T½ = 8.02 d): λ = 0.693/8.02 ≈ 0.0864 per day. After 24.1 days N = N₀·e^(−0.0864 × 24.1) ≈ 0.125·N₀ remains, one eighth. You can check the result: 24.1 d is exactly three half-lives, and (1/2)³ = 1/8. For neat multiples of the half-life the power form N = N₀·(1/2)^(t/T½) is usually quicker than the exponential.
What exactly does the half-life mean?+
The half-life T½ is the time after which, on average, half of the originally present nuclei have decayed. It is linked to the decay constant via T½ = ln(2)/λ and is characteristic for each nuclide: C-14 has 5,730 years, iodine-131 about 8 days, polonium-214 only fractions of a millisecond. The multiplicative character is important: after each further half-life the stock halves again; after two, 25% remains, after four, 6.25%; the substance is never exactly "gone". For a single nucleus T½ is only a probability statement: it decays with 50% probability within one half-life, but it does not age; the probability always stays the same.
How does C-14 dating work?+
Living organisms constantly exchange carbon with the atmosphere and thus maintain a known C-14 fraction. With death the exchange stops, and the incorporated C-14 decays with T½ = 5,730 years while the stable C-12 remains. Measuring the ratio today, the remaining fraction reveals the age: t = −ln(N/N₀)/λ. Example: if a sample still shows 25% of the original C-14, two half-lives have passed, about 11,460 years. In practice the method works up to about 50,000 years; after that too little C-14 is left. For older finds one uses longer-lived clocks such as potassium-argon (T½ = 1.25 billion years) with the same calculation principle.
What is the difference between the decay constant and the activity?+
The decay constant λ is the decay probability per nucleus and unit time, a fixed material property with unit 1/s. The activity A, by contrast, is the actual number of decays per second in a concrete sample: A = λ·N, measured in becquerels (1 Bq = 1 decay/s). It therefore depends on the sample size and decreases exponentially with time just like N: A(t) = A₀·e^(−λt). Example: 10¹⁵ nuclei with λ = 0.0864 d⁻¹ = 10⁻⁶ s⁻¹ have A = 10⁹ Bq. In measurement problems you almost always work with the activity, because counters register decays per time; the decay law holds for A and N in identical form.
Can you predict when a single nucleus will decay?+
No. Radioactive decay is a fundamentally random quantum process. For a single nucleus only a probability can be stated: per unit time it decays with probability λ, regardless of how long it has already existed. Nuclei do not age; an "old" nucleus is as ready to decay as a freshly created one. Only for very many nuclei does chance turn into a precise law: with 10²⁰ particles the relative fluctuations are vanishingly small, and N(t) = N₀·e^(−λt) describes the stock practically exactly. It is the same principle as rolling dice: a single roll is unpredictable, the average of millions of rolls very accurate. No external influence, temperature, pressure or chemical bonding, changes λ measurably.
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How do you calculate with Radioactive Decay Law?
Here is how to work through a typical Radioactive Decay Law (N(t) = N₀·e^(−λt)) task step by step:
- 1
Task
After how many half-lives is 6.25% of a sample left?
Solution path
6.25% = 1/16 = (1/2)⁴, so after 4 half-lives.
- 2
Task
A C-14 sample shows 25% activity (T½ = 5,730 a). How old is it?
Solution path
25% = (1/2)², i.e. 2 half-lives: t = 2 × 5,730 = 11,460 years.