Physics · Nuclear physics

Radioactive Decay Law

The decay law describes the exponential decrease of the undecayed atomic nuclei of a radioactive substance.

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Formula

LaTeX: N(t) = N_0 \cdot e^{-\lambda t}
N(t), N₀ dimensionless (number of nuclei) · λ in 1/s · t in seconds [s]
Diagram: a falling exponential curve N over t; dashed lines mark N₀/2 at the half-life T½.tNN₀N₀/2
The number of undecayed nuclei falls exponentially; after the half-life T½ half of them remain.

Variables & units – Radioactive Decay Law

SymbolMeaningUnit
N(t)Number of undecayed nuclei at time tdimensionless
N₀Initial number of nuclei (t = 0)dimensionless
λDecay constant (substance-specific)1/s
tTimes

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.

  1. 1Ansatz: dN/dt = −λ·N (decay rate proportional to the stock).
  2. 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.

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?

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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

Common notations & search queries

N(t)=N0*e^(-lambda*t)N=N0e^-λtZerfallsgesetz FormelHalbwertszeit Formelradioaktiver Zerfall berechnenZerfallskonstanteradioactive decay lawexponentieller Zerfall

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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.

Retain Radioactive Decay Law for exams

Create a curated FSRS exam set for N(t) = N₀·e^(−λt): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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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. 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. 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.