Radioactive Decay Law

Equivalent forms

N(t) = N_0 (1/2)^{t/T_{1/2}}

A pure exponential: the cleanest stochastic law in all of physics.

Symbol	Name	SI unit	Dimension	Range
$N$	Remaining nucleioutput Number of undecayed nuclei at time t	count	1	0 – 1000000000000000
$N₀$	Initial nuclei Number of nuclei at t = 0	count	1	1 – 1000000000000000
$λ$	Decay constant Probability of decay per unit time	1/s	T⁻¹	1e-10 – 1
$t$	Time Elapsed time	s	T	0 – 1000000

Unit systems

Symbol	SI	CGS	Practical
$N$	count	count	count
$N₀$	count	count	count
$λ$	s⁻¹	s⁻¹	yr⁻¹
$t$	s	s	yr

Where it holds

Valid for large numbers of identical, non-interacting radioactive nuclei where quantum statistical averaging holds. Breaks down for very small sample sizes where stochastic fluctuations dominate.

Dimensional analysis

N \to [1]

(dimensionless count)

N_{0} \cdot e^{-\lambda t} \to [1] \cdot e^{[T^{-1}]\cdot [T]} = [1] \cdot e^{[1]} = [1]

Both sides are dimensionless counts. The exponent

\lambda t

is dimensionless

[T^{-1}\cdot T = 1]

.

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Discovery

Ernest Rutherford & Frederick Soddy · 1902

Rutherford and Soddy discovered that radioactive substances transmute into other elements at a rate proportional to the amount present.

Try this

Why do we trust carbon dating on a 5000-year-old mummy?

A sample has 10¹² radioactive nuclei and a decay constant of 1.2×10⁻⁴ s⁻¹. How many remain after 1 hour?

Research status: stable

Real-world applications

Carbon-14 dating of archaeological artifacts $(T_{1}/_{2} \approx 5730$ years)
Medical diagnostics using technetium- $99m (T_{1}/_{2} \approx 6$ hours)
Nuclear waste management and storage timeline planning
Smoke detectors using americium-241 alpha decay

Common misconceptions

Individual nuclei do not 'age' — a nucleus that has survived 10 half-lives is just as likely to decay in the next second as a freshly created one.
The decay law does not predict when a specific nucleus will decay, only the statistical behavior of a large ensemble.
Radioactive decay cannot be sped up or slowed down by temperature, pressure, or chemical bonding under normal conditions.

Experimental verification

Rutherford and Soddy (1902) measured thorium emanation activity dropping exponentially. Modern experiments confirm the law using Geiger counters, scintillation detectors, and mass spectrometry across half-lives from nanoseconds (e.g.,

^{8}Be)

to billions of years (e.g.,

^{238}U)

.

Derivation

Each nucleus has a constant probability

\lambda dt

of decaying in time interval dt, independent of all others.

The rate of change is

dN/dt = -\lambda N

.

Separating variables:

dN/N = -\lambda dt

.

Integrating from 0 to t:

\ln (N/N_{0}) = -\lambda t

, giving

N(t) = N_{0} e^{-\lambda t}

.

Limiting cases

t \to 0

⟶

N \to N_{0}

At the start, no nuclei have had time to decay.

t \to \infty

⟶

N \to 0

Given infinite time, every nucleus will eventually decay.

\lambda \to 0

⟶

N \to N_{0}

A zero decay constant means the isotope is stable and never decays.

What if…

What if the decay constant doubles?

The nuclei decay twice as fast — the half-life halves, and far fewer nuclei remain at any given time.

What if we start with only 10 nuclei?

The smooth exponential becomes a poor approximation. Decay events are discrete and stochastic; you'd see random step-downs rather than a smooth curve.

What if decay constant were zero?

The exponential becomes $e^{0} = 1$ for all t, meaning $N(t) = N_{0}$ forever — the isotope is completely stable.

1

Nuclei remaining after 1 hour

Given ·

N₀:: 1000000000000
λ:: 0.00012
t:: 3600

Find · N

Steps

Step 1: Compute exponent: $\lambda t = 1.2 \times 10^{-4} \times 3600 = 0.432$ .
Step 2: Evaluate exponential: $e^{-0.432} = 0.6493$ .
Step 3: $N = 10^{12} \times 0.6493 = 6.49 \times 10^{11}$ nuclei remain.

Result ·

N = 10^{12} \times e^{-1.2\times 10^{-4} \times 3600} = 10^{12} \times e^{-0.432} = 6.49 \times 10^{11}

2

Carbon-14 dating a 5730-year-old sample

Given ·

N₀:: 10000000000
λ:: 0.000121
t:: 5730

Find · N

Steps

Step 1: Decay constant for C-14: $\lambda = \ln 2 / T_{1}/_{2} = 0.6931 / 5730\,\mathrm{yr} = 1.21 \times 10^{-4} yr^{-1}$ .
Step 2: Exponent: $\lambda t = 1.21 \times 10^{-4} \times 5730 = 0.6931 = \ln 2$ .
Step 3: $e^{-\ln 2} = 0.5$ , so $N = 10^{10} \times 0.5 = 5.0 \times 10^{9}$ — exactly half remain after one half-life.

Result ·

N = 10^{10} \times e^{-(\ln 2/5730yr) \times 5730yr} = 10^{10} \times e^{-\ln 2} = 5.0 \times 10^{9}

Half-Life Relation→Geiger–Nuttall Law→Nuclear Binding Energy→