Radioactive Decay Law

Equivalent forms

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

t_{1/2} = \ln 2 / \lambda

Pure randomness at the nuclear level produces a perfectly deterministic macroscopic curve.

Symbol	Name	SI unit	Dimension	Range
$N$	Number of nucleioutput Remaining undecayed nuclei at time t	count	1	0 – 1e+24
$N_0$	Initial number of nuclei Population at t = 0	count	1	1 – 1e+24
$λ$	Decay constant Probability per unit time of decay	1/s	T⁻¹	1e-12 – 1
$t$	Time Elapsed time since t = 0	s	T	0 – 1000000000000

Unit systems

Symbol	SI	CGS	Imperial
$N$	count	count	count
$N_0$	count	count	count
$λ$	1/s	1/s	1/yr
$t$	s	s	yr

Where it holds

Valid for any large ensemble of identical, statistically independent unstable nuclei. Breaks down for small N (Poisson fluctuations dominate) and for daughter chains where Bateman equations are needed.

Dimensional analysis

$[\lambda t]$ dimensionless: $T^{-1} \cdot T = 1 \checkmark$ ; both sides of $N = N_0\,\mathrm{e}^{-\lambda t}$ are pure counts.

Discovery

Ernest Rutherford and Frederick Soddy · 1902

Rutherford and Soddy showed radioactive substances transmute exponentially, founding nuclear physics.

Try this

How can we date a 5000-year-old artifact from the carbon inside it?

A sample contains 25% of its original Carbon-14. Given t½ = 5730 yr, how old is it?

Research status: stable

Real-world applications

Carbon-14 dating of organic materials up $to \sim 50$ ,000 years.
U-Pb and K-Ar dating of rocks (deep time geology).
Medical imaging: $^{99}$ ᵐ $Tc (t_{1}$ ᐟ $_{2} = 6\,\mathrm{hr})$ for SPECT, $^{18}F (110 \min )$ for PET.
Radiotherapy dosing and nuclear reactor fuel burn-up calculations.

Common misconceptions

Half-life is NOT the average lifetime — ⟨ $\tau$ ⟩ $= 1/\lambda = t_{1}$ ᐟ $_{2}/\ln 2 \approx 1.44\,\mathrm{t}_{1}$ ᐟ $_{2}$ .
Decay is genuinely random for each nucleus; only the bulk follows the smooth curve.
Decay constants are essentially independent of temperature, pressure, and chemistry (electron-capture isotopes show tiny exceptions).

Experimental verification

Rutherford'

s \alpha

-counting experiments confirmed exponential behavior. Modern liquid scintillation counters and HPGe

\gamma

-spectroscopy verify the law over dozens of half-lives. Radiocarbon dating (Libby, 1949 Nobel) cross-checks with tree-ring dendrochronology to <50-yr accuracy.

Derivation

Each nucleus has constant decay probability

\lambda per

unit time.

For N nuclei,

dN/dt = -\lambda N

.

Separating variables:

dN/N = -\lambda dt \to \ln N = -\lambda t + C

.

Applying

N(0) = N_0

gives

N(t) = N_0\,\mathrm{e}^{-\lambda t}

.

Half-life satisfies

\tfrac{1}{2} = e^{-\lambda t_{1}

ᐟ

_{2}} \to t_{1}

ᐟ

_{2} = (\ln 2)/\lambda

.

Limiting cases

t = 0

⟶

N = N_0

Initial population — no decays yet.

t = t_{1}

ᐟ

_{2}

⟶

N = N_0/2

Half-life: defines the timescale of the decay process.

t \to \infty

⟶

N \to 0

All nuclei eventually decay, but never literally zero in finite time (statistical asymptote).

\lambda t

≪ 1⟶

N \approx N_0(1 - \lambda t)

Linear decay regime for small times — useful for activity measurements.

What if…

What if you doubled the sample mass?

N_0 doubles, so the activity at every time doubles, but the half-life and shape of the curve are unchanged.

What if you cooled radioactive material to 1 K?

Decay rate is essentially unchanged. Nuclear decay depends on nuclear physics, not thermal energy $(\sim meV vs MeV$ barriers).

What if a daughter is also radioactive?

You get the Bateman equations: secular equilibrium when $t_{1}$ ᐟ $_{2}$ (parent) ≫ $t_{1}$ ᐟ $_{2}$ (daughter), so daughter activity matches parent — basis of radiogenerators (e.g., $^{99}Mo \to ^{99}$ ᵐTc).

1

Carbon-14 dating of a 25% sample

Given ·

N/N 0:: 0.25
t half:: 5730

Find · age t

Steps

Take log: $\log _{2}(0.25) = -2 = -t/5730$
$t = 2 \times 5730 = 11460\,\mathrm{yr}$
Two half-lives have elapsed.

Result ·

0.25 = (1/2)^(t/5730) \to t/5730 = 2 \to t = 11460\,\mathrm{yr}

2

Activity of 1 g of Co-60 (t₁ᐟ₂ = 5.27 yr)

Given ·

m:: 0.001
M:: 0.06
t half:: 5.27

Find · Activity

A = \lambda N

Steps

Convert $t_{1}$ ᐟ $_{2} to$ seconds: $5.27 \times 3.156\times 10^{7} \approx 1.66\times 10^{8} s$
$\lambda = 0.693 / 1.66\times 10^{8} \approx 4.17\times 10^{-9} s^{-1}$
$N = (1\times 10^{-3} / 0.060) \times 6.022\times 10^{23} \approx 1.00\times 10^{22}$
$A = \lambda N \approx 4.18\times 10^{13}$ decays/ $s \approx 1130\,\mathrm{Ci}$ — extremely radioactive.

Result ·

\lambda = \ln 2 / (5.27\times 3.156\times 10^{7}) \approx 4.17\times 10^{-9} s^{-1}

;

N \approx 1.0\times 10^{22} \to A \approx 4.2\times 10^{13} Bq

Mass-Energy Equivalence→