Half-Life Relation

Equivalent forms

\lambda = \frac{\ln 2}{T_{1/2}}

ln 2 — the transcendental fingerprint of every decaying thing in the universe.

Symbol	Name	SI unit	Dimension	Range
$T₁/₂$	Half-lifeoutput Time for half the sample to decay	s	T	1e-10 – 100000000000000000
$λ$	Decay constant Probability of decay per unit time	1/s	T⁻¹	1e-18 – 1

Unit systems

Symbol	SI	CGS	Practical
$T₁/₂$	s	s	yr
$λ$	s⁻¹	s⁻¹	yr⁻¹

Where it holds

Valid for any process described by first-order exponential decay with constant decay probability per unit time. This includes radioactive decay, first-order chemical reactions, and RC circuit discharge.

Dimensional analysis

T_{1}/_{2} \to [T]

\ln 2 / \lambda \to [1] / [T^{-1}] = [T]

Both sides are

[T] =

seconds.

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Discovery

Ernest Rutherford · 1907

Rutherford coined the term 'half-life' while studying thorium decay, noticing activity halved in a predictable time.

Try this

How is carbon-14 used to date ancient bones?

An isotope has a decay constant λ = 2.31×10⁻² s⁻¹. What is its half-life?

Research status: stable

Real-world applications

Radiometric age dating (choosing isotopes with half-lives appropriate to the timescale being measured)
Nuclear medicine dosimetry (calculating when a radiotracer falls below therapeutic or safe levels)
Nuclear reactor fuel cycle planning
Forensic science (determining time since contamination events)

Common misconceptions

After two half-lives, 1/4 of the sample remains — not zero. The sample never fully decays in finite time.
Half-life is not the average lifetime. The mean lifetime $\tau = 1/\lambda = T_{1}/_{2}/\ln 2 \approx 1.443 \times T_{1}/_{2}$ .
Half-life does not depend on the initial amount of material — it is an intrinsic property of the isotope.

Experimental verification

Half-lives have been measured for thousands of isotopes, from

2.6 \times 10^{-21} s

(hydrogen-7) to

2.2 \times 10^{24} yr

(tellurium-128), all consistent with

T_{1}/_{2} = \ln 2/\lambda

. Precision measurements use decay curve fitting to extract

\lambda and

verify the relationship.

Derivation

Starting from

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

,

set N(T_{1}/_{2}) = N_{0}/2

.

Then

N_{0}/2 = N_{0} e^{-\lambda T_{1}/_{2}}

.

Dividing both sides by

N_{0}

:

1/2 = e^{-\lambda T_{1}/_{2}}

.

Taking the natural log:

-\ln 2 = -\lambda T_{1}/_{2}

.

Therefore

T_{1}/_{2} = \ln 2/\lambda \approx 0.6931/\lambda

.

Limiting cases

\lambda \to 0^{+}

⟶

T_{1}/_{2} \to \infty

A nearly-zero decay constant means the isotope is nearly stable and takes an astronomically long time to halve.

\lambda \to \infty

⟶

T_{1}/_{2} \to 0

An extremely large decay constant means the isotope decays almost instantaneously.

\lambda = \ln 2

⟶

T_{1}/_{2} = 1\,\mathrm{s}

When

\lambda

equals ln2 per second, the half-life is exactly one second — a useful reference point.

What if…

What if the decay constant is halved?

The half-life doubles — the isotope takes twice as long to lose half its nuclei, making it more persistent.

What if we need 99% of the sample to decay?

We need about 6.64 half-lives, since $(1/2)^n = 0.01$ gives $n = \log _{2}(100) \approx 6.64$ .

What if half-life is shorter than our measurement time resolution?

The isotope decays too quickly to observe directly. Physicists infer its existence from resonance widths: $\Gamma = \hbar /\tau = \hbar \ln 2/T_{1}/_{2}$ .

1

Half-life from a known decay constant

Given ·

λ:: 0.0231

Find ·

T_{1}/_{2}

Steps

Step 1: Recall $\ln 2 = 0.6931$ .
Step 2: $T_{1}/_{2} = 0.6931 / 0.0231 = 30.0\,\mathrm{s}$ .
Step 3: This is consistent with a short-lived nuclear isomer's decay timescale.

Result ·

T_{1}/_{2} = \ln 2 / \lambda = 0.6931 / 0.0231 = 30.0\,\mathrm{s}

2

Decay constant of Carbon-14

Given ·

T₁/₂ yr:: 5730

Find ·

\lambda

Steps

Step 1: Convert $T_{1}/_{2} to$ seconds: $5730\,\mathrm{yr} \times 3.156 \times 10^{7} s/yr = 1.808 \times 10^{11} s$ .
Step 2: $\lambda = 0.6931 / 1.808 \times 10^{11} = 3.83 \times 10^{-12} s^{-1}$ .
Step 3: Equivalently, $\lambda = 1.21 \times 10^{-4} yr^{-1}$ , useful for dating calculations in years.

Result ·

\lambda = \ln 2 / T_{1}/_{2} = 0.6931 / (5730 \times 3.156 \times 10^{7}) = 3.83 \times 10^{-12} s^{-1}

Radioactive Decay Law→Geiger–Nuttall Law→Nuclear Binding Energy→