Playground
Visualize how half-life relates to the decay constant: adjust λ and watch the sample halve repeatedly.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Half-lifeoutput Time for half the sample to decay | s | T | 1e-10 – 100000000000000000 | |
| Decay constant Probability of decay per unit time | 1/s | T⁻¹ | 1e-18 – 1 |
Deep dive
Derivation
Starting from N(t) = N₀ e^{−λt}, set N(T₁/₂) = N₀/2. Then N₀/2 = N₀ e^{−λT₁/₂}. Dividing both sides by N₀: 1/2 = e^{−λT₁/₂}. Taking the natural log: −ln2 = −λT₁/₂. Therefore T₁/₂ = ln2/λ ≈ 0.6931/λ.
Experimental verification
Half-lives have been measured for thousands of isotopes, from 2.6 × 10⁻²¹ s (hydrogen-7) to 2.2 × 10²⁴ yr (tellurium-128), all consistent with T₁/₂ = ln2/λ. Precision measurements use decay curve fitting to extract λ and verify the relationship.
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 τ = 1/λ = T₁/₂/ln2 ≈ 1.443 × T₁/₂.
- Half-life does not depend on the initial amount of material — it is an intrinsic property of the isotope.
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)
Worked examples
Half-life from a known decay constant
Given:
- λ:
- 0.0231
Find: T₁/₂
Solution
T₁/₂ = ln2 / λ = 0.6931 / 0.0231 = 30.0 s
Decay constant of Carbon-14
Given:
- T₁/₂_yr:
- 5730
Find: λ
Solution
λ = ln2 / T₁/₂ = 0.6931 / (5730 × 3.156 × 10⁷) = 3.83 × 10⁻¹² s⁻¹
Scenarios
What if…
- scenario:
- What if the decay constant is halved?
- answer:
- The half-life doubles — the isotope takes twice as long to lose half its nuclei, making it more persistent.
- scenario:
- What if we need 99% of the sample to decay?
- answer:
- We need about 6.64 half-lives, since (1/2)^n = 0.01 gives n = log₂(100) ≈ 6.64.
- scenario:
- What if half-life is shorter than our measurement time resolution?
- answer:
- The isotope decays too quickly to observe directly. Physicists infer its existence from resonance widths: Γ = ℏ/τ = ℏ ln2/T₁/₂.
Limiting cases
- condition:
- λ → 0⁺
- result:
- T₁/₂ → ∞
- explanation:
- A nearly-zero decay constant means the isotope is nearly stable and takes an astronomically long time to halve.
- condition:
- λ → ∞
- result:
- T₁/₂ → 0
- explanation:
- An extremely large decay constant means the isotope decays almost instantaneously.
- condition:
- λ = ln2
- result:
- T₁/₂ = 1 s
- explanation:
- When λ equals ln2 per second, the half-life is exactly one second — a useful reference point.
Context
Ernest Rutherford · 1907
Rutherford coined the term 'half-life' while studying thorium decay, noticing activity halved in a predictable time.
Hook
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?
Dimensions:
- lhs:
- T₁/₂ → [T]
- rhs:
- ln2 / λ → [1] / [T⁻¹] = [T]
- check:
- Both sides are [T] = seconds. ✓
Validity: 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.