Playground
Watch nuclei decay in real time: a grid of dots disappears exponentially while the decay curve plots below.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Remaining nucleioutput Number of undecayed nuclei at time t | count | 1 | 0 – 1000000000000000 | |
| Initial nuclei Number of nuclei at t = 0 | count | 1 | 1 – 1000000000000000 | |
| Decay constant Probability of decay per unit time | 1/s | T⁻¹ | 1e-10 – 1 | |
| Time Elapsed time | s | T | 0 – 1000000 |
Deep dive
Derivation
Each nucleus has a constant probability λ dt of decaying in time interval dt, independent of all others. The rate of change is dN/dt = −λN. Separating variables: dN/N = −λ dt. Integrating from 0 to t: ln(N/N₀) = −λt, giving N(t) = N₀ e^{−λt}.
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., ⁸Be) to billions of years (e.g., ²³⁸U).
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.
Real-world applications
- Carbon-14 dating of archaeological artifacts (T₁/₂ ≈ 5730 years)
- Medical diagnostics using technetium-99m (T₁/₂ ≈ 6 hours)
- Nuclear waste management and storage timeline planning
- Smoke detectors using americium-241 alpha decay
Worked examples
Nuclei remaining after 1 hour
Given:
- N₀:
- 1000000000000
- λ:
- 0.00012
- t:
- 3600
Find: N
Solution
N = 10¹² × e^{−1.2×10⁻⁴ × 3600} = 10¹² × e^{−0.432} = 6.49 × 10¹¹
Carbon-14 dating a 5730-year-old sample
Given:
- N₀:
- 10000000000
- λ:
- 0.000121
- t:
- 5730
Find: N
Solution
N = 10¹⁰ × e^{−(ln2/5730yr) × 5730yr} = 10¹⁰ × e^{−ln2} = 5.0 × 10⁹
Scenarios
What if…
- scenario:
- What if the decay constant doubles?
- answer:
- The nuclei decay twice as fast — the half-life halves, and far fewer nuclei remain at any given time.
- scenario:
- What if we start with only 10 nuclei?
- answer:
- The smooth exponential becomes a poor approximation. Decay events are discrete and stochastic; you'd see random step-downs rather than a smooth curve.
- scenario:
- What if decay constant were zero?
- answer:
- The exponential becomes e⁰ = 1 for all t, meaning N(t) = N₀ forever — the isotope is completely stable.
Limiting cases
- condition:
- t → 0
- result:
- N → N₀
- explanation:
- At the start, no nuclei have had time to decay.
- condition:
- t → ∞
- result:
- N → 0
- explanation:
- Given infinite time, every nucleus will eventually decay.
- condition:
- λ → 0
- result:
- N → N₀
- explanation:
- A zero decay constant means the isotope is stable and never decays.
Context
Ernest Rutherford & Frederick Soddy · 1902
Rutherford and Soddy discovered that radioactive substances transmute into other elements at a rate proportional to the amount present.
Hook
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?
Dimensions:
- lhs:
- N → [1] (dimensionless count)
- rhs:
- N₀ · e^{−λt} → [1] · e^{[T⁻¹]·[T]} = [1] · e^{[1]} = [1]
- check:
- Both sides are dimensionless counts. The exponent λt is dimensionless [T⁻¹·T = 1]. ✓
Validity: 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.