Radioactive decay
Radioactive decay follows an exponential law because each undecayed nucleus has approximately the same probability per unit time of decaying. The fraction lost per equal time interval is therefore roughly constant, while the absolute number lost becomes smaller as the sample shrinks.
The exponential law
N(t) = N₀e-λt and A(t) = A₀e-λt
The decay constant λ has units of inverse time. It describes a probability rate, not a countdown attached to each nucleus.
Half-life and decay constant are equivalent descriptions
Setting N(T1/2)=N₀/2 gives T1/2 = ln2/λ. A large λ means rapid decay and therefore a short half-life.
Branching and decay chains need extensions
Some nuclides can decay by more than one route. The total decay constant is the sum of the partial decay constants, and branch fractions describe the relative probabilities. In a decay chain, daughter nuclei are simultaneously produced and removed, so a single exponential for the daughter is generally insufficient.
Measurements contain counting statistics
Observed counts fluctuate. For independent radioactive events, Poisson statistics often provide a useful approximation, with a standard deviation of about √N counts. Background must be measured and subtracted with its own uncertainty rather than treated as an exact number.
One exponential is a model with conditions. The simple law assumes a population of the same radionuclide with a constant decay probability per unit time. It describes the expected number of parent nuclei, not a perfectly smooth sequence of detector clicks. Mixtures, daughter activity and changing measurement efficiency can produce curves that are not single exponentials even though each individual radionuclide still obeys its own decay law. When fitting measurements, always identify which population the exponential is supposed to represent and whether background has been removed.
Worked examples
1. Fraction after three half-lives
Solution
(1/2)³ = 1/8 = 12.5% remains on average.
2. Decay constant
Solution
For T½=10 d, λ=ln2/10=0.0693 d⁻1.
3. Continuous decay
Solution
If N₀=1000, λ=0.20 h⁻1 and t=5 h, N=1000e⁻1≈368.