Radioactivity Kinetics

Let $N(t)$ be the number of undecayed radioactive nuclei at time $t$. The fundamental empirical observation is that the rate of decay is proportional to $N$:\n\[-\frac{dN}{dt} = \lambda N\]\nThe negative sign indicates $N$ is decreasing. The proportionality constant $\lambda$ is the decay constant, unique to each radioactive nuclide, with SI unit s$^{-1}$. A large $\lambda$ means rapid decay; a small $\lambda$ means slow decay. This is exactly analogous to a first-order chemical reaction rate law $r = k[A]$.

Rearrange the differential equation and integrate both sides:\n\[\frac{dN}{N} = -\lambda\, dt \implies \int_{N_0}^{N} \frac{dN'}{N'} = -\lambda \int_0^t dt'\]\n\[\ln N - \ln N_0 = -\lambda t \implies \ln\frac{N}{N_0} = -\lambda t\]\nExponentiating both sides gives the Radioactive Decay Law:\n\[N(t) = N_0\, e^{-\lambda t}\]\nwhere $N_0$ is the number of radioactive nuclei at $t = 0$. Because mass and activity are both proportional to $N$, the same exponential law applies to mass $m(t) = m_0 e^{-\lambda t}$ and activity $A(t) = A_0 e^{-\lambda t}$.

The half-life is the time required for exactly half the radioactive nuclei to decay. Set $N = N_0/2$:\n\[\frac{N_0}{2} = N_0 e^{-\lambda t_{1/2}} \implies \frac{1}{2} = e^{-\lambda t_{1/2}}\]\n\[\ln 2 = \lambda t_{1/2} \implies \boxed{t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}}\]\nAfter each successive half-life, the quantity remaining is halved: $N_0 \to N_0/2 \to N_0/4 \to N_0/8 \to \cdots$. After $n$ half-lives, $N = N_0 / 2^n$. Half-life is the most intuitively accessible measure of decay rate and varies enormously across nuclides — from microseconds ($^{212}\text{Po}$, $t_{1/2} = 0.3\,\mu\text{s}$) to billions of years ($^{238}\text{U}$, $t_{1/2} = 4.47 \times 10^9$ yr).

The mean life $\tau$ is the statistical average time a single nucleus survives before decaying. Each nucleus that decays at time $t$ contributes a lifetime of $t$. The number decaying in interval $dt$ is $|dN| = \lambda N_0 e^{-\lambda t}\,dt$. Averaging over all nuclei:\n\[\tau = \frac{1}{N_0}\int_0^{\infty} t\,\lambda N_0 e^{-\lambda t}\,dt = \lambda \int_0^{\infty} t e^{-\lambda t}\,dt = \lambda \cdot \frac{1}{\lambda^2} = \frac{1}{\lambda}\]\nSo $\tau = 1/\lambda$, and since $\lambda = \ln 2 / t_{1/2}$:\n\[\tau = \frac{t_{1/2}}{\ln 2} = \frac{t_{1/2}}{0.693} \approx 1.443\, t_{1/2}\]\nThe mean life is always slightly longer than the half-life. At $t = \tau$, the number of nuclei remaining is $N_0 e^{-1} \approx 0.368\, N_0$.

Activity $A$ is the number of disintegrations per second: $A = \lambda N$. The SI unit is the becquerel (Bq), where $1\,\text{Bq} = 1\,\text{disintegration per second}$. An older unit, the curie (Ci), is defined as $1\,\text{Ci} = 3.7 \times 10^{10}\,\text{Bq}$, originally chosen as the activity of 1 gram of $^{226}\text{Ra}$. Since $A = \lambda N$ and $N$ decreases exponentially, activity also falls as $A(t) = A_0 e^{-\lambda t}$. In nuclear medicine, the activity of a radiopharmaceutical must be carefully scheduled relative to its half-life so the dose reaches the patient at the right moment.

• N(t) — Number of undecayed nuclei at time t (dimensionless (count)): The remaining number of radioactive nuclei at elapsed time t.
• N_0 — Initial number of nuclei (dimensionless (count)): The number of radioactive nuclei present at t = 0.
• \lambda — Decay constant (s$^{-1}$ (or yr$^{-1}$, min$^{-1}$, etc.)): The probability per unit time that a single nucleus will decay; unique to each nuclide.
• t — Time (s (or any consistent time unit)): Elapsed time since t = 0.
• t_{1/2} — Half-life (s (or yr, min, etc.)): The time for half the nuclei to decay; inversely proportional to $\lambda$.
• \tau — Mean (average) life (s (or yr, min, etc.)): The average time a nucleus exists before decaying; equals 1/$\lambda$ and is about 1.443 times the half-life.

Problem. A freshly prepared sample of iodine-131 ($^{131}\text{I}$), used in thyroid cancer treatment, has an initial activity of $A_0 = 8.0 \times 10^8$ Bq. The half-life of $^{131}\text{I}$ is 8.02 days. (a) Calculate the decay constant $\lambda$ in s$^{-1}$. (b) Find the activity after 24.06 days. (c) Calculate the mean life $\tau$ in days.

Answer. λ = 1.00 × 10⁻⁶ s⁻¹; Activity after 24.06 days = 1.0 × 10⁸ Bq; Mean life τ = 11.57 days

• First-Order Reactions and Half-Life in Chemical Kinetics — Radioactive decay is mathematically identical to a first-order chemical reaction; the same integrated rate law and half-life formula apply, making the analogy a powerful bridge between nuclear and chemical kinetics.
• Alpha, Beta, and Gamma Decay — Understanding what kind of radiation is emitted (α, β, γ) and how nuclear equations are balanced is the qualitative complement to the quantitative decay-rate mathematics studied here.
• Nuclear Binding Energy and Mass Defect — The stability of a nucleus — and hence the value of λ — is determined by its binding energy per nucleon. This concept explains why some nuclides decay rapidly while others are stable for billions of years.
• Radiometric Dating (Carbon-14 and Uranium-Lead) — Radiometric dating is the most important applied use of the radioactive decay law, directly using the equation N(t) = N₀e^(−λt) to calculate the age of archaeological and geological samples.