The Exponential Decay Law

In Chemistry you used half-lives: after each one, half of the remaining nuclei are left. Calculus shows why the pattern is exponential, and it adds two useful quantities: the decay constant and the activity. The key fact is that a nucleus has no memory: an old uranium nucleus is exactly as likely to decay in the next second as a new one.

Key idea

Each nucleus decays with the same probability $\lambda\,dt$ in a short time $dt$, so $$\frac{dN}{dt} = -\lambda N \quad\Longrightarrow\quad N(t) = N_0e^{-\lambda t}.$$ The half-life is $T_{1/2} = \frac{\ln 2}{\lambda}$, the mean life is $\tau = \frac{1}{\lambda}$, and the activity (decays per second, in becquerels, Bq) is $$A = \lambda N = A_0e^{-\lambda t}.$$

Why it works

With $N$ nuclei, each decaying with probability $\lambda\,dt$, the expected number of decays in $dt$ is $\lambda N\,dt$, so $dN = -\lambda N\,dt$. Separate the variables: $\frac{dN}{N} = -\lambda\,dt$, so $\ln N = -\lambda t + C$ and $N = N_0e^{-\lambda t}$, the exponential whose derivative is proportional to itself. The half-life comes from $\frac{1}{2} = e^{-\lambda T_{1/2}}$: take logs, $T_{1/2} = \frac{\ln 2}{\lambda}$. A big $\lambda$ means a short half-life: they are inverses, not the same thing. And the activity is the decay rate, $-\frac{dN}{dt} = \lambda N$: a sample with many nuclei but a long half-life can have a small activity.

400 nuclei, each with a 2-second half-life, decaying at random. No one can say which nucleus goes next, yet the count follows $N_0e^{-\lambda t}$ closely: about half are left after each half-life.

Worked examples

Worked example: Carbon-14's decay constant

Carbon-14 has a half-life of 5730 years. Find its decay constant per year and per second.

$\lambda = \frac{\ln 2}{5730\ \text{yr}} = 1.21 \times 10^{-4}\ \text{yr}^{-1}$.
One year is $3.156 \times 10^7$ s: $\lambda = \frac{1.21 \times 10^{-4}}{3.156 \times 10^7} = 3.83 \times 10^{-12}\ \text{s}^{-1}$.

Answer: $1.21 \times 10^{-4}$ per year, or $3.83 \times 10^{-12}$ per second

Worked example: The activity of a gram of radium

Radium-226 has a half-life of 1600 years and a molar mass of 226 g/mol. What is the activity of 1.00 g of it?

Number of nuclei: $N = \frac{1.00}{226} \times 6.022 \times 10^{23} = 2.66 \times 10^{21}$.
$\lambda = \frac{\ln 2}{(1600)(3.156 \times 10^7\ \text{s})} = 1.37 \times 10^{-11}\ \text{s}^{-1}$.
$A = \lambda N = 3.66 \times 10^{10}$ Bq.

Answer: $3.7 \times 10^{10}$ decays per second, which is how the old unit, the curie, was defined

Worked example: How much is left?

Iodine-131 has a half-life of 8.02 days. What fraction of a sample remains after 30.0 days?

$\lambda t = \frac{\ln 2}{8.02} \times 30.0 = 2.59$.
$\frac{N}{N_0} = e^{-2.59} = 0.075$. (Check: 30 days is 3.74 half-lives, and $0.5^{3.74} = 0.075$.)

Answer: About 7.5%

Each nucleus decays on its own, at random. So the number of decays counted in a short time jumps around a little from one count to the next, scattered about the average $\lambda N\,\Delta t$. With the enormous numbers of nuclei in any real sample, the smooth exponential describes it extremely well, as the simulation above shows even for 400 nuclei.

Common mistake

Mixing up the decay constant with the half-life, or the activity with the number of nuclei. $\lambda$ has units of 1/time and is large when the half-life is short: $\lambda = \ln 2/T_{1/2}$. The activity counts decays per second ($A = \lambda N$), not nuclei. Check the units: $A$ in Bq (per second) needs $\lambda$ in s⁻¹, so convert years to seconds first.

The Exponential Decay Law

Worked examples

Practice

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Sources and further reading