22.4.1 Exponential Nature of Nuclear Decay

Law of Radioactive Decay

Experiment shows that the rate of decay of a radioactive sample is directly proportional to the number of undecayed nuclei $N$ present at that instant:

$-\frac{\Delta N}{\Delta t}\propto N$

Introducing the decay constant $\lambda$ as the constant of proportionality:

$-\frac{\Delta N}{\Delta t}=\lambda N$

The negative sign indicates that $N$ is decreasing with time.

The Decay Constant ($\lambda$)

The decay constant $\lambda$ is defined as the fraction of the total number of atoms that decay per unit time. It represents the probability that a given nucleus will decay in one second.

• SI unit: ${\text{s}}^{-1}$ (per second)
• A large $\lambda$ means rapid decay (short-lived isotope).
• A small $\lambda$ means slow decay (long-lived isotope).
• $\lambda$ is a constant for a given nuclide — it does not change with temperature, pressure, or chemical state.

Activity ($A$)

The activity $A$ of a radioactive source is the number of disintegrations (decays) per second:

$A=\lambda N$

• SI unit: Becquerel (Bq), where $1\text{ Bq}=1\text{ decay per second}$.
• Activity decreases over time as $N$ decreases.
• An older unit is the Curie (Ci): $1\text{ Ci}=3.7\times 10^{10}\text{ Bq}$.

Exponential Nature of Decay

Solving the differential equation $\frac{dN}{dt}=-\lambda N$ gives the exponential decay equation:

$\boxed{N=N_0{e}^{-\lambda t}}$

where:

• $N_0$ = initial number of undecayed nuclei at $t=0$
• $N$ = number of undecayed nuclei at time $t$
• $\lambda$ = decay constant
• $e$ = base of natural logarithm ($\approx 2.718$)

The term $e^{-\lambda t}$ represents the fraction of nuclei remaining undecayed at time $t$.

Since $A=\lambda N$, the activity also decays exponentially:

$A=A_0{e}^{-\lambda t}$

where $A_0=\lambda N_0$ is the initial activity.

Key Feature

As time increases, $N$ decreases exponentially — the rate of decay slows down because there are fewer undecayed nuclei remaining. This is the exponential nature of radioactive decay.

Worked Example

Problem: A radioactive sample initially contains $N_0=8.0\times 10^{20}$ nuclei and has a decay constant $\lambda =2.0\times 10^{-3}{\text{ s}}^{-1}$. Find:

(a) The initial activity $A_0$.

(b) The number of undecayed nuclei after $t=500\text{ s}$.

Solution:

(a) $A_0=\lambda N_0=(2.0\times 10^{-3})(8.0\times 10^{20})=1.6\times 10^{18}\text{ Bq}$

(b) $N=N_0{e}^{-\lambda t}=8.0\times 10^{20}\times e^{-(2.0\times 10^{-3})(500)}$

$N=8.0\times 10^{20}\times e^{-1.0}=8.0\times 10^{20}\times 0.368\approx 2.94\times 10^{20}$

Summary Table