When a nucleus undergoes radioactive decay it transforms into a different nucleus, emitting particles or radiation. Two fundamental quantities used to describe a radioactive sample are the activity and the decay constant.
The activity of a sample containing N radioactive nuclei is given by
$$A = \lambda N$$where:
The number of nuclei remaining after a time $t$ follows the exponential law
$$N(t) = N_0 e^{-\lambda t}$$Differentiating with respect to time gives the rate of change of $N$:
$$\frac{dN}{dt} = -\lambda N(t)$$The negative sign indicates a decrease in $N$. The magnitude of this rate is the activity:
$$A(t) = -\frac{dN}{dt} = \lambda N(t)$$The half‑life $t_{1/2}$ is the time required for half of the original nuclei to decay. It is related to the decay constant by
$$t_{1/2} = \frac{\ln 2}{\lambda}$$Thus, knowing either $t_{1/2}$ or $\lambda$ allows you to calculate the other.
| Question | Solution |
|---|---|
| 1 | $$A = \lambda N = (5.0 \times 10^{-4}\,\text{s}^{-1})(2.0 \times 10^{20}) = 1.0 \times 10^{17}\,\text{Bq}$$ |
| 2 | $$N = \frac{A}{\lambda} = \frac{3.0 \times 10^{6}\,\text{Bq}}{2.0 \times 10^{-3}\,\text{s}^{-1}} = 1.5 \times 10^{9}\,\text{nuclei}$$ |
| Quantity | Symbol | SI Unit | Typical Symbol |
|---|---|---|---|
| Activity | $A$ | becquerel (Bq) | $\text{s}^{-1}$ |
| Decay constant | $\lambda$ | second⁻¹ (s⁻¹) | $\text{s}^{-1}$ |
| Number of nuclei | $N$ | dimensionless (count) | – |
| Half‑life | $t_{1/2}$ | second (s) | – |