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Answer: \(N = N_0 e^{-\lambda t}\); 1 Bq = 1 disintegration per second.

Law of radioactive decay: the number of nuclei disintegrating per unit time is directly proportional to the number of undecayed nuclei present at that time.

If \(N\) nuclei are present at time \(t\) and \(dN\) of them decay in time \(dt\),

\[-\frac{dN}{dt} \propto N \Rightarrow \frac{dN}{dt} = -\lambda N\]

where \(\lambda\) is the decay constant. Rearranging and integrating, with \(N = N_0\) at \(t = 0\):

\[\int_{N_0}^{N}\frac{dN}{N} = -\lambda\int_0^t dt \Rightarrow \ln\frac{N}{N_0} = -\lambda t\]\[\therefore N = N_0 e^{-\lambda t}\]

The number of undecayed nuclei decreases exponentially with time.

One becquerel (Bq): the activity of a radioactive sample in which one nucleus disintegrates per second. 1 Bq = 1 decay/s (1 curie = \(3.7 \times 10^{10}\) Bq).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report
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