Governing Equations
Fundamental decay law:
$$N(t)=N_0\,e^{-\lambda t},\quad A(t)=\lambda N(t)=A_0\,e^{-\lambda t}$$Half-life and decay constant: $T_{1/2}=\dfrac{\ln 2}{\lambda}$
Carbon-14 dating: $t=-\dfrac{\ln(A/A_0)}{\lambda}$
Decay chain (Bateman equation): $\dfrac{dN_2}{dt}=\lambda_1 N_1 - \lambda_2 N_2$
Analytical solution: $N_2(t)=N_1(0)\dfrac{\lambda_1}{\lambda_2-\lambda_1}\!\left(e^{-\lambda_1 t}-e^{-\lambda_2 t}\right)$
Student 🧑🎓: Why does the decay law show an exponential, not a linear, decrease?
Professor 🎓: Because radioactive decay is a purely statistical process — each nucleus decays independently with a constant probability λ per unit time, regardless of how long it has already existed or how many other nuclei surround it. If you have N nuclei, the rate of decay at any moment is dN/dt = −λN, which is a differential equation whose solution is N(t) = N₀e^(−λt). The more nuclei you have, the more decays happen per second; as the number drops, so does the rate — a self-reinforcing relationship that naturally produces an exponential curve. This is fundamentally different from, say, a radioactive source that runs out at a fixed rate regardless of how much remains.