Lessons · modern

Radioactive decay: the rate is the slope of the count

Regime 2 — calculus does what algebra cannot. The acceleration isn't constant, so the algebra formulas don't apply; calculus is the only road in, and SymPy proves the closed form solves the equation of motion.

Regime 2 · calculus does more Math machine-derived & checked by SymPy 1 modeling assumptions (author-asserted)

A sample of $N_0 = 1000$ nuclei decays with decay constant $\lambda = 0.5$ s⁻¹. Each nucleus has the same fixed chance of decaying per second, so the number decaying per second is proportional to how many are left: $dN/dt = -\lambda N$ . Solving this gives the exponential $N(t) = N_0 e^{-\lambda t}$ , and the half-life $t_{1/2} = \ln 2 / \lambda$ falls out of it. Stacked $N$ – $t$ over $dN/dt$ – $t$ , the decay rate is the slope of the count — the same exponential machine as a discharging capacitor, in the nucleus.

Half-life

t_{1/2} = \ln 2 / \lambda

1.386 s

Mean lifetime

\tau = 1/\lambda

2 s

Initial decay rate

\lambda N_0

500 1/s

Nuclei left after 5 s

82.08 1

The decay rate is proportional to the number remaining

\frac{dN}{dt} = -\lambda N

Each nucleus has the same fixed chance per second of decaying, so the number decaying per second is proportional to how many are left. That is a first-order differential equation — the rate $dN/dt$ depends on $N$ itself.

check $dN/dt$ is the slope; check $dN/dt = -\lambda N$ ; $t_{1/2} = \ln 2/\lambda$ ; $1/e$ left after $\tau = 1/\lambda$

✓ The decay rate is the slope of $N$ : the lower panel is $\dfrac{dN}{dt}$ exactly, the derivative of the curve above. [structural]
✓ Each nucleus decays independently, so the rate is proportional to how many remain: $\dfrac{dN}{dt} = -\lambda N$ . [structural]
✓ The half-life is $t_{1/2} = \dfrac{\ln 2}{\lambda}$ : after it, exactly half remain, $N(t_{1/2}) = N_0/2$ . [structural]
✓ After one mean lifetime $\tau = 1/\lambda$ , a fraction $1/e$ of the sample remains. [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “A radioactive sample decays at a steady rate, losing the same number of nuclei every second until it is gone.”

It does not decay at a steady rate. The number lost per second is proportional to how many remain, $dN/dt = -\lambda N$ , so the rate is fastest at the start and slows as the sample shrinks — it never quite reaches zero. That is exactly why decay is exponential and is described by a constant half-life rather than a constant lifetime: every half-life removes half of whatever is left, not a fixed number. Watch the panels: where $N$ is steepest (at the start) the $dN/dt$ curve is largest in magnitude; as $N$ flattens, the rate fades toward zero. The activity a detector reads is $|dN/dt| = \lambda N$ .

Modeling assumptions — author-asserted, disclosed not discharged

A large number of identical nuclei, each decaying independently with a constant probability per unit time, so the count can be treated as a smooth continuous $N(t)$ .

The stacked graph, fully annotated

A static rendering (Matplotlib) at the default parameters — the interactive version is in the Graph tab above.

Formulas used

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