Physics · Ch 8 — Atomic and Nuclear Physics

Law of radioactive decay

8.6.4

Law of radioactive decay

A real radioactive sample contains an enormous number of nuclei, and not all of them decay simultaneously - each individual nucleus decays at a random, unpredictable moment (like the toss of a coin), but the collective behaviour of the whole sample can still be predicted precisely on a statistical basis.

Deriving the decay law. At any instant, the rate of decay $dN/dt$ (the number of decays per unit time) is found experimentally to be directly proportional to the number of undecayed nuclei $N$ present at that same instant:

\frac{dN}{dt}=-\lambda N\qquad (8.32)

Here $\lambda$ , called the decay constant, is a positive quantity characteristic of the particular radioactive species, and the negative sign reflects that $N$ is decreasing with time. Rearranging as $dN/N=-\lambda\,dt$ and integrating from the initial condition $N=N_0$ at $t=0$ up to $N$ at time $t$ ,

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

and exponentiating both sides gives the law of radioactive decay:

N=N_0e^{-\lambda t}\qquad (8.35)

This shows that the number of undecayed nuclei falls off exponentially, and - since an exponential never actually reaches zero - it also implies that, strictly, an infinite time would be needed for every single nucleus in the sample to decay.

Activity. A more directly measurable quantity is the activity (or decay rate) $R=-dN/dt$ , the number of decays occurring per second (a positive quantity). Differentiating the decay law gives $R=\lambda N_0e^{-\lambda t}$ , i.e.

$R=R_0e^{-\lambda t},\qquad R_0=\lambda N_0\qquad (8.37)$ …

Figure 8.26Law of radioactive decay

What this figure shows. This figure plots the number of undecayed nuclei N on the vertical axis against time t on the horizontal axis, drawing the smooth exponential decay curve N = N0 e^(-lambda t) that starts at N0 when t = 0 and falls off ever more slowly, never quite touching zero. The horizontal axis is marked at the successive half-life intervals T(1/2), 2T(1/2), 3T(1/2) and 4T(1/2), with dashed guide lines showing that the population has dropped to N0/2, N0/4, N0/8 and N0/16 respectively at those points, visually demonstrating that each equal …

Misc Example 8.12Carbon-14 nuclei remaining after 22,920 years

Worked out. This worked example finds how many carbon-14 nuclei remain undecayed after 22,920 years, starting from an initial 10,000 atoms and given the carbon-14 half-life of 5730 years. The elapsed time corresponds to n = t/T(1/2) = 22920/5730 = 4 half-lives exactly, so using N = N0 (1/2)^n gives N = 10000 times (1/2)^4 = 10000/16 = 625 nuclei remaining undecaye …

Misc Example 8.13Activity of a nitrogen-13 sample

Worked out. This four-part worked example concerns 2.6 micrograms of pure $^{13}_{7}\text{N}$ , half-life 10 minutes. (a) Since 13 g of nitrogen-13 contains Avogadro's number of atoms, the initial number of nuclei is N0 = (6.02e23/13) times 2.6e-6 g ≈ 1.204e16 atoms. (b) The decay constant is λ = ln2/T(1/2) = 0.6931/600 s = 1.155e-3 per second, giving an initial activity R0 = λN0 ≈ 1.39e13 decays per second (Bq), which converts to about 3.75e2 curie using 1 Ci = 3.7e10 Bq. (c) After 2 hours (7200 s, which is 12 half-lives), the activity has fallen by a factor of (1/2)^12 ≈ 2.4e-4, giving R ≈ 0.9 Ci by either the exponential formula or the half-life-count method, and both methods are shown to agree. …

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