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The decay law

Radiation and the atom · level 8, lesson 4 of 6

A question to start

Nobody can say when one particular radioactive nucleus will decay. Yet a gram of plutonium-238 gives out heat so steadily that a rover can run on it for years. How can random events be so predictable?

The idea

Each unstable nucleus has the same chance of decaying in each second, whatever its age. With huge numbers of nuclei, the same share decays in each equal stretch of time. The time for half of them to decay is the half-life \(T\):

\[N = N_0 \left(\tfrac12\right)^{t/T} = N_0\, e^{-\lambda t}, \qquad \lambda = \frac{\ln 2}{T} = \frac{0.693}{T}.\]

\(\lambda\) is the decay constant: the chance that one nucleus decays in a second. The activity \(A\), the number of decays each second, is

\[A = \lambda N.\]

Activity is measured in becquerels (Bq): one decay a second.

Isotope Decays to Half-life
Carbon-14 nitrogen-14 5,730 years
Plutonium-238 uranium-234 88 years
Uranium-235 lead-207 (in steps) 704 million years
Potassium-40 calcium-40 or argon-40 1.25 billion years
Uranium-238 lead-206 (in steps) 4.5 billion years
01234 1/21/41/81/16half-lives gone by

Worked examples

1. A rover's heat. Plutonium-238 after 14 years: \((\tfrac12)^{14/88} = 0.90\). The heat is still 90% of the start.

2. Activity of a gram. \(N = 6.02\times10^{23} / 238 = 2.53\times10^{21}\) nuclei. \(T = 88 \times 3.156\times10^7 = 2.78\times10^9\) s, so \(\lambda = 0.693 / 2.78\times10^9 = 2.5\times10^{-10}\) per s. \(A = 2.5\times10^{-10} \times 2.53\times10^{21} = 6.3\times10^{11}\) Bq.

3. Its power. Suppose each decay releases about 5.6 MeV, \(9.0\times10^{-13}\) J. Power \(= 6.3\times10^{11} \times 9.0\times10^{-13} \approx 0.57\) W per gram.

Try it

  1. What share of carbon-14 is left after 11,460 years?
  2. Find the decay constant of carbon-14, per year.
  3. How long until plutonium-238 gives out a quarter of its starting heat?
  4. Why does a sample with a short half-life have a high activity?
Answers
  1. Two half-lives: a quarter.
  2. \(0.693 / 5{,}730 = 1.21\times10^{-4}\) per year.
  3. Two half-lives: 176 years.
  4. \(\lambda = 0.693/T\) is large, so \(A = \lambda N\) is large for the same number of nuclei.

Common mistake

"After two half-lives, it is all gone." After one half-life, half is left; after two, a quarter; after ten, about a thousandth. It never quite reaches zero.

Where it leads