Radioactive Half-Life, Decay Constant, and Activity

Radioactive decay is a first-order process described by N = N0e−λt, where λ is the decay constant, the half-life is t½ = ln 2 / λ = 0.693/λ, and the activity A = λN is the number of decays per second, measured in becquerels (1 Bq = 1 decay/s). In brief: iodine-131 has a half-life of 8.02 days, so after 24 days (three half-lives) only about 12.5 % of the original amount is left, and 1.00 g of radium-226 has an activity of 3.66 × 1010 Bq.

Each radioactive nucleus decays at random, and it is not possible to predict when, but a large number of nuclei decay in a regular way. The half-life tells how long a radioactive material remains dangerous or useful, from fractions of a second to billions of years. This article defines the quantities, gives the equations, shows a table of half-lives, works through calculations of remaining amount, activity and time, and describes how the units developed.

What is the law of radioactive decay?

The probability that a given nucleus decays in a short time is constant and independent of its age, the temperature or the chemical state. The rate of decay is therefore proportional to the number of nuclei present: −dN/dt = λN. This is the same as the first-order rate law in chemical kinetics. Integration gives:

N = N0 e−λt, or ln(N/N0) = −λt

where N0 is the number of nuclei at t = 0, N the number remaining at time t, and λ the decay constant (in s−1, y−1, …). The mean life τ = 1/λ is the average time a nucleus lives, and is 1.44 times the half-life.

What is the half-life?

The half-life t½ is the time in which half of the nuclei in a sample decay:

t½ = ln 2 / λ = 0.693 / λ

After n half-lives the fraction left is (½)n. The half-life is a property of the nuclide and cannot be changed by temperature, pressure or chemical reactions (with a few tiny exceptions in electron capture). The half-life is the same, whatever the amount of the substance.

n = 0 1 n = 1 ½ n = 2 ¼ n = 3 ⅛
Exponential decay of iodine-131 with half-lives marked 0 8 16 24 0 0.25 0.5 0.75 1 Time (days) Fraction of I-131 left 50% 25% 12.6%
Decay of iodine-131 (half-life 8.02 days). The amount, and so the activity, halves every 8.02 days: 50% after one half-life, 25% after two and about 12.6% after 24 days, as in Example 1.

What is activity?

The activity A is the number of decays per second: A = −dN/dt = λN = A0e−λt. Like the number of nuclei, the activity falls by half in every half-life.

Half-lives of important nuclides

Half-lives and uses
Nuclide Half-life Main decay Importance
99mTc6.01 hγMedical imaging
222Rn3.82 dαIndoor radon hazard
131I8.02 dβ−Thyroid treatment, fallout
60Co5.27 yβ−, γRadiotherapy, sterilization
3H12.3 yβ−Tracer, fusion fuel
137Cs30.1 yβ−, γFission product, gauges
14C5730 yβ−Radiocarbon dating
239Pu24,100 yαReactor product, weapons
235U7.04 × 108 yαFission fuel
238U4.47 × 109 yαDating of rocks

Worked examples

Example 1: the fraction remaining

What fraction of an iodine-131 sample (t½ = 8.02 days) is left after 24.0 days? Number of half-lives = 24.0 / 8.02 = 2.99. Fraction = (½)2.99 = 0.126, which is about 12.6 %. Using the decay constant: λ = 0.693 / 8.02 = 0.0864 d−1, N/N0 = e−0.0864 × 24.0 = e−2.074 = 0.126.

Example 2: the activity of radium

Find the activity of 1.00 g of radium-226 (t½ = 1600 y, M = 226.03 g/mol).

  1. t½ = 1600 × 3.156 × 107 s = 5.05 × 1010 s, so λ = 0.693 / 5.05 × 1010 = 1.37 × 10−11 s−1.
  2. N = (1.00 / 226.03) × 6.022 × 1023 = 2.66 × 1021 nuclei.
  3. A = λN = 1.37 × 10−11 × 2.66 × 1021 = 3.66 × 1010 Bq, or about 0.99 Ci.

This is the origin of the definition of the curie.

Example 3: the activity of uranium-238

For 1.00 kg of uranium-238 (t½ = 4.47 × 109 y): N = (1000 / 238.05) × 6.022 × 1023 = 2.53 × 1024. λ = 0.693 / (4.47 × 109 × 3.156 × 107) = 4.91 × 10−18 s−1. A = 4.91 × 10−18 × 2.53 × 1024 = 1.24 × 107 Bq (12.4 MBq). The long half-life explains the low specific activity.

Example 4: the time needed for a given decay

A sample of phosphorus-32 (t½ = 14.3 d) has an activity of 5.0 MBq. How long until it is 0.50 MBq? A/A0 = 0.10, so t = ln(10) / λ = 2.303 / (0.693/14.3) = 2.303 / 0.04847 = 47.5 days. After 30 days, A = 5.0 × (½)30/14.3 = 5.0 × 0.234 = 1.17 MBq.

Example 5: determining the half-life from measurements

A sample gives 1200 counts per minute at the start and 750 counts per minute 6.0 hours later (background subtracted). λ = ln(1200/750) / 6.0 = 0.470 / 6.0 = 0.0783 h−1, so t½ = 0.693 / 0.0783 = 8.85 hours.

Example 6: the number of atoms

An activity of 1.00 MBq of technetium-99m (t½ = 6.01 h = 2.164 × 104 s) corresponds to N = A / λ = A × t½ / 0.693 = 1.00 × 106 × 2.164 × 104 / 0.693 = 3.12 × 1010 atoms, a mass of only 5.1 × 10−12 g. Such tiny amounts of a short-lived nuclide give a high activity.

What happens with decay chains?

The daughter of a decay is often itself radioactive, and the decay continues in a series until a stable nuclide is reached. The uranium-238 series ends with lead-206 after 14 steps (8 alpha and 6 beta decays), and the thorium-232 series ends with lead-208. When the half-life of the parent is much longer than that of the daughter, the system reaches secular equilibrium, in which the activities of parent and daughter become equal. This is the reason why radium and radon are found with uranium ores. The 8 alpha decays can be checked: 238 − 4 × 8 = 206, and the 6 beta decays restore the atomic number: 92 − 2 × 8 + 6 = 82.

How did the decay law develop?

Henri Becquerel discovered radioactivity in 1896, when he found that uranium salts darken photographic plates. Marie and Pierre Curie introduced the term radioactivity and discovered polonium and radium in 1898. Ernest Rutherford and Frederick Soddy showed in 1902 that radioactive decay follows an exponential law, and that atoms of one element are transformed into atoms of another, which was an unexpected finding at the time. Rutherford introduced the half-life in 1907. The becquerel became the SI unit in 1975, and the gray and the sievert, for absorbed dose, in 1975 and 1979.

Which mistakes are common?