Coaching Centre
Radioactive decay and half-life
Why some nuclei are unstable, the band of stability on the N–Z chart, alpha, beta-minus, beta-plus and gamma decay with balanced nuclear equations, properties of the radiations, decay series, the exponential decay law N = N₀e^(−λt), the decay constant and half-life, activity, and modelling half-life in the laboratory.
Every second, about 4000 potassium-40 nuclei in your body decay. You can't stop them, speed them up or predict which one will go next. Yet if you have enough of them, you can predict with great precision how many will be left in a billion years. Radioactive decay is random for one nucleus and almost perfectly predictable for many.
Why are some nuclei unstable?
A nucleus is held together by the strong nuclear force, which attracts all nucleons (protons and neutrons) to their nearest neighbours but acts only over about $10^{-15}$ m. Protons also repel each other electrostatically, and that force reaches across the whole nucleus. Neutrons add strong-force "glue" without adding repulsion.
- For light nuclei, the stable ratio is about $N/Z = 1$ (for example $^{12}$C, $^{16}$O).
- Heavier nuclei need extra neutrons to dilute the growing repulsion: $N/Z$ rises to about 1.5 for lead-208, the heaviest stable nucleus.
- Above $Z = 82$, no nucleus is stable; all eventually decay.
A nucleus off the band of stability decays spontaneously, in a way that moves it towards the band. The type of decay depends on where it is.
Types of decay
In every nuclear equation, mass number (A, top) and charge (Z, bottom) are conserved.
Alpha decay happens in heavy nuclei (mostly $Z > 82$) that have too many nucleons. The nucleus emits a helium nucleus, 2 protons and 2 neutrons:
$$^{226}_{\ 88}\text{Ra} \rightarrow {}^{222}_{\ 86}\text{Rn} + {}^{4}_{2}\text{He}$$
Beta-minus decay happens in nuclei with too many neutrons (below the band). Inside the nucleus a neutron changes into a proton, emitting an electron and an electron antineutrino. A goes unchanged; Z goes up by 1:
$$^{1}_{0}n \rightarrow {}^{1}_{1}p + {}^{\ 0}_{-1}e + \bar{\nu}_e, \qquad {}^{14}_{\ 6}\text{C} \rightarrow {}^{14}_{\ 7}\text{N} + {}^{\ 0}_{-1}e + \bar{\nu}_e$$
Beta-plus decay happens in nuclei with too many protons (above the band). A proton changes into a neutron, emitting a positron and an electron neutrino. Z goes down by 1:
$$^{1}_{1}p \rightarrow {}^{1}_{0}n + {}^{\ 0}_{+1}e + \nu_e, \qquad {}^{18}_{\ 9}\text{F} \rightarrow {}^{18}_{\ 8}\text{O} + {}^{\ 0}_{+1}e + \nu_e$$
(Proton-rich nuclei can also capture one of their own inner electrons instead, called electron capture, with the same effect on Z.) Fluorine-18 is the positron emitter used in PET scans; it is made in cyclotrons, including at ANSTO and in major hospitals.
Gamma decay often follows alpha or beta decay, when the new nucleus is left in an excited state. The nucleus drops to a lower energy level, emitting a high-energy photon. A and Z don't change:
$$^{99\text{m}}_{\ 43}\text{Tc} \rightarrow {}^{99}_{43}\text{Tc} + \gamma$$
Technetium-99m, made at Lucas Heights in Sydney and sent to hospitals across Australia, is used in about 80% of nuclear medicine scans.
Why the neutrino? In beta decay the electrons come out with a whole range of energies, up to a maximum. If only two particles were produced, energy and momentum conservation would fix the electron's energy. In 1930 Wolfgang Pauli proposed that a third, neutral, almost massless particle carries off the missing energy. Neutrinos were finally detected in 1956.
Properties of the radiations
| Property | Alpha (α) | Beta-minus (β⁻) | Gamma (γ) |
|---|---|---|---|
| Nature | Helium nucleus $^4_2$He | Electron (β⁺: positron) | High-energy photon |
| Charge | +2e | −1e (β⁺: +1e) | 0 |
| Mass | 4 u | 1/1836 u | 0 |
| Typical speed | about 5% of c | up to 99% of c | c |
| Ionising ability | Very high | Medium | Low |
| Penetration | Stopped by a sheet of paper or a few cm of air | Stopped by a few mm of aluminium | Reduced by several cm of lead or metres of concrete |
| In electric and magnetic fields | Deflected (small curvature, massive) | Deflected the opposite way (large curvature) | Not deflected |
Alpha emitters are harmless outside the body (the dead outer layer of skin stops them) but very dangerous inside it, for example radon gas breathed into the lungs.
Decay series
A heavy nucleus often decays into another unstable nucleus, which decays again, and so on until a stable one is reached. Uranium-238 goes through 14 decays before ending as lead-206. Each alpha decay moves the nucleus down 2 in Z and 2 in N; this pushes it below the band of stability (too many neutrons), so beta-minus decays follow to bring it back.
Uranium-238 ($Z = 92$) decays through a series to lead-206 ($Z = 82$). How many alpha and beta-minus decays occur?
Only alpha decays change A, each by 4. $238 - 206 = 32$, so $32/4 = 8$ alpha decays.
8 alpha decays lower Z by 16, to $92 - 16 = 76$. But lead has $Z = 82$, so Z must rise by 6: there are 6 beta-minus decays.
- Press Next decay and watch uranium-238 move on the N–Z grid. Which direction does an alpha decay move it? A beta-minus decay?
- Read the equation for each step and check that A and Z balance.
- Find the longest-lived nucleus in the chain (other than U-238) and the shortest-lived. Compare uranium-234 (about 246 000 years) with polonium-214 (164 μs).
- Radon-222 is a gas. Why does that make it the main radiation hazard from uranium in rocks and soil?
- Step to the end. Check the counts: 8 alpha and 6 beta-minus decays.
The decay law and half-life
Decay is random and spontaneous: each nucleus has the same fixed probability of decaying per second, unaffected by temperature, pressure or chemical bonding, and with no memory of how long it has existed. That probability per unit time is the decay constant, $\lambda$ (units s−1, or y−1 etc.). For a large number of nuclei $N$, the rate of decay is proportional to how many are left, which gives exponential decay:
$$N_t = N_0 e^{-\lambda t}$$
The half-life, $t_{1/2}$, is the time for half the nuclei to decay. Put $N_t = N_0/2$: $\tfrac12 = e^{-\lambda t_{1/2}}$, so
$$\lambda = \frac{\ln 2}{t_{1/2}}$$
The activity $A$ is the number of decays per second, measured in becquerels (1 Bq = 1 decay per second). Activity is proportional to the number of nuclei, $A = \lambda N$, so it follows the same law: $A_t = A_0 e^{-\lambda t}$. Use whichever matches the data you're given: number of nuclei, mass of the isotope, activity or count rate all decay with the same half-life.
A patient is given iodine-131 (half-life 8.02 days) with an activity of 400 MBq. What is the activity after 30.0 days?
$\lambda = \dfrac{\ln 2}{8.02} = 0.0864$ day−1 (keep time in days throughout)
$A = A_0e^{-\lambda t} = 400 \times e^{-0.0864 \times 30.0} = 400 \times e^{-2.593} = 400 \times 0.0748 = 29.9$ MBq
Charcoal from an ancient campfire has a carbon-14 activity per gram that is 21% of that in living wood. Carbon-14 has a half-life of 5730 years. How old is the charcoal?
$\lambda = \dfrac{\ln 2}{5730} = 1.21 \times 10^{-4}$ y−1
$\dfrac{A}{A_0} = e^{-\lambda t}$, so $t = -\dfrac{\ln(A/A_0)}{\lambda} = -\dfrac{\ln 0.21}{1.21 \times 10^{-4}} = \dfrac{1.561}{1.21 \times 10^{-4}} = 1.29 \times 10^{4}$ years
About 12 900 years old. Radiocarbon dating works back to about 50 000 years. Dating of sites such as Madjedbebe in Arnhem Land, using radiocarbon and other methods, shows Aboriginal people have lived in Australia for at least 50 000 to 65 000 years.
Test yourself
Thorium-234 ($Z = 90$) undergoes beta-minus decay. The product is
β⁻: A stays 234, Z rises by 1 to 91 (protactinium).
A nucleus lies above the band of stability on an N–Z chart (with N on the vertical axis). It is most likely to decay by
Above the band means too many neutrons; β⁻ converts a neutron into a proton.
A radioisotope has a half-life of 6.0 hours. What fraction remains after 1 day?
24 h = 4 half-lives: $(1/2)^4 = 1/16$.
The decay constant of cobalt-60 (half-life 5.27 years) is
$\lambda = 0.693/5.27 = 0.132$ y⁻¹.
Which radiation is deflected most by a magnetic field?
Beta particles have a far smaller mass, so a much smaller radius ($r = mv/qB$), despite higher speed. Gamma is uncharged.
Label the decay modes.
Match each property to the radiation.
The decay constant is related to half-life by $\lambda = $ $/ t_{1/2}$. Activity is measured in . In β⁺ decay, a proton becomes a and a is emitted. Decay is random, but large samples follow an decay law.
A sample of phosphorus-32 has an activity of $5.0 \times 10^{6}$ Bq. After 20.0 days its activity is $1.9 \times 10^{6}$ Bq. Calculate the half-life of phosphorus-32 and the time for the activity to fall to $1.0 \times 10^{5}$ Bq.
$\dfrac{A}{A_0} = e^{-\lambda t}$: $\lambda = -\dfrac{\ln(1.9/5.0)}{20.0} = \dfrac{0.968}{20.0} = 0.0484$ day⁻¹. $t_{1/2} = \dfrac{\ln 2}{0.0484} = 14.3$ days. For $1.0 \times 10^{5}$ Bq: $t = -\dfrac{\ln(1.0 \times 10^{5}/5.0 \times 10^{6})}{0.0484} = \dfrac{\ln 50}{0.0484} = \dfrac{3.912}{0.04838} = 80.9$ days.
Analyse the spontaneous decay of unstable nuclei by explaining how the position of a nucleus relative to the band of stability determines its mode of decay. Use a balanced equation for each mode you describe.
A nucleus is held together by the short-range strong force between nucleons, opposed by the long-range electrostatic repulsion between protons. Stable nuclei lie in a band where N/Z is about 1 for light nuclei, rising to about 1.5 for the heaviest, because extra neutrons add attraction without repulsion. Nuclei off the band decay in a way that moves them towards it.
Nuclei with too many neutrons (above the band) undergo β⁻ decay, converting a neutron to a proton: $^{14}_{\ 6}\text{C} \rightarrow {}^{14}_{\ 7}\text{N} + {}^{\ 0}_{-1}e + \bar{\nu}_e$. Nuclei with too many protons (below the band) undergo β⁺ decay: $^{18}_{\ 9}\text{F} \rightarrow {}^{18}_{\ 8}\text{O} + {}^{\ 0}_{+1}e + \nu_e$. Very heavy nuclei (Z > 82) are too large for the strong force to overcome repulsion and emit alpha particles, reducing both Z and N by 2: $^{238}_{\ 92}\text{U} \rightarrow {}^{234}_{\ 90}\text{Th} + {}^4_2\text{He}$. A nucleus left in an excited state after any of these emits a gamma ray without changing A or Z: $^{99\text{m}}_{\ 43}\text{Tc} \rightarrow {}^{99}_{43}\text{Tc} + \gamma$.