Coaching Centre
Radioactivity and nuclear stability
Why some nuclei are unstable, the band of stability and the neutron-to-proton ratio, alpha, beta and gamma radiation, balancing nuclear equations, decay series, half-life and uses of radioisotopes in Australia.
At Lucas Heights in southern Sydney, ANSTO's OPAL reactor fires neutrons into targets of uranium-235. One of the fission products, molybdenum-99, is extracted and shipped to hospitals around Australia, where it decays into technetium-99m. Tc-99m gives off gamma rays that pass out through the body to a camera, lighting up bones, hearts and tumours, and it is almost gone within a day. ANSTO estimates that about one in two Australians will have a nuclear medicine procedure during their life. All of this depends on understanding which nuclei are unstable and how they decay.
In Year 9 you met alpha, beta and gamma radiation and half-life. This lesson explains why a nucleus is unstable, how to predict the way it will decay, and how to write and balance nuclear equations.
Why are some nuclei unstable?
Protons in a nucleus repel each other electrostatically. They are held together by the strong nuclear force, which attracts all nucleons (protons and neutrons) but acts only over very short distances, about the width of a few nucleons. Neutrons add strong-force attraction without adding repulsion, so they act as nuclear glue.
- For light nuclei, stability needs about one neutron per proton: N/Z ≈ 1 (for example ¹²C, ¹⁶O, ⁴⁰Ca).
- As Z increases, repulsion grows faster than the short-range attraction, so heavier nuclei need extra neutrons. The stable ratio rises to about 1.5 near lead (²⁰⁸Pb: 126/82 = 1.54).
- Beyond lead (Z = 82) no nucleus is truly stable. Bismuth-209 was long listed as stable, but in 2003 it was found to undergo alpha decay with a half-life of about 2 × 10¹⁹ years, a billion times the age of the universe.
Plotting N against Z for all known stable nuclides gives a narrow band of stability. Where an unstable nuclide sits relative to the band predicts how it will decay.
Types of decay
| Emission | Identity and symbol | Change to nucleus | Who does it | Penetration | Ionising ability |
|---|---|---|---|---|---|
| Alpha, α | helium nucleus, $^{4}_{2}\text{He}$ | A − 4, Z − 2 | heavy nuclei, Z > 82 | a few cm of air; stopped by paper or skin | high |
| Beta minus, β⁻ | electron, $^{0}_{-1}\text{e}$ | A same, Z + 1 (n → p) | neutron-rich nuclei, above the band | about a metre of air; stopped by a few mm of aluminium | moderate |
| Beta plus, β⁺ | positron, $^{0}_{+1}\text{e}$ | A same, Z − 1 (p → n) | proton-rich nuclei, below the band | annihilates with an electron almost at once, making two gamma rays | moderate |
| Gamma, γ | high-energy photon, $\gamma$ | none: the nucleus loses energy | nuclei left in an excited state | reduced by thick lead or concrete | low |
Proton-rich nuclei can also undergo electron capture: the nucleus captures one of the atom's own inner electrons, turning a proton into a neutron (Z − 1, A unchanged). Potassium-40 decays mostly by β⁻ but sometimes by electron capture; the argon-40 it produces is used to date volcanic rock.
Balancing nuclear equations
In every nuclear equation, the total mass number and the total atomic number are the same on both sides. Use this to identify an unknown particle or nuclide, then use the periodic table to name the element from its Z.
(a) Write the equation for the alpha decay of radium-226. (b) Identify X in $^{234}_{90}\text{Th} \rightarrow X + {}^{0}_{-1}\text{e}$.
(a) Radium has Z = 88. Alpha decay removes A = 4 and Z = 2: the product has A = 222, Z = 86, which is radon.
$$^{226}_{88}\text{Ra} \rightarrow {}^{222}_{86}\text{Rn} + {}^{4}_{2}\text{He}$$
(b) Mass: 234 = A + 0, so A = 234. Charge: 90 = Z − 1, so Z = 91, protactinium. X is $^{234}_{91}\text{Pa}$.
Uranium-238 decays through a series of α and β⁻ emissions to stable lead-206. How many of each are emitted?
Only α changes A: 238 − 206 = 32, so 32 ÷ 4 = 8 α particles.
8 α would lower Z by 16, from 92 to 76. Lead has Z = 82, so Z must rise by 6: 6 β⁻ particles.
- Step to C-14. It sits above the band with N/Z = 1.33. Predict its decay mode, then check the equation.
- Find F-18, used in PET scans. Why does it emit positrons rather than electrons?
- Compare K-40 (N/Z = 1.11) with Ca-40 (N/Z = 1.00). Why is the calcium stable but the potassium not?
- Try Tc-99m. Its mode is γ only. What makes Tc-99m different from the Tc-99 it becomes?
- Switch on U-238 decay chain and step through all 14 decays. Check the counter reaches 8 α and 6 β⁻.
Half-life
The half-life, t½, is the time for half the radioactive nuclei in a sample to decay. It is fixed for each nuclide and does not depend on temperature, pressure or chemical form. After n half-lives the fraction remaining is $\left(\tfrac{1}{2}\right)^n$:
$$m = m_0\left(\tfrac{1}{2}\right)^{t/t_{1/2}}$$
A hospital receives 80 MBq of technetium-99m (t½ = 6.0 h) at 7:00 am. What activity remains at 7:00 pm?
12 hours = 12.0 ÷ 6.0 = 2 half-lives.
$80 \times \left(\tfrac{1}{2}\right)^2 = 20$ MBq. A six-hour half-life is long enough to do a scan and short enough that little remains in the patient by the next day.
Radioisotopes in use, and their effects
| Radioisotope | Emission, half-life | Use and why it suits |
|---|---|---|
| Technetium-99m | γ, 6.0 h | Diagnostic imaging: gamma escapes the body; short half-life limits dose. |
| Iodine-131 | β⁻ and γ, 8.0 days | Treating thyroid cancer: the thyroid absorbs iodine, so beta radiation is delivered where it is needed. |
| Lutetium-177 | β⁻, 6.6 days | Targeted therapy for prostate and neuroendocrine cancers; produced at OPAL. |
| Fluorine-18 | β⁺, 110 min | PET scans: the positrons annihilate, giving pairs of gamma rays the scanner locates. |
| Cobalt-60 | β⁻ and γ, 5.3 years | Sterilising medical equipment and food with penetrating gamma rays. |
| Americium-241 | α, 432 years | Smoke detectors: alpha particles ionise air in a chamber; smoke interrupts the current. Alpha cannot escape the casing. |
| Carbon-14 | β⁻, 5730 years | Dating organic material up to about 50 000 years old. |
Ionising radiation damages living cells by knocking electrons out of molecules, including DNA. Inside the body, α emitters are the most dangerous because all their energy is deposited in a few cells; outside the body, γ is the main hazard because it penetrates. Exposure is reduced by time (minimise it), distance (increase it) and shielding. In the environment, long-lived fission products such as caesium-137 and strontium-90 (which behaves like calcium and is taken up into bone) are the main concern after accidents such as Chernobyl (1986) and Fukushima (2011). The British nuclear tests at Maralinga in South Australia in the 1950s and 1960s contaminated the lands of the Anangu people; clean-up and a return of the land took decades.
Test yourself
Which equation correctly shows the β⁻ decay of strontium-90?
β⁻ raises Z by 1 (38 → 39, yttrium) and leaves A unchanged.
Nitrogen-13 has 7 protons and 6 neutrons. It lies below the band of stability. Its most likely decay mode is
Proton-rich (N/Z < 1), so a proton becomes a neutron: Z 7 → 6, carbon-13.
Thorium-232 decays to lead-208. How many α and β⁻ particles are emitted?
(232 − 208) ÷ 4 = 6 α; Z would fall 12 to 78, but lead is 82, so 4 β⁻.
Why do heavy stable nuclei have more neutrons than protons?
Extra neutrons offset the growing proton–proton repulsion.
A 64 g sample of a radioisotope decays to 4 g in 20 days. Its half-life is
64 → 32 → 16 → 8 → 4 is 4 half-lives in 20 days, so t½ = 5 days.
Label the band of stability diagram.
Predict the decay mode of each nuclide.
An alpha particle is a nucleus. In β⁻ decay a changes into a proton. A positron is the of the electron. Emission of a gamma ray changes neither the mass number nor the . The time for half of a radioactive sample to decay is its .
Iodine-131 ($^{131}_{53}\text{I}$, N/Z = 1.47) is used to treat thyroid cancer, while iodine-127 is the only stable isotope of iodine. Explain why iodine-131 is unstable, predict its mode of decay with a balanced nuclear equation, and explain why its radiation and half-life (8.0 days) make it suitable for this use.
I-131 has 78 neutrons, four more than the stable I-127 (74 neutrons). Its N/Z ratio is too high, so it lies above the band of stability and is neutron-rich.
It undergoes β⁻ decay, in which a neutron becomes a proton, lowering N/Z: $^{131}_{53}\text{I} \rightarrow {}^{131}_{54}\text{Xe} + {}^{0}_{-1}\text{e}$.
The thyroid absorbs iodine, so the isotope concentrates in the tumour. Beta particles travel only a few millimetres in tissue, so their energy kills thyroid cells while sparing most other organs. An 8-day half-life gives enough time to deliver a useful dose, but after about ten half-lives (80 days) less than 0.1% remains.