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Nuclear stability and modes of decay (VCE Physics Unit 1)

Syllabus dot point

“Describe the structure of atomic nuclei, the strong nuclear force, and the modes of radioactive decay (alpha, beta-minus, beta-plus, gamma), and write balanced nuclear equations”

7 min read

Quick answer

The short-range strong force binds protons and neutrons, and the neutron-to-proton ratio sets stability. Unstable nuclei decay by alpha (−4 A, −2 Z), beta-minus (+1 Z), beta-plus (−1 Z) or gamma (no change), and every equation must conserve mass number A and atomic number Z.

Jump to a section
  1. What this dot point is asking
  2. The nucleus
  3. Stability
  4. Alpha decay
  5. Beta-minus decay
  6. Beta-plus decay
  7. Gamma emission
  8. Conservation in nuclear equations
  9. VCAA exam style
  10. Common traps
  11. In one sentence
  12. Examples in context
  13. Try this
  14. Exam-style questions

What this dot point is asking

VCAA wants you to describe the nucleus (protons, neutrons, the strong force, stability), identify the four classical decay modes, and write balanced nuclear equations that conserve mass number and atomic number.

The nucleus

Atomic nuclei contain protons (ZZ) and neutrons (NN), together called nucleons. The mass number A=Z+NA = Z + N. Notation: ZAX^A_Z X where XX is the chemical symbol.

The strong nuclear force binds nucleons. It is short-range (about 11 fm), attractive, and stronger than the electromagnetic repulsion between protons over very short distances.

Stability

Stable nuclei occupy a narrow "valley of stability" on a NN vs ZZ chart. For light nuclei (Z<20Z < 20), stable isotopes have N≈ZN \approx Z. For heavier nuclei, more neutrons are needed to dilute the proton-proton repulsion: stable isotopes have N/Z≈1.5N/Z \approx 1.5 near uranium.

Unstable nuclei decay toward the valley of stability through one or more decay events.

Alpha decay

The nucleus emits an alpha particle (24^4_2He). Common for heavy nuclei (A>200A > 200).

ZAX→ Z−2A−4Y+ 24He^A_Z X \to \,^{A-4}_{Z-2}Y + \,^4_2\text{He}

Alpha particles are highly ionising but have low penetration (stopped by paper).

Beta-minus decay

A neutron decays into a proton plus an electron plus an electron antineutrino. The electron is emitted as the beta particle.

ZAX→ Z+1AY+ −10e+νˉe^A_Z X \to \,^A_{Z+1}Y + \,^0_{-1}e + \bar{\nu}_e

Common for neutron-rich nuclei. The antineutrino carries away part of the kinetic energy, producing the continuous beta energy spectrum.

Beta-plus decay

A proton decays into a neutron plus a positron plus an electron neutrino. Common for proton-rich nuclei.

ZAX→ Z−1AY+ +10e+νe^A_Z X \to \,^A_{Z-1}Y + \,^0_{+1}e + \nu_e

The positron annihilates with an electron, producing two 0.5110.511 MeV gamma photons in opposite directions. This is the basis of PET imaging.

Gamma emission

A nucleus in an excited state drops to a lower state by emitting a gamma photon. AA and ZZ are unchanged.

ZAY∗→ ZAY+γ^A_Z Y^* \to \,^A_Z Y + \gamma

Often follows alpha or beta decay (the daughter is left in an excited state).

Conservation in nuclear equations

Always conserve:

  • Mass number AA (total nucleons).
  • Atomic number ZZ (total charge).
  • Lepton number (electron + neutrino vs positron + antineutrino).

Energy, momentum and angular momentum are also conserved, but VCE Year 11 questions focus on AA and ZZ.

VCAA exam style

Year 11 SAC tasks include:

  • Identifying the decay mode from a parent and daughter pair.
  • Writing balanced equations.
  • Identifying stable vs unstable isotopes from a chart of nuclides.

Common traps

Wrong ZZ direction in beta decay
Beta-minus increases ZZ (extra proton). Beta-plus decreases ZZ.
Treating gamma emission as changing the element
Gamma keeps both AA and ZZ unchanged.
Forgetting the (anti)neutrino
For full marks, include it. The antineutrino in beta-minus accompanies the electron; the neutrino in beta-plus accompanies the positron.

In one sentence

Atomic nuclei contain protons and neutrons bound by the short-range strong force, with stability determined by the neutron-to-proton ratio; unstable nuclei decay by alpha (−4A-4 A, −2Z-2 Z), beta-minus (+1Z+1 Z, neutron →\to proton plus electron plus antineutrino), beta-plus (−1Z-1 Z, proton →\to neutron plus positron plus neutrino) or gamma (no AA/ZZ change) emission, and every decay equation must conserve mass number and atomic number.

Examples in context

Example 1. Carbon-14 beta-minus decay used at the AMS facility, ANSTO. Carbon-14 decays by beta-minus emission: 614C→714N+e−+νˉe^{14}_{6}\text{C} \to {^{14}_{7}\text{N}} + e^- + \bar{\nu}_e. ANSTO's accelerator mass spectrometer at Lucas Heights counts 14^{14}C atoms directly in samples as small as a milligram, which is how archaeologists date Lake Mungo charcoal. A 11 mg modern carbon sample contains 5×10195 \times 10^{19} carbon atoms, of which 6×1076 \times 10^7 are 14^{14}C. Decay rate A=λNA = \lambda N with λ=ln⁡(2)/T1/2=3.83×10−12\lambda = \ln(2)/T_{1/2} = 3.83 \times 10^{-12} s−1^{-1} gives an activity of 2.3×10−42.3 \times 10^{-4} Bq (0.23 mBq), so low that decay counting would take days to collect a few counts, while AMS counts the atoms directly in minutes.

Example 2. Positron emission tomography at the Peter MacCallum Cancer Centre. Fluorine-18 used in PET scans decays by beta-plus emission: 918F→818O+e++νe^{18}_{9}\text{F} \to {^{18}_{8}\text{O}} + e^+ + \nu_e with a half-life of 110110 minutes. The emitted positron annihilates with a tissue electron, producing back-to-back 511511 keV gamma photons detected in coincidence. The Royal Melbourne and Peter Mac receive 18^{18}F from a Clayton cyclotron at start-of-day; after a 3030 min transport, activity falls to A/A0=(1/2)30/110=0.83A/A_0 = (1/2)^{30/110} = 0.83. Balanced equation p→n+e++νp \to n + e^+ + \nu inside the nucleus reduces atomic number by one without changing mass number, consistent with the change from F (Z=9Z=9) to O (Z=8Z=8).

Try this

Q1. Write the balanced nuclear equation for the alpha decay of 238^{238}U. [2 marks]

  • Cue. 92238U→90234Th+24He^{238}_{92}\text{U} \to {^{234}_{90}\text{Th}} + {^{4}_{2}\text{He}}. Mass number drops by 4, atomic number by 2.

Q2. A nucleus undergoes beta-minus decay followed by an alpha decay. If the parent is 60^{60}Co (Z=27Z = 27), determine the mass number and atomic number of the final nucleus. [4 marks]

  • Cue. 2760Co→2860Ni+e−+νˉ^{60}_{27}\text{Co} \to {^{60}_{28}\text{Ni}} + e^- + \bar{\nu} then 2860Ni→2656Fe+α^{60}_{28}\text{Ni} \to {^{56}_{26}\text{Fe}} + \alpha. Final: A=56A = 56, Z=26Z = 26.

Q3. Refer to 18^{18}F beta-plus decay used at Peter Mac. (a) Write the balanced nuclear equation. (b) Explain the origin of the 511511 keV photons detected. (c) Determine the activity of a 400400 MBq sample after 44 hours given a 110110 min half-life. [2+2+2 marks]

  • Cue. (a) 918F→818O+e++νe^{18}_{9}\text{F} \to {^{18}_{8}\text{O}} + e^+ + \nu_e. (b) Positron annihilation with an electron produces two 511511 keV photons. (c) 400×(1/2)240/110=88400 \times (1/2)^{240/110} = 88 MBq.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Year 11 SAC-style4 marks
Carbon-14 (614^{14}_6C) undergoes beta-minus decay. (a) Write the balanced equation. (b) Identify the daughter nucleus and explain the conservation laws applied.
Show worked answer →

(a) Balanced equation.

614C→714N+−10e+νˉe^{14}_{6}\text{C} \to ^{14}_{7}\text{N} + ^0_{-1}e + \bar{\nu}_e

(b) Daughter and conservation.

Daughter nucleus: nitrogen-14.

Conservation: mass number AA (14=14+014 = 14 + 0) and atomic number ZZ (6=7+(−1)6 = 7 + (-1)) are both conserved. Charge is conserved (the antineutrino is neutral). Energy and momentum are conserved (shared between the beta particle and the antineutrino, explaining the continuous beta energy spectrum).

Markers reward conservation of AA and ZZ in the equation, the daughter named explicitly, and inclusion of the antineutrino.

Practise this

Sources & how we know this

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