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Nuclear physics and radioactivity: QCE Physics Unit 1 Year 11

Syllabus dot point

“Atomic nucleus, isotopes, types of radioactive decay (alpha, beta, gamma), half-life, fission and fusion”

10 min read

Quick answer

Nuclei contain protons and neutrons; unstable nuclei undergo alpha (emit He nucleus), beta-minus (neutron to proton plus electron plus antineutrino) or gamma (photon) decay; the half-life formula N=N0(1/2)t/T1/2N = N_0 (1/2)^{t/T_{1/2}} describes statistical decay over time, fission of heavy nuclei (uranium) and fusion of light nuclei (hydrogen) release energy by converting mass to energy (E=mc2E = mc^2).

Jump to a section
  1. What this dot point is asking
  2. Atomic structure
  3. Radioactive decay
  4. Nuclear equations
  5. Half-life
  6. Fission
  7. Fusion
  8. Examples in context
  9. Try this
  10. Exam-style questions

What this dot point is asking

QCAA wants Year 11 students to describe atomic structure, recognise radioactive decay types, and apply the half-life formula.

Atomic structure

Nucleus: protons (+e, 1.67×10−271.67 \times 10^{-27} kg) and neutrons (0 charge, similar mass).

Notation ZAX^A_Z X: ZZ atomic number (protons), AA mass number (protons + neutrons), N=A−ZN = A - Z neutrons.

Isotopes. Same ZZ, different NN. E.g., 12^{12}C, 13^{13}C, 14^{14}C.

Radioactive decay

Alpha
Emit helium nucleus 24^4_2He. Mass number drops 4, atomic number drops 2. Range: cm in air; stopped by paper.
Beta-minus
Neutron →\to proton + electron + antineutrino. Atomic number increases by 1; mass number unchanged. Range: metres in air; stopped by aluminium.
Gamma
High-energy photon from excited nucleus. Mass and atomic numbers unchanged. Highly penetrating; lead/concrete shielding.

Nuclear equations

Conservation: mass number and charge conserved on both sides.

Examples:

92238U→90234Th+24He^{238}_{92} \text{U} \to ^{234}_{90} \text{Th} + ^4_2 \text{He} (alpha).

614C→714N+−10e+νˉe^{14}_6 \text{C} \to ^{14}_7 \text{N} + ^0_{-1} e + \bar{\nu}_e (beta-minus).

Half-life

N=N0(12)t/T1/2N = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}

Half-life is statistical; random for individual atom.

Common: C-14 (5,730 yr, carbon dating), I-131 (8 days, medical), U-238 (4.5 Gyr).

Fission

Heavy nucleus splits: 235U+n→141Ba+92Kr+3n^{235}\text{U} + n \to ^{141}\text{Ba} + ^{92}\text{Kr} + 3n, releasing ∼200\sim 200 MeV.

Chain reaction possible (more than one neutron per fission triggers next).

Fusion

Light nuclei combine: 2H+3H→4He+n^2\text{H} + ^3\text{H} \to ^4\text{He} + n, ∼17.6\sim 17.6 MeV.

Powers the sun. Controlled fusion remains research goal.

Common errors
Non-conservation in nuclear equations
Both mass number and charge must balance.
Confusing decay types
Alpha: heavy, slow. Beta: fast electron. Gamma: photon.
Half-life as deterministic
Individual decay is random.

Examples in context

Example 1. ANSTO Mt Cotton's archived monazite samples include thorium-232232 (half-life 1.4×1010 years1.4 \times 10^{10} \text{ years}), which decays in a long chain ending at lead-208208. Each alpha step lowers nucleon number by 44 and proton number by 22; each beta-minus step leaves nucleon number unchanged and raises proton number by 11. Geologists use the parent-to-daughter ratio to date Queensland sapphire-bearing basalt fields north of Anakie, applying N=N0(1/2)t/T1/2N = N_0 (1/2)^{t/T_{1/2}}. The QCAA Unit 1 nuclear summary expects students to balance such decay equations.

Example 2. Royal Brisbane Hospital cyclotron production of fluorine-1818 (T1/2=110 minT_{1/2} = 110 \text{ min}) supplies PET scanners. A batch leaving the cyclotron with 20 GBq20 \text{ GBq} at 07000700 has fallen to 20×(1/2)2.18=4.4 GBq20 \times (1/2)^{2.18} = 4.4 \text{ GBq} by the patient injection at 11001100. The isotope decays by positron (β+\beta^+) emission, the positrons annihilate with electrons and produce paired 511 keV511 \text{ keV} gammas detected coincidentally to localise tumours. The same dot-point isotope concept (same Z, different N) is the basis of all medical radio-tracers.

Try this

Q1. Define isotope and identify the decay particle that does not change the mass number. [2 marks]

  • Cue. Isotope: same Z, different N; beta decay and gamma decay leave A unchanged.

Q2. Complete the alpha decay equation 88226Ra→?Rn+α^{226}_{88}\text{Ra} \rightarrow ?\text{Rn} + \alpha. State the mass and charge of the alpha particle. [3 marks]

  • Cue. 86222Rn^{222}_{86}\text{Rn}; alpha =24He= ^{4}_{2}\text{He}, mass ≈4.00 u\approx 4.00 \text{ u}, charge +2e+2e.

Q3. A 40 MBq40 \text{ MBq} source of cobalt-6060 (T1/2=5.3 yearsT_{1/2} = 5.3 \text{ years}) is purchased for radiotherapy. (a) Calculate the activity remaining after 15.9 years15.9 \text{ years}. (b) Cobalt-6060 decays by beta-minus then gamma. Write the daughter nuclide and explain how the gamma originates. (c) Discuss one regulatory implication in Queensland of selecting an isotope with this half-life. [3+3+2 marks; ISMG: Knowledge and conceptual understanding, Evaluation]

  • Cue. (a) Three half-lives, 5.0 MBq5.0 \text{ MBq}; (b) 60Ni^{60}\text{Ni} excited state emits gammas; (c) ARPANSA storage requirements.

Exam-style questions

Questions in the style of QCAA 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 has half-life 5,7305,730 years. (a) Atoms remaining after 11,46011,460 years from 101010^{10} initial? (b) Write the beta-minus decay equation.
Show worked answer →

(a) 11460/5730=211460/5730 = 2 half-lives. N=1010(1/2)2=2.5×109N = 10^{10} (1/2)^2 = 2.5 \times 10^9 atoms.

(b) 614C→714N+−10e+νˉe^{14}_6 \text{C} \to ^{14}_7 \text{N} + ^0_{-1} e + \bar{\nu}_e.

Charge: 6 = 7 + (-1). Mass: 14 = 14 + 0. Conserved.

Markers reward half-life calculation and conservation in nuclear equation.

Practise this

Sources & how we know this

ExamExplained