Topic 04 | Chemical sciences

Atomic model evolution & radioactivity

Year 9 (Levels 9-10 band): how the model of the atom changed from Dalton to the quantum model, isotopes, alpha/beta/gamma decay, and half-life calculations.

55-75 min Printable practice Answer key Challenge included
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Read the explanation, work through the examples, then complete the core practice before printing.

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What you will learn

Worked example 0 Real-world example: smoke alarm in your ceiling

A household smoke alarm contains a tiny amount of americium-241, an alpha emitter with half-life about 432 years.

  1. Alpha particles ionise air molecules in a small chamber, allowing a tiny current to flow.
  2. Smoke particles disrupt the ionisation and reduce the current — the alarm triggers.
  3. Because the half-life is long, the source’s activity barely changes over the alarm’s 10-year lifetime.
  4. Alpha particles travel only a few cm in air and are stopped by the plastic case, so external exposure is negligible.

Key idea: the right isotope for a job matches the type of radiation, the activity, and the half-life to the application.

1. How the model of the atom changed

YearScientistModelKey evidence
1803Daltonsolid indivisible sphereconstant proportions in compounds
1897Thomson”plum pudding” — electrons in a positive cloudcathode-ray tube showed negatively charged particles
1911Rutherfordtiny dense positive nucleus, electrons around itgold-foil experiment: most alpha particles passed through, a few deflected
1913Bohrelectrons in fixed circular orbits (shells)hydrogen line spectrum
1920s+Schrodinger / quantumelectrons in orbitals (probability clouds)wave-particle duality, Heisenberg uncertainty
Daltonsolid sphereThomsonplum puddingRutherfordnuclearBohrshells
Four key atomic models in sequence. Each was the best fit to the evidence of its time, and each was later revised when new evidence appeared.

2. Structure of an atom

ParticleChargeRelative massLocation
Proton+1+11nucleus
Neutron001nucleus
Electron1-1118360\dfrac{1}{1836} \approx 0orbitals around nucleus

Notation: ZAX^{A}_{Z}\text{X}, where ZZ is the atomic number (protons) and AA is the mass number (protons + neutrons).

3. Isotopes

Isotopes are atoms of the same element (same ZZ) with different numbers of neutrons (different AA).

All isotopes of an element have the same chemistry (because chemistry depends on electrons) but different nuclear properties.

Worked example 1 Identifying particles

For 92238U^{238}_{92}\text{U}, state the number of protons, neutrons, and electrons (neutral atom).

  1. Protons: Z=92Z = 92.
  2. Neutrons: AZ=23892=146A - Z = 238 - 92 = 146.
  3. Electrons: 92 (neutral).

4. Radioactive decay

Unstable (radioactive) nuclei spontaneously emit radiation and become more stable. Three common decay types.

Types of radioactive decay

Alpha decay ($\alpha$)

A helium-4 nucleus is emitted; ZZ decreases by 2, AA decreases by 4.

ZAXZ2A4Y+24α^{A}_{Z}\text{X} \to \,^{A-4}_{Z-2}\text{Y} + \,^{4}_{2}\alpha
Beta decay ($\beta^{-}$)

A neutron converts to a proton + electron; the electron is emitted. ZZ increases by 1, AA is unchanged.

ZAXZ+1AY+10β^{A}_{Z}\text{X} \to \,^{A}_{Z+1}\text{Y} + \,^{0}_{-1}\beta
Gamma emission ($\gamma$)

A high-energy photon is released, often after alpha or beta decay. AA and ZZ are unchanged.

Penetration and shielding:

RadiationChargeStopped by
Alpha+2+2a sheet of paper or a few cm of air
Beta1-1a few mm of aluminium
Gamma00dense lead or thick concrete
Worked example 2 Writing a decay equation

Uranium-238 undergoes alpha decay. Write the nuclear equation and identify the daughter nucleus.

  1. Alpha decay: AA drops by 4, ZZ drops by 2.
  2. 92238U90234Th+24α^{238}_{92}\text{U} \to \,^{234}_{90}\text{Th} + \,^{4}_{2}\alpha.
  3. Daughter: thorium-234.
Worked example 3 Beta decay of carbon-14

Carbon-14 undergoes beta-minus decay. Write the equation.

  1. AA unchanged, ZZ increases by 1.
  2. 614C714N+10β^{14}_{6}\text{C} \to \,^{14}_{7}\text{N} + \,^{0}_{-1}\beta.
  3. Daughter: nitrogen-14.

Key idea: check that mass numbers and atomic numbers balance on both sides.

5. Half-life

The half-life t1/2t_{1/2} of a radioactive isotope is the time for half of the nuclei in a sample to decay. It is a constant for each isotope.

Remaining nuclei after time $t$
N=N0(12)t/t1/2N = N_0 \left(\dfrac{1}{2}\right)^{t / t_{1/2}}

where N0N_0 is the initial number, NN is the number after time tt, and t1/2t_{1/2} is the half-life.

Worked example 4 Counting half-lives

Iodine-131 has a half-life of 8 days. A hospital has 240 mg of I-131. How much remains after 24 days?

  1. Number of half-lives: 248=3\dfrac{24}{8} = 3.
  2. Remaining: 240×(12)3=240×18=30240 \times \left(\dfrac{1}{2}\right)^3 = 240 \times \dfrac{1}{8} = 30 mg.
Worked example 5 Carbon dating

A wooden artefact has 14\dfrac{1}{4} of the carbon-14 of a living sample. Estimate its age. (t1/2t_{1/2} for C-14 =5730= 5730 years.)

  1. 14=(12)2\dfrac{1}{4} = \left(\dfrac{1}{2}\right)^2, so t=2×t1/2=2×5730=11460t = 2 \times t_{1/2} = 2 \times 5730 = 11\,460 years.
  2. The artefact is about 11 460 years old.

Key idea: exponential decay means repeated halving. Each half-life cuts the amount to half of what was there.


Practice: Year 9

Fluency

Atoms and isotopes

    1. List the atomic models in order and name the scientist most associated with each.
    2. Describe Rutherford’s gold-foil experiment and the conclusion he drew.
    3. For 1735Cl^{35}_{17}\text{Cl}, state the number of protons, neutrons, and electrons.
    4. Define isotope. Give one example.
    5. Why do isotopes of the same element have the same chemistry?
    6. Write the symbol for: (a) 6 protons, 7 neutrons; (b) 20 protons, 22 neutrons.
Fluency

Decay equations

    1. Write the alpha decay equation for radium-226 (88226Ra^{226}_{88}\text{Ra}).
    2. Write the beta-minus decay equation for strontium-90 (3890Sr^{90}_{38}\text{Sr}).
    3. Americium-241 decays by alpha emission. Write the equation and name the daughter.
    4. State which radiation (alpha, beta, gamma) is (a) stopped by paper, (b) an electron, (c) a photon.
    5. A nucleus loses 2 alpha and 1 beta particle. How do AA and ZZ change overall?
Reasoning

Half-life

    1. A sample starts with 1600 atoms. After 3 half-lives, how many remain?
    2. Technetium-99m has half-life 6 hours. Starting with 80 mg, how much remains after 24 hours?
    3. A sample has dropped to 116\dfrac{1}{16} of its original activity. How many half-lives have passed?
    4. Cobalt-60 has half-life 5.27 years. Starting with 100 g, how much remains after 10.54 years?
    5. A patient is injected with 40 MBq of an isotope with half-life 2 hours. What is the activity after 6 hours?
Problem solving

Applications and reasoning

    1. Why is a long half-life (thousands of years) suitable for radioactive dating but unsuitable for medical imaging?
    2. An archaeologist finds that a bone contains 18\dfrac{1}{8} of the C-14 in a living bone. Estimate the age. (t1/2=5730t_{1/2} = 5730 y.)
    3. Explain why gamma radiation is used in cancer radiotherapy but alpha sources are not used externally.
    4. Suggest a reason Bohr’s model, though successful for hydrogen, failed to predict the spectra of larger atoms.

Challenge

Reasoning

Harder reasoning

    1. Rutherford’s team fired alpha particles at a thin gold foil. Most passed straight through, but a tiny fraction bounced back. Using this evidence, argue why the “plum pudding” model had to be replaced.
    2. A decay chain: 92238U^{238}_{92}\text{U} eventually becomes 82206Pb^{206}_{82}\text{Pb} via multiple alpha and beta decays. If the total change in AA is 32-32 and in ZZ is 10-10, how many alpha and how many beta-minus decays are involved? Show your working.
    3. A patient receives a technetium-99m scan with activity 800 MBq at injection. If the effective half-life in the body is 4 hours, what activity remains after 16 hours? Comment on why Tc-99m is chosen for imaging.
    4. The number NN of undecayed nuclei follows N=N0(1/2)t/t1/2N = N_0 (1/2)^{t/t_{1/2}}. Rearrange to express tt in terms of NN, N0N_0, and t1/2t_{1/2}, and use this to estimate the age of a rock in which NN0=0.3\dfrac{N}{N_0} = 0.3 for an isotope with half-life 1.3×1091.3 \times 10^9 years.
Answers

Answer key

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Year 9 answers

Fluency

Atoms and isotopes

    1. Dalton (solid sphere) -> Thomson (plum pudding) -> Rutherford (nuclear) -> Bohr (shells) -> Schrodinger/quantum (orbitals).
    2. Alpha particles were fired at a thin gold foil. Most passed straight through, a few were deflected, and a very small fraction bounced back. Conclusion: the atom is mostly empty space with a tiny, dense, positively charged nucleus at its centre.
    3. Protons 17, neutrons 3517=1835 - 17 = 18, electrons 17.
    4. Isotopes are atoms of the same element (same number of protons) with different numbers of neutrons. Example: carbon-12 and carbon-14.
    5. Chemistry depends on electrons (and their arrangement). Isotopes have the same electron configuration because they have the same number of protons and electrons.
    6. (a) 613C^{13}_{6}\text{C} (carbon-13), (b) 2042Ca^{42}_{20}\text{Ca} (calcium-42).
Fluency

Decay equations

    1. 88226Ra86222Rn+24α^{226}_{88}\text{Ra} \to \,^{222}_{86}\text{Rn} + \,^{4}_{2}\alpha.
    2. 3890Sr3990Y+10β^{90}_{38}\text{Sr} \to \,^{90}_{39}\text{Y} + \,^{0}_{-1}\beta.
    3. 95241Am93237Np+24α^{241}_{95}\text{Am} \to \,^{237}_{93}\text{Np} + \,^{4}_{2}\alpha. Daughter: neptunium-237.
    4. (a) alpha, (b) beta-minus (an electron emitted from the nucleus), (c) gamma (high-energy photon).
    5. Two alpha: AA drops by 8, ZZ drops by 4. One beta: ZZ rises by 1, AA unchanged. Overall: ΔA=8\Delta A = -8, ΔZ=3\Delta Z = -3.
Reasoning

Half-life

    1. 1600×(1/2)3=1600/8=2001600 \times (1/2)^3 = 1600 / 8 = 200 atoms.
    2. Half-lives in 24 h: 24/6=424/6 = 4. Remaining: 80×(1/2)4=80/16=580 \times (1/2)^4 = 80/16 = 5 mg.
    3. (1/2)n=1/16=(1/2)4(1/2)^n = 1/16 = (1/2)^4, so 4 half-lives.
    4. 10.54/5.27=210.54 / 5.27 = 2 half-lives. Remaining: 100×(1/2)2=25100 \times (1/2)^2 = 25 g.
    5. 6/2=36/2 = 3 half-lives. Remaining activity: 40×(1/2)3=540 \times (1/2)^3 = 5 MBq.
Problem solving

Applications and reasoning

    1. Long half-lives work for dating because activity changes measurably over thousands/millions of years. For medical imaging the source should decay quickly after the scan so the patient is not exposed to radiation for long — a short half-life (hours) is better.
    2. 1/8=(1/2)31/8 = (1/2)^3, so 3 half-lives. Age =3×5730=17190= 3 \times 5730 = 17\,190 years.
    3. Gamma rays penetrate tissue to reach tumours and can be aimed precisely. Alpha particles have very short range in tissue, so external alpha cannot reach tumours; their high ionising power also makes them dangerous inside the body but ineffective from outside.
    4. Bohr’s model assumed fixed circular orbits and a single electron. In multi-electron atoms, electron-electron repulsion and subshell structure (s, p, d) affect spectra in ways Bohr could not predict. The quantum model with orbitals handles these correctly.
Reasoning

Challenge

    1. If atoms were uniform positive “puddings”, alpha particles (positive, fast) should have deflected only slightly. The back-scattering of a few shows there is a concentrated positive mass inside — the nucleus. Most passing straight through shows the atom is mostly empty space. This contradicts the plum pudding and supports Rutherford’s nuclear model.
    2. Each alpha: ΔA=4\Delta A = -4, ΔZ=2\Delta Z = -2. Each beta: ΔA=0\Delta A = 0, ΔZ=+1\Delta Z = +1. Let aa alphas and bb betas. Then 4a=32a=84a = 32 \Rightarrow a = 8, and 2a+b=1016+b=10b=6-2a + b = -10 \Rightarrow -16 + b = -10 \Rightarrow b = 6. So 8 alpha and 6 beta-minus decays.
    3. 16/4=416/4 = 4 half-lives. Remaining: 800×(1/2)4=50800 \times (1/2)^4 = 50 MBq. Tc-99m has a short half-life (about 6 hours physically, shorter effectively), giving enough time for imaging but minimal radiation dose afterwards; it also emits gamma rays detectable outside the body.
    4. N/N0=(1/2)t/t1/2t=t1/2log(N/N0)log(1/2)N/N_0 = (1/2)^{t/t_{1/2}} \Rightarrow t = t_{1/2} \cdot \dfrac{\log(N/N_0)}{\log(1/2)}. For N/N0=0.3N/N_0 = 0.3 and t1/2=1.3×109t_{1/2} = 1.3 \times 10^9 y: t=1.3×109×log0.3log0.5=1.3×109×1.7372.26×109t = 1.3 \times 10^9 \times \dfrac{\log 0.3}{\log 0.5} = 1.3 \times 10^9 \times 1.737 \approx 2.26 \times 10^9 years.

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