Atomic model evolution & radioactivity

Fluency

Dalton (solid sphere) -> Thomson (plum pudding) -> Rutherford (nuclear) -> Bohr (shells) -> Schrodinger/quantum (orbitals).
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.
Protons 17, neutrons $35 - 17 = 18$ , electrons 17.
Isotopes are atoms of the same element (same number of protons) with different numbers of neutrons. Example: carbon-12 and carbon-14.
Chemistry depends on electrons (and their arrangement). Isotopes have the same electron configuration because they have the same number of protons and electrons.
(a) $^{13}_{6}\text{C}$ (carbon-13), (b) $^{42}_{20}\text{Ca}$ (calcium-42).

Fluency

$^{226}_{88}\text{Ra} \to \,^{222}_{86}\text{Rn} + \,^{4}_{2}\alpha$ .
$^{90}_{38}\text{Sr} \to \,^{90}_{39}\text{Y} + \,^{0}_{-1}\beta$ .
$^{241}_{95}\text{Am} \to \,^{237}_{93}\text{Np} + \,^{4}_{2}\alpha$ . Daughter: neptunium-237.
(a) alpha, (b) beta-minus (an electron emitted from the nucleus), (c) gamma (high-energy photon).
Two alpha: $A$ drops by 8, $Z$ drops by 4. One beta: $Z$ rises by 1, $A$ unchanged. Overall: $\Delta A = -8$ , $\Delta Z = -3$ .

Reasoning

$1600 \times (1/2)^3 = 1600 / 8 = 200$ atoms.
Half-lives in 24 h: $24/6 = 4$ . Remaining: $80 \times (1/2)^4 = 80/16 = 5$ mg.
$(1/2)^n = 1/16 = (1/2)^4$ , so 4 half-lives.
$10.54 / 5.27 = 2$ half-lives. Remaining: $100 \times (1/2)^2 = 25$ g.
$6/2 = 3$ half-lives. Remaining activity: $40 \times (1/2)^3 = 5$ MBq.

Problem solving

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.
$1/8 = (1/2)^3$ , so 3 half-lives. Age $= 3 \times 5730 = 17\,190$ years.
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.
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

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.
Each alpha: $\Delta A = -4$ , $\Delta Z = -2$ . Each beta: $\Delta A = 0$ , $\Delta Z = +1$ . Let $a$ alphas and $b$ betas. Then $4a = 32 \Rightarrow a = 8$ , and $-2a + b = -10 \Rightarrow -16 + b = -10 \Rightarrow b = 6$ . So 8 alpha and 6 beta-minus decays.
$16/4 = 4$ half-lives. Remaining: $800 \times (1/2)^4 = 50$ 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.
$N/N_0 = (1/2)^{t/t_{1/2}} \Rightarrow t = t_{1/2} \cdot \dfrac{\log(N/N_0)}{\log(1/2)}$ . For $N/N_0 = 0.3$ and $t_{1/2} = 1.3 \times 10^9$ y: $t = 1.3 \times 10^9 \times \dfrac{\log 0.3}{\log 0.5} = 1.3 \times 10^9 \times 1.737 \approx 2.26 \times 10^9$ years.

Year 9 answers