Atomic Structure (Classical)

Q1. (4 marks) Any three postulates (1 mark each, max 3):

• All matter is made of tiny indivisible particles called atoms.
• Atoms of a given element are identical in mass and properties; atoms of different elements differ.
• Compounds form when atoms combine in fixed, whole-number ratios.
• Atoms are neither created nor destroyed in a chemical reaction (rearranged only).

Limitation (1 mark): Atoms are divisible (contain electrons, protons, neutrons) / cannot explain isotopes or isobars / does not account for subatomic particles.

Why: Tests recall of postulates and awareness that later discoveries contradicted indivisibility.

Q2. (3 marks) — 1 mark each:

• Electron: J.J. Thomson — cathode ray (discharge) tube; determined charge-to-mass ratio $e/m$; rays deflected by electric/magnetic fields toward positive plate.
• Proton: Goldstein — anode/canal rays (positive rays) in a perforated cathode discharge tube.
• Neutron: Chadwick — bombardment of beryllium with $\alpha$-particles producing neutral penetrating radiation.

Q3. (3 marks) — 1 mark each (definition + example):

• Isotopes: same $Z$, different $A$ (different number of neutrons). Example: $^{1}_1\text{H}, ^{2}_1\text{H}$.
• Isobars: same $A$, different $Z$. Example: $^{40}_{18}\text{Ar}, ^{40}_{20}\text{Ca}$.
• Isotones: same number of neutrons $(A-Z)$, different $Z$. Example: $^{14}_6\text{C}, ^{16}_8\text{O}$ (each 8 neutrons).

Q4. (4 marks) Description (1): A thin gold foil was bombarded with a beam of $\alpha$-particles; scattering was detected on a fluorescent (ZnS) screen. Observations (1 each, max 2):

• Most $\alpha$-particles passed straight through undeflected → atom is mostly empty space.
• A few were deflected at large angles / very few bounced back → a small, dense, positively charged nucleus. Conclusion (1): The atom has a tiny, dense, positively charged nucleus containing most of the mass, with electrons around it (nuclear model).

Q5. (3 marks) $\bar{A} = \frac{(35)(75.77) + (37)(24.23)}{100}$ $= \dfrac{2651.95 + 896.51}{100} = \dfrac{3548.46}{100}$ (1 for setup, 1 for computation) $\bar{A} = 35.48\ \text{u} \quad (1)$

Why: Weighted mean by fractional abundance.

Q6. (4 marks) (a) $r_n = 0.529 \dfrac{n^2}{Z}\ \text{Å}$; $r_3 = 0.529 \times \dfrac{9}{1} = 4.761\ \text{Å}$ (1 formula + 1 answer) (b) $E_n = -13.6 \dfrac{Z^2}{n^2}\ \text{eV}$; $E_2 = -\dfrac{13.6}{4} = -3.4\ \text{eV}$ (1 formula + 1 answer)

Q7. (5 marks) Coulomb force provides centripetal force (1): $\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r^2} = \frac{mv^2}{r}\ \Rightarrow\ mv^2 = \frac{Ze^2}{4\pi\varepsilon_0 r}\quad(1)$ Quantization (1): $mvr = \dfrac{nh}{2\pi} \Rightarrow v = \dfrac{nh}{2\pi mr}$. Substitute $v$ (1): $m\left(\frac{nh}{2\pi mr}\right)^2 = \frac{Ze^2}{4\pi\varepsilon_0 r} \Rightarrow \frac{n^2h^2}{4\pi^2 m r^2} = \frac{Ze^2}{4\pi\varepsilon_0 r}$ Solve for $r$ (1): $\boxed{r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m Z e^2}} = 0.529\frac{n^2}{Z}\ \text{Å}$

Q8. (4 marks) — 0.5 per correct series naming (2.5), region (1.5):

• $n_1=1$: Lyman
• $n_1=2$: Balmer
• $n_1=3$: Paschen
• $n_1=4$: Brackett
• $n_1=5$: Pfund

Lyman lies in the ultraviolet (UV) region; Balmer lies in the visible region. (Naming all five = 2.5; regions correct = 1.5)

Q9. (4 marks) $\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = 1.097\times10^7\left(\frac{1}{2^2} - \frac{1}{4^2}\right)\quad(1)$ $= 1.097\times10^7\left(\dfrac{1}{4} - \dfrac{1}{16}\right) = 1.097\times10^7 \times \dfrac{3}{16}$ (1) $= 1.097\times10^7 \times 0.1875 = 2.0569\times10^6\ \text{m}^{-1}$ (1) $\lambda = \frac{1}{2.0569\times10^6} = 4.862\times10^{-7}\ \text{m} \approx 486\ \text{nm}\quad(1)$

Q10. (2 marks) — 1 mark each (any two):

• Fails to explain spectra of multi-electron atoms.
• Cannot explain fine structure / splitting of spectral lines (Zeeman, Stark effects).
• Violates Heisenberg's uncertainty principle (fixed orbit + defined momentum).
• Cannot explain chemical bonding / shapes of molecules.

[
  {"claim": "Average atomic mass of Cl = 35.48 u", "code": "avg = (35*75.77 + 37*24.23)/100; result = abs(avg - 35.4846) < 0.01"},
  {"claim": "r3 for hydrogen = 4.761 Angstrom", "code": "r3 = 0.529*(3**2)/1; result = abs(r3 - 4.761) < 0.001"},
  {"claim": "E2 for hydrogen = -3.4 eV", "code": "E2 = -13.6*(1**2)/(2**2); result = abs(E2 - (-3.4)) < 0.001"},
  {"claim": "Balmer 4->2 wavelength approx 486 nm", "code": "R=1.097e7; inv_lam = R*(Rational(1,4)-Rational(1,16)); lam = 1/inv_lam; result = abs(float(lam)*1e9 - 486) < 2"}
]