Level 3 — ProductionAtomic Structure (Classical)

Atomic Structure (Classical)

⏱ 45 minutes60 marksprintable — key stays hidden on paper

Level: 3 — Production (from-scratch derivations, explain-out-loud, code-from-memory) Time limit: 45 minutes Total marks: 60

Constants (use as needed): $e = 1.602\times10^{-19}\,\text{C}$ , $m_e = 9.109\times10^{-31}\,\text{kg}$ , $\varepsilon_0 = 8.854\times10^{-12}\,\text{C}^2\text{N}^{-1}\text{m}^{-2}$ , $h = 6.626\times10^{-34}\,\text{J·s}$ , $R_H = 1.097\times10^{7}\,\text{m}^{-1}$ , $1\,\text{eV}=1.602\times10^{-19}\,\text{J}$ .

Q1. [12 marks] — Full Bohr derivation from scratch. Starting only from (a) the Coulomb attraction providing centripetal force and (b) the angular-momentum quantization postulate $mvr = \dfrac{nh}{2\pi}$ , derive from first principles:

(i) the expression for the radius $r_n$ of the $n$ -th orbit of a hydrogen-like ion (charge $+Ze$ ). [6]
(ii) the total energy $E_n$ of the electron in that orbit. [6]

Show every algebraic step and clearly state which physics each equation encodes.

Q2. [10 marks] — Rydberg formula derivation + spectral prediction.

(i) Using $E_n = -13.6\,\dfrac{Z^2}{n^2}$ eV, derive the Rydberg formula $\dfrac{1}{\lambda}=R Z^2\!\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)$ , expressing $R$ in terms of fundamental constants (symbolically). [5]
(ii) Compute the wavelength (in nm) of the $n_2=3 \to n_1=2$ line of hydrogen ( $H_\alpha$ , Balmer). State which series it belongs to. [5]

Q3. [10 marks] — Explain out loud. In your own words, structured as a clear exposition:

(i) Describe Rutherford's gold-foil experiment: setup, the key observations, and the reasoning that led from each observation to a specific structural conclusion. [6]
(ii) State two limitations of the Bohr model and, for each, name the phenomenon it fails to explain. [4]

Q4. [10 marks] — Code from memory. Write a short Python function atomic_mass(isotopes) where isotopes is a list of (mass, fractional_abundance) tuples, that returns the average atomic mass. Then:

(i) Give the function. [4]
(ii) Hand-trace it to compute the atomic mass of chlorine given $^{35}\text{Cl}$ (mass 34.969, abundance 75.77%) and $^{37}\text{Cl}$ (mass 36.966, abundance 24.23%). Show the arithmetic. [6]

Q5. [10 marks] — Classification + reasoning. For each pair, state whether they are isotopes, isobars, or isotones, and justify using $Z$ , $A$ , and neutron number $N$ :

(i) $^{40}_{18}\text{Ar}$ and $^{40}_{20}\text{Ca}$ [3]
(ii) $^{14}_{6}\text{C}$ and $^{14}_{7}\text{N}$ — additionally give $N$ for each and confirm your label. [4]
(iii) $^{39}_{19}\text{K}$ and $^{40}_{19}\text{K}$ [3]

Q6. [8 marks] — Discovery chain, explain out loud. Trace the experimental logic that established the sub-atomic particles: for electron (Thomson), proton (Goldstein), and neutron (Chadwick), state in one line each the experiment used and the crucial deduction. Then explain why Dalton's postulate "atoms are indivisible" is a limitation in light of these discoveries. [8]

Answer keyMark scheme & solutions

Q1 (12 marks)

(i) Radius — [6]

Coulomb force = centripetal force (physics: electrostatic attraction supplies the circular-motion force): $\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r^2}=\frac{mv^2}{r} \quad\Rightarrow\quad mv^2=\frac{Ze^2}{4\pi\varepsilon_0 r}\quad(1)$ [2]
Quantization postulate (physics: only orbits with quantized angular momentum are allowed): $mvr=\frac{nh}{2\pi}\Rightarrow v=\frac{nh}{2\pi m r}\quad(2)$ [1]
Substitute (2) into (1): $m\left(\frac{nh}{2\pi mr}\right)^2=\frac{Ze^2}{4\pi\varepsilon_0 r}\Rightarrow \frac{n^2h^2}{4\pi^2 m r^2}=\frac{Ze^2}{4\pi\varepsilon_0 r}$ [2]
Solve for $r$ : $\boxed{r_n=\frac{\varepsilon_0 n^2 h^2}{\pi m Z e^2}=0.529\,\frac{n^2}{Z}\ \text{Å}}$ [1]

(ii) Energy — [6]

Kinetic energy from (1): $KE=\tfrac12 mv^2=\dfrac{Ze^2}{8\pi\varepsilon_0 r}$ [1]
Potential energy (electrostatic): $PE=-\dfrac{Ze^2}{4\pi\varepsilon_0 r}$ (negative: bound/attractive) [1]
Total $E=KE+PE=\dfrac{Ze^2}{8\pi\varepsilon_0 r}-\dfrac{Ze^2}{4\pi\varepsilon_0 r}=-\dfrac{Ze^2}{8\pi\varepsilon_0 r}$ [2]
Insert $r_n$ : $\boxed{E_n=-\frac{m Z^2 e^4}{8\varepsilon_0^2 n^2 h^2}=-13.6\,\frac{Z^2}{n^2}\ \text{eV}}$ [2]

Q2 (10 marks)

(i) — [5] Photon energy for transition $n_2\to n_1$ : $\Delta E=E_{n_2}-E_{n_1}=13.6\,Z^2\!\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\text{eV}$ [2] Using $\Delta E=hc/\lambda$ : [1] $\frac{1}{\lambda}=\frac{\Delta E}{hc}=\frac{m e^4}{8\varepsilon_0^2 h^3 c}Z^2\!\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ [1] So $R=\dfrac{m e^4}{8\varepsilon_0^2 h^3 c}$ . [1]

(ii) — [5] $\frac{1}{\lambda}=R_H\left(\frac1{2^2}-\frac1{3^2}\right)=1.097\times10^7\left(0.25-0.1111\right)=1.097\times10^7\times0.13889$ [2] $\frac1\lambda=1.5236\times10^6\,\text{m}^{-1}\Rightarrow \lambda=6.563\times10^{-7}\text{m}=656.3\ \text{nm}$ [2] Belongs to the Balmer series ( $n_1=2$ ), visible red line $H_\alpha$ . [1]

Q3 (10 marks)

(i) — [6]

Setup: thin gold foil bombarded by $\alpha$ -particles; a fluorescent (ZnS) screen detects scattering angles. [1]
Observation 1: most $\alpha$ pass straight through → atom is mostly empty space. [2]
Observation 2: some deflected at large angles → concentrated positive charge. [1]
Observation 3: very few (~1 in 20000) bounce back (>90°) → nucleus is tiny, dense, and carries essentially all the mass and positive charge. [2]
Conclusion: nuclear model — small dense positive nucleus with electrons around it. [1]

(ii) — [4] (any two, 2 each: 1 limitation + 1 phenomenon)

Fails for multi-electron atoms (only works for one-electron systems, e.g. H, He⁺). [2]
Cannot explain fine structure / spectral line splitting (nor Zeeman/Stark effects); violates Heisenberg uncertainty by assuming fixed orbits. [2]

Q4 (10 marks)

(i) — [4]

def atomic_mass(isotopes):
    return sum(mass * abundance for mass, abundance in isotopes)

(1 mark correct signature/loop, 1 product term, 1 sum, 1 return.) If abundances are given in %, divide by 100 first.

(ii) — [6] Convert %: $0.7577$ and $0.2423$ . [1] $34.969\times0.7577=26.4959$ [2] $36.966\times0.2423=8.9569$ [2] $\text{Atomic mass}=26.4959+8.9569=35.453\ \text{amu}$ [1]

Q5 (10 marks)

(i) [3] Same $A=40$ , different $Z$ (18 vs 20) → isobars. [3]

(ii) [4] Same $A=14$ , different $Z$ (6 vs 7) → isobars. $N(^{14}C)=14-6=8$ ; $N(^{14}N)=14-7=7$ . Different $N$ confirms not isotones; same $A$ confirms isobars. [4]

(iii) [3] Same $Z=19$ , different $A$ (39 vs 40) → isotopes. [3]

Q6 (8 marks)

Electron (Thomson): cathode-ray tube; rays deflected by electric/magnetic fields → measured charge-to-mass ratio $e/m$ ; particles negative, common to all matter. [2]
Proton (Goldstein): canal/anode rays in perforated-cathode tube; positive rays, lightest from hydrogen → the proton. [2]
Neutron (Chadwick): bombardment of beryllium with $\alpha$ -particles gave a neutral, penetrating radiation of mass ≈ proton → the neutron. [2]
Limitation of Dalton: he postulated atoms indivisible/uncreatable; discovery of electrons, protons, neutrons shows atoms have internal sub-structure, so the "indivisible" postulate is false. [2]

[
  {"claim":"H-alpha (3->2) wavelength ~656.3 nm", "code":"R=1.097e7; inv=R*(1/2**2-1/3**2); lam=1/inv*1e9; result = abs(lam-656.3)<1.0"},
  {"claim":"Chlorine average atomic mass = 35.453 amu", "code":"m=34.969*0.7577+36.966*0.2423; result = abs(m-35.453)<0.01"},
  {"claim":"Bohr radius coefficient 0.529 Å from constants", "code":"e0=8.854e-12; h=6.626e-34; me=9.109e-31; e=1.602e-19; r=e0*h**2/(pi*me*e**2); result = abs(r*1e10-0.529)<0.01"},
  {"claim":"Balmer fraction 1/4 - 1/9 = 0.138889", "code":"val=Rational(1,4)-Rational(1,9); result = val==Rational(5,36)"}
]