Physics • Year 12 • Module 8 • Lesson 13
Quantum Mechanics and the Atom
Apply your understanding of de Broglie wavelengths, quantum numbers, and the uncertainty principle to real data, real scenarios, and a source critique.
1. Interpret graph — de Broglie wavelength vs particle speed
The graph below shows the de Broglie wavelength of an electron (λ) plotted against its speed (v). Use the data to answer the questions. 7 marks
Figure 1. De Broglie wavelength (λ) of an electron vs speed (v). Calculated from λ = h/(mev), h = 6.63×10−34 J·s, me = 9.11×10−31 kg. Illustrative data.
1.1 Describe the shape of the curve and explain why the de Broglie wavelength decreases as electron speed increases. 2 marks
1.2 Use the graph to estimate the de Broglie wavelength of an electron travelling at 3.0×106 m/s. Then verify your answer using λ = h/(mev) with h = 6.63×10−34 J·s and me = 9.11×10−31 kg. Show your working. 3 marks
1.3 The Davisson–Germer experiment used electrons accelerated to approximately 54 eV of kinetic energy. Using your knowledge of the relationship between kinetic energy and momentum, predict qualitatively whether the resulting de Broglie wavelength would be larger or smaller than at 2.0×106 m/s. Justify your answer. 2 marks
2. Compare Bohr orbits and quantum orbitals across five criteria
Complete the two-column table below. For each feature, write a concise description that contrasts the two models. 10 marks (1 per cell)
| Feature | Bohr orbit (1913) | Quantum orbital (1926) |
|---|---|---|
| Electron trajectory | ||
| Position of electron | ||
| Basis for quantisation | ||
| Energy levels for hydrogen | ||
| Scope of model |
3. Predict and justify — confining a particle to a tiny box
A physicist confines an electron within a very small region of space of width 1.0×10−10 m (roughly the diameter of a hydrogen atom). Using the Heisenberg uncertainty principle Δx Δp ≥ ℏ/2, with ℏ = 1.055×10−34 J·s: 5 marks
3.1 Calculate the minimum uncertainty in the electron’s momentum (Δp). Show full working. 2 marks
3.2 Predict what happens to the minimum momentum uncertainty if the box size is halved to 5.0×10−11 m. Explain this result in terms of wave behaviour. 2 marks
3.3 Explain why this result means that electrons cannot be “at rest” inside an atom. 1 mark
4. Data table — electrons in the n = 3 shell
The table below lists allowed quantum states for electrons in the n = 3 shell. Some cells are missing. Complete the table and answer the sub-questions. 8 marks
| n | l | Subshell | Allowed ml values | ms values | Max electrons |
|---|---|---|---|---|---|
| 3 | 0 | 3s | 0 | ± ½ | 2 |
| 3 | 1 | ± ½ | |||
| 3 | 2 | ± ½ |
4.1 Complete the three missing subshell names, the ml values, and the maximum electrons columns above. 6 marks
4.2 State the total maximum number of electrons in the n = 3 shell and explain why, using the Pauli exclusion principle. 2 marks
Q1.1 — Shape of the curve (2 marks)
The curve is a hyperbola [1]. As speed increases, λ decreases because λ = h/(mev) — momentum is directly proportional to speed, and wavelength is inversely proportional to momentum. Doubling the speed halves the wavelength [1].
Q1.2 — Estimate and calculate λ at 3.0×106 m/s (3 marks)
Graph estimate: λ ≈ 2.4×10−10 m [1]. Calculation: p = mev = 9.11×10−31 × 3.0×106 = 2.733×10−24 kg·m/s [1]. λ = h/p = 6.63×10−34 / 2.733×10−24 = 2.43×10−10 m ≈ 0.243 nm [1].
Q1.3 — Davisson–Germer electrons (2 marks)
A 54 eV electron has KE = ½mv² = 54×1.6×10−19 = 8.64×10−18 J, giving v = √(2KE/m) = √(2×8.64×10−18/9.11×10−31) ≈ 4.36×106 m/s [1]. This is faster than 2.0×106 m/s, so by λ = h/(mv), the wavelength is smaller than at 2.0×106 m/s — roughly λ ≈ 1.67×10−10 m [1].
Q2 — Compare and contrast table
Electron trajectory: Bohr: precise circular orbit at fixed radius; Quantum: no defined trajectory — the electron exists as a probability cloud. Position of electron: Bohr: exact position on the orbit at any time; Quantum: only probability of finding the electron at any point (|ψ|²). Basis for quantisation: Bohr: quantisation postulated (L = nℏ) without physical justification; Quantum: quantisation emerges naturally from standing wave condition (2πr = nλ). Energy levels for hydrogen: Bohr: En = −13.6/n² eV (correct result); Quantum: same En = −13.6/n² eV derived from Schrödinger equation. Scope: Bohr: works only for hydrogen (one-electron atoms); Quantum: applies to all atoms and multi-electron systems.
Q3.1 — Minimum momentum uncertainty (2 marks)
Δpmin = ℏ/(2Δx) = 1.055×10−34 / (2×1.0×10−10) [1] = 5.28×10−25 kg·m/s [1].
Q3.2 — Halving the box (2 marks)
Δp doubles to approximately 1.055×10−24 kg·m/s [1]. This makes physical sense: confining a wave into a smaller space requires a shorter wavelength (higher frequency components), which corresponds to greater momentum uncertainty — a fundamental consequence of the wave nature of matter [1].
Q3.3 — Why electrons cannot be at rest (1 mark)
If an electron were at rest inside an atom, its momentum would be exactly zero, giving Δp = 0. By Δx Δp ≥ ℏ/2, Δx would then be infinite — the electron would have completely undefined position. Since an atom has finite size, the electron must always have non-zero momentum uncertainty, meaning it is never at rest [1].
Q4.1 — Completed table (6 marks)
n=3, l=1 (3p): ml values = −1, 0, +1 (3 values); max electrons = 6. n=3, l=2 (3d): ml values = −2, −1, 0, +1, +2 (5 values); max electrons = 10. Award 1 mark per correct subshell name, 1 per correct ml set, 1 per max electrons.
Q4.2 — Total n=3 electrons + Pauli (2 marks)
Total = 2 + 6 + 10 = 18 electrons [1]. By the Pauli exclusion principle, no two electrons can share the same four quantum numbers. Each unique (n, l, ml, ms) combination allows exactly one electron, and counting all valid combinations for n = 3 gives 18 states [1].