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1.2 Uncertainty Principle, de Broglie Relation and Wave–Particle Duality

★★★ 2021–2025: 7 Mains parts linked: IFoS 2 parts / 13 m and Geo-Scientist 2 / 10 m directly on this topic, plus Civil Services 2024 Q2(a), 2023 Q1(a) and Geo-Scientist 2025 Q1(l), where “state the uncertainty principle” or the zero-point energy is part of a particle-in-a-box question. Short and numerical: one formula, one line of working, the meaning.

Syllabus lines. Civil Services: “Heisenberg’s uncertainty principle, Schrödinger wave equation (time independent)”. IFoS: “Quantum theory, Heisenberg’s uncertainty principle”. Geo-Scientist: “Uncertainty relation … Comparison with free particle eigenfunctions and eigenvalues.”

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CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes

1.2.1 Wave–particle duality ★★

  • Light behaves as particles (photoelectric effect, Einstein 1905): electrons leave a metal only if ν exceeds a threshold ν₀, at once, however dim the light; their maximum kinetic energy rises linearly with ν and not with intensity. Each photon carries hν and gives it to one electron.SN
(1.2.1)
  • Na (Φ = 2.28 eV): ν₀ = Φ/h = 5.51 × 10¹⁴ Hz (λ₀ = 544 nm); with 400 nm light (hν = 3.10 eV) KEmax = 0.82 eV. Compton effect: X-rays scattered by electrons come out with a longer wavelength, Δλ = (h/mec)(1 − cos θ) = 2.43 pm × (1 − cos θ): photons carry momentum h/λ.SN
  • Matter behaves as waves: Davisson and Germer (1927) diffracted electrons from a nickel crystal; G.P. Thomson passed them through thin foils. The diffraction angles fit λ = h/p. Electron microscopes use this. Light and matter both show wave and particle behaviour: wave–particle duality.SN
0481216-4-224ν / 1014  HzKEmax  / eVν0 Cs (Φ = 2.14 eV)−Φ = −2.14 eVν0 Zn (Φ = 4.30 eV)−Φ = −4.30 eVparallel lines: slope = h for every metalbelow ν0: no electrons, however bright
Fig. 1.2.1 Photoelectric effect: KEmax against ν is a straight line of slope h; each metal has its own threshold ν₀ = Φ/h.
Draw it in 60 s: two parallel lines rising to the right, starting on the ν-axis at ν0(Cs) and ν0(Zn); extend them dashed back to cut the y-axis at −Φ; write “slope = h”.

1.2.2 de Broglie relation ★★★

(1.2.2)
  • de Broglie (1924): any particle of momentum p has a wavelength h/p. For an electron through 100 V: λ = 0.123 nm, about an interatomic spacing, so crystals diffract electrons. An electron of λ = 0.10 nm moves at 7.27 × 10⁶ m s⁻¹ (KE 150 eV).SN
  • A 0.15 kg ball at 40 m s⁻¹: λ = 1.1 × 10⁻³⁴ m, far smaller than any slit or nucleus: wave behaviour of everyday objects is never seen.SN
  • Bohr orbits as standing waves: a stable orbit must hold a whole number of wavelengths, nλ = 2πr. With λ = h/mv this gives mvr = nh/2π: Bohr’s quantised angular momentum. For H, n = 1: λ = 2πa₀ = 0.332 nm, v = 2.19 × 10⁶ m s⁻¹.SN
nucleus2πr = 4λ ✓wave joins itself: standing wavenucleus2πr = 4.5λ ✗next lap out of step: cancels
Fig. 1.2.2 Only a whole number of wavelengths fits round an orbit; a fractional number cancels itself on the next lap.
Draw it in 60 s: two circles; round the first draw a wave with 4 crests that joins; round the second 4½ crests with a gap at the start; write nλ = 2πr ⇒ mvr = nh/2π.

Solved PYQ IFoS 2024 Q1(a) 8m: show how the Schrödinger equation for a free particle gives the de Broglie relation.

  1. Free particle: V = 0, so , i.e. with .
  2. Solutions: , waves running to +x and −x; any E ≥ 0 is allowed (no boundary, no quantisation).SN
  3. Momentum: : an eigenfunction with p = ħk; and E = p²/2m.
  4. eikx repeats when kx grows by 2π, so kλ = 2π. Then p = ħk = (h/2π)(2π/λ): λ = h/p, the de Broglie relation.SN

1.2.3 Heisenberg uncertainty principle ★★★

(1.2.3)
  • Statement: it is impossible to know both the position and the momentum of a particle along the same direction exactly at the same time; the product of their uncertainties is at least ħ/2. No limit applies to x and py. It is a property of matter, not of poor instruments CSE 2024 Q2(a) 10m CSE 2023 Q1(a) 10m.SN
  • Why: a particle of definite p is a pure wave eikx spread over all space (Δx = ∞). To localise it we add waves of many wavelengths into a packet, and every added wavelength is another momentum. Narrow packet ⇔ wide spread of p. Formally, x̂ and p̂x do not commute (see 1.1.6).SN
  • Heisenberg’s γ-ray microscope: to see an electron sharply we need short-wavelength light, but such a photon has a large momentum h/λ and kicks the electron (Compton recoil) by an unknown amount.
pure wave eikx Δx = ∞Δp = 0ħk0 broad packetΔx largeΔp smallħk0 narrow packetΔx smallΔp largeħk0 ψ(x): positionspread of momentum
Fig. 1.2.3 Localising a particle costs momentum certainty.
Draw it in 60 s: left: a long wave, a broad packet, a narrow packet; right of each: a single spike, a narrow hump, a wide hump; label Δx and Δp.

Solved PYQ Geo 2021 Q1(l) 5m: Δx of an electron in an atom = 0.4 Å. Find Δv.

  1. Δx·mΔv ≥ h/4π ⇒ = 1.45 × 10⁶ m s⁻¹.
  2. This is 0.66 of the electron’s own speed in the first Bohr orbit (2.19 × 10⁶ m s⁻¹): an electron held to atomic size has no defined path, so we speak of orbitals, not orbits. For a 1 g body with the same Δx, Δv = 1.3 × 10⁻²¹ m s⁻¹: negligible.SN
  3. Some books write ΔxΔp ≈ h (Δv = 1.8 × 10⁷) or ≥ ħ (2.9 × 10⁶ m s⁻¹): state the form you use.
  • No electron in the nucleus: Δx = 10⁻¹⁴ m gives Δp ≥ 5.3 × 10⁻²¹ kg m s⁻¹, Δv ≥ 5.8 × 10⁹ m s⁻¹ > c (non-relativistic): an electron cannot be confined there.SN
  • Zero-point energy: a particle in a box of length L has Δx ≤ L, so Δp ≥ ħ/2L > 0 and its kinetic energy cannot be zero. The exact lowest level is ; E = 0 would need ψ = 0 everywhere, which is not acceptable Geo 2025 Q1(l) 5m. Worked in 1.3 Particle in a box.SN
  • ★ Beyond the syllabus: for any two operators ΔAΔB ≥ ½|⟨[Â, B̂]⟩|; ΔA = (⟨A²⟩ − ⟨A⟩²)1/2. A Gaussian packet is the only shape that reaches the equality ΔxΔp = ħ/2.

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