recall and use λ = h / p

Wave‑Particle Duality (Cambridge 9702 – 22.3)

Learning Objective

Recall and apply the de Broglie relation

\$\lambda = \frac{h}{p}\$

where λ is the wavelength associated with a particle, h = 6.626 × 10⁻³⁴ J·s is Planck’s constant and p is the linear momentum of the particle.

Context within the Syllabus

  • 22.1 introduced the photon energy‑frequency relation E = hf and the wave‑speed relation c = fλ.

  • 22.2 covered the photo‑electric effect, establishing the particle nature of light.

  • 22.3 extends the wave‑particle duality to *all* free particles by the de Broglie hypothesis.

  • Later topics (e.g. electron microscopes, neutron scattering) use the de Broglie wavelength as a practical tool – an “application of physics” required by the syllabus.

Key Concepts

Concept box – What does λ represent?

λ is the wavelength of the particle’s *matter wave* (the probability‑amplitude wave described by the Schrödinger equation). It is not a physical size of the particle; it characterises the spatial periodicity of the particle’s quantum‑mechanical wavefunction.

  • De Broglie post‑ulate: every free particle of momentum p has an associated wave with wavelength λ = h/p.

  • Momentum:

    • Non‑relativistic: 

      p = mv

    • Relativistic (v ≳ 0.1 c): 

      p = \gamma mv,\qquad \gamma = \frac{1}{\sqrt{1-v^{2}/c^{2}}}

  • Photons: p = E/c and E = hf ⇒ λ = c/f = h/p, showing the formula works for massless particles.

  • Domain of validity: the simple form λ = h/mv is accurate when v ≲ 0.1 c. For higher speeds the γ‑factor must be included.

  • Assumption: the particle is free (no external potential) when the relation is applied.

  • Diffraction condition (single‑ or double‑slit): 

    d\sin\theta = n\lambda\;(n=0,\pm1,\pm2,\dots)

  • Fringe spacing on a screen (distance L from the slits): 

    \Delta y = \frac{\lambda L}{d}

  • Diffraction is observable only when λ is comparable with the characteristic dimension of the aperture (typically λ ≈ d/10 or larger).

Derivation of the de Broglie Wavelength

  1. Photon relations (already known from 22.1):

    • E = hf

    • p = E/c

    • c = fλ

  2. Combine them:

    \$p = \frac{hf}{c} = \frac{h}{\lambda}\$

  3. Re‑arrange to obtain the de Broglie hypothesis for any particle:

    \$\lambda = \frac{h}{p}\$

  4. Relativistic justification (optional) – using the energy‑momentum relation

    \$E^{2}=p^{2}c^{2}+m^{2}c^{4}\;\;\Longrightarrow\;\;\lambda=\frac{h}{\gamma mv}\$

    where \(\gamma\) accounts for relativistic mass increase.

Worked Examples

1. Photon (visible light)

Find the wavelength of a photon of frequency \(f = 6.0\times10^{14}\,\text{Hz}\) (≈ green light).

  1. λ = c/f = \(\frac{3.00\times10^{8}}{6.0\times10^{14}} = 5.0\times10^{-7}\,\text{m} = 500\,\text{nm}\).

  2. Momentum from de Broglie: \(p = h/λ = \frac{6.626\times10^{-34}}{5.0\times10^{-7}} = 1.33\times10^{-27}\,\text{kg·m·s}^{-1}\).

  3. Check with \(p = E/c\): \(E = hf = 3.98\times10^{-19}\,\text{J}\); \(p = E/c = 1.33\times10^{-27}\,\text{kg·m·s}^{-1}\) – identical.

2. Electron accelerated through 150 V (non‑relativistic)

  1. Kinetic energy gained: \(E_k = eV = (1.602\times10^{-19})(150) = 2.40\times10^{-17}\,\text{J}\).

  2. Momentum: \(p = \sqrt{2meEk}= \sqrt{2(9.11\times10^{-31})(2.40\times10^{-17})}=6.62\times10^{-24}\,\text{kg·m·s}^{-1}\).

  3. De Broglie wavelength: \(\lambda = h/p = 1.00\times10^{-10}\,\text{m}=0.10\,\text{nm}\).

  4. Speed: \(v = p/m_e = 7.3\times10^{6}\,\text{m·s}^{-1}\) (≈ 2.4 % c – non‑relativistic regime).

3. Relativistic electron (5 MeV)

  1. Rest energy: \(E0 = mec^{2}=511\,\text{keV}\). Total energy \(E = E_0 + 5\,\text{MeV}=5.511\,\text{MeV}\).

  2. Momentum: \(p = \frac{\sqrt{E^{2}-E_0^{2}}}{c}= \frac{\sqrt{(5.511)^{2}-(0.511)^{2}}\,\text{MeV}}{c}=9.8\times10^{-22}\,\text{kg·m·s}^{-1}\).

  3. De Broglie wavelength: \(\lambda = h/p = 6.8\times10^{-13}\,\text{m}=0.68\,\text{pm}\).

  4. γ‑factor: \(\gamma = E/E_0 = 5.511/0.511 \approx 10.8\). Using the relativistic form \(p=\gamma mv\) gives the same result.

4. Neutron moving at \(2.0\times10^{5}\,\text{m·s}^{-1}\)

  1. Momentum: \(p = mv = (1.675\times10^{-27})(2.0\times10^{5}) = 3.35\times10^{-22}\,\text{kg·m·s}^{-1}\).

  2. Wavelength: \(\lambda = h/p = 1.98\times10^{-12}\,\text{m}=2.0\,\text{pm}\).

Comparison of Typical de Broglie Wavelengths

ParticleMass (kg)Speed (m s⁻¹)Momentum p (kg·m s⁻¹)λ (m)Diffraction observable?
(λ ≈ d/10)
Electron (150 V)9.11 × 10⁻³¹7.3 × 10⁶6.6 × 10⁻²⁴1.0 × 10⁻¹⁰Yes – requires crystal lattice spacing ≈ 0.1 nm.
Electron (5 MeV, relativistic)9.11 × 10⁻³¹≈ 0.98 c9.8 × 10⁻²²6.8 × 10⁻¹³Yes – observable with high‑resolution electron microscopes.
Proton (1 MeV)1.67 × 10⁻²⁷4.4 × 10⁷7.3 × 10⁻²⁰9.1 × 10⁻¹⁵Yes – needs atomic‑scale slits or crystal planes.
Neutron (thermal, 2 × 10⁵ m s⁻¹)1.68 × 10⁻²⁷2.0 × 10⁵3.35 × 10⁻²²2.0 × 10⁻¹²Yes – neutron interferometers use crystal spacings ≈ 0.2 nm.
Macroscopic ball (0.1 kg, 1 m s⁻¹)1.0 × 10⁻¹11.0 × 10⁻¹6.6 × 10⁻³³No – λ far smaller than any realistic aperture.

Experimental Evidence (Cambridge‑required)

  1. Davisson–Germer electron diffraction (1927) – Electrons accelerated through 54 V produced a diffraction pattern from a nickel crystal. Measured λ ≈ 1.7 × 10⁻¹⁰ m, matching λ = h/p.

  2. Neutron double‑slit interferometer (1974) – Thermal neutrons (λ ≈ 1.8 Å) passed through a silicon crystal double slit, giving clear interference fringes that obeyed \(d\sin\theta=n\lambda\).

  3. C₆₀ fullerene interference (1999) – Massive neutral molecules (≈ 720 u) were diffracted by nanometre‑scale gratings. Observed fringe spacing corresponded to λ ≈ 2.5 pm, confirming the de Broglie relation for complex particles.

  4. Applications (syllabus illustration) – Electron microscopes (λ ≈ 0.005 nm) and neutron scattering techniques rely on matter‑wave diffraction to resolve atomic‑scale structures.

Common Misconceptions

  • “Only light behaves as a wave.” – De Broglie extends wave behaviour to *all* particles; the effect is simply too small to notice for macroscopic objects.

  • “λ is the physical size of the particle.” – λ is the wavelength of the particle’s probability‑amplitude wave, not a literal dimension.

  • “Higher speed gives a longer wavelength.” – Since \(p=mv\) (or \(p=\gamma mv\)), increasing speed raises momentum, which *reduces* λ (λ ∝ 1/p).

  • “Any particle can be diffracted.” – Diffraction is observable only when λ is comparable with the size of the aperture or crystal spacing; for macroscopic objects λ is astronomically small.

Practice Questions

  1. Calculate the de Broglie wavelength of a neutron moving at \(v = 2.0\times10^{5}\,\text{m·s}^{-1}\). (mn = 1.675 × 10⁻²⁷ kg.)

  2. A beam of electrons has kinetic energy 100 eV. They are incident on a double‑slit apparatus with slit separation \(d = 0.5\,\mu\text{m}\) and screen distance \(L = 1.0\,\text{m}\).

    • Determine the electron wavelength.

    • Calculate the first‑order fringe spacing \(\Delta y\).

    • Comment on whether an interference pattern would be visible.

  3. Explain, using the de Broglie relation, why a 0.1 kg ball moving at 1 m s⁻¹ does not show observable diffraction, even if it passes through a slit 10⁻⁶ m wide.

  4. For a photon of wavelength 400 nm, compute its momentum and verify that \(\lambda = h/p\).

Suggested Diagram

Double‑slit arrangement for matter‑wave interference. Labels: slit separation d, distance to screen L, de Broglie wavelength λ, diffraction angle θ (where \(d\sin\theta=n\lambda\)), and fringe spacing \(\Delta y = \lambda L/d\).