Playground
Adjust momentum and watch the matter wave shrink: higher momentum means shorter wavelength, probing smaller scales.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Wavelengthoutput Quantum mechanical wavelength of the particle | m | L | 1e-20 – 0.000001 | |
| Planck constant Planck's constant | J·s | M·L²·T⁻¹ | 6.62607015e-34 – 6.62607015e-34 | |
| Momentum Relativistic momentum of the particle | kg·m/s | M·L·T⁻¹ | 1e-25 – 1e-15 |
Deep dive
Derivation
De Broglie postulated that matter exhibits wave-particle duality analogous to light. For a photon, E = hf and p = E/c = h/λ, giving λ = h/p. De Broglie extended this to massive particles by using the relativistic momentum p = γmv, so λ = h/(γmv). The formula λ = h/p holds universally regardless of whether p is classical or relativistic.
Experimental verification
Confirmed by Davisson and Germer (1927) who observed electron diffraction from a nickel crystal. Modern electron microscopes routinely use 200 keV electrons with λ ≈ 2.5 pm. Neutron diffraction in crystallography uses thermal neutrons with λ ≈ 1 Å.
Common misconceptions
- The formula λ = h/p is exact and not an approximation — it works at any speed as long as you use the correct relativistic momentum.
- The de Broglie wavelength is not a physical size of the particle; it describes the spatial periodicity of the quantum probability amplitude.
- Doubling kinetic energy does NOT halve the wavelength — the relationship between KE and wavelength is nonlinear, especially relativistically.
Real-world applications
- Electron microscopy resolution limits
- Neutron diffraction for crystal structure determination
- Particle accelerator design (resolving power)
- Quantum tunneling probability estimation
Worked examples
Wavelength of a 1 GeV/c proton
Given:
- p:
- 1 GeV/c = 5.34 × 10⁻¹⁹ kg·m/s
- h:
- 6.62607015e-34
Find: λ
Solution
λ = h/p = 6.626×10⁻³⁴ / 5.34×10⁻¹⁹ ≈ 1.24 × 10⁻¹⁵ m = 1.24 fm
Wavelength of a 100 keV electron
Given:
- KE:
- 100 keV
- m_e:
- 9.1093837015 × 10⁻³¹ kg
- h:
- 6.62607015e-34
Find: λ
Solution
λ = h/p ≈ 3.70 × 10⁻¹² m = 3.70 pm
Scenarios
What if…
- scenario:
- What if momentum is doubled?
- answer:
- Wavelength is halved exactly (λ = h/p is inversely proportional). The resolving power of the particle doubles.
- scenario:
- What if we use a heavier particle (e.g., muon instead of electron) at the same kinetic energy?
- answer:
- The heavier particle has more momentum at the same KE, so its wavelength is shorter. Muons at the same KE have ~14× shorter wavelength than electrons.
- scenario:
- What if the particle approaches the speed of light?
- answer:
- As v → c, momentum p = γmv → ∞, so λ → 0. The particle's wave nature becomes negligible and it behaves classically at extremely high energies.
Limiting cases
- condition:
- p → ∞
- result:
- λ → 0
- explanation:
- Extremely high momentum particles have vanishingly small wavelengths, behaving classically.
- condition:
- p → 0
- result:
- λ → ∞
- explanation:
- As momentum approaches zero, the wavelength diverges — the particle becomes completely delocalized.
- condition:
- p = mc (Compton scale)
- result:
- λ = h/(mc)
- explanation:
- At the Compton momentum, the wavelength equals the Compton wavelength — the boundary between particle and field descriptions.
Context
Louis de Broglie · 1924
De Broglie proposed that matter has wave-like properties in his PhD thesis, unifying particle and wave descriptions.
Hook
How small can we 'see' with a particle accelerator?
A proton has momentum 1 GeV/c. What is its de Broglie wavelength?
Dimensions:
- lhs:
- λ → [L]
- rhs:
- h/p → [M·L²·T⁻¹] / [M·L·T⁻¹] = [L]
- check:
- Both sides are [L] = meters. ✓
Validity: Valid for all free particles with well-defined momentum. Applies at any speed (non-relativistic and relativistic) since p itself can be the relativistic momentum γmv. Breaks down when the particle cannot be treated as free (strong external potentials).