De Broglie Wavelength

Equivalent forms

\lambda = \frac{h}{mv}

\lambda = \frac{h}{\sqrt{2mE_k}}

One equation unifies waves and particles for everything that exists.

Symbol	Name	SI unit	Dimension	Range
$λ$	Wavelengthoutput De Broglie wavelength of the particle	m	L	1e-15 – 0.000001
$p$	Momentum Linear momentum of the particle	kg·m/s	M·L·T⁻¹	1e-30 – 1e-20

Unit systems

Symbol	SI	CGS	Imperial
$λ$	m	cm	ft
$p$	kg·m/s	g·cm/s	slug·ft/s

Where it holds

Valid for all matter, from electrons to bowling balls. Use the relativistic momentum

p = \gamma mv

when v approaches c. For non-localized states use wave-packet treatment.

Dimensional analysis

$[\lambda ] = [h]/[p] = (M\cdot L^{2}\cdot T^{-1})/(M\cdot L\cdot T^{-1}) = L \checkmark$

Discovery

Louis de Broglie · 1924

In his PhD thesis, de Broglie proposed wave-particle duality for matter; confirmed by Davisson-Germer electron diffraction in 1927.

Try this

If electrons are waves, why don't baseballs diffract?

Find the de Broglie wavelength of an electron accelerated through 100 V.

Research status: stable

Real-world applications

Electron microscopes — $\lambda \sim 0.005\,\mathrm{nm}$ gives nanometer resolution, far better than light.
Neutron diffraction for crystallography (sensitive to light atoms like H).
Atom interferometers for ultra-precise gravity and inertial sensing.
Scanning tunneling microscopy relies on the wave nature of electrons.

Common misconceptions

$\lambda is$ not the size of the particle — it's the spacing of probability oscillations of its wavefunction.
The wave is not a 'physical' wave in 3D space; it's a probability amplitude $(|\psi |^{2}$ gives probability density).
Matter waves don't violate locality — interference patterns build up statistically over many particles.

Experimental verification

Davisson and Germer (1927) accidentally diffracted electrons off a nickel crystal and observed the predicted angular pattern. G.P. Thomson independently confirmed it the same year. Since then: neutron diffraction (1940s), atom interferometry, and even

C_{60}

buckyball interference (Zeilinger, 1999) confirm matter waves.

Derivation

De Broglie generalized the photon relation

p = h/\lambda

(which Compton had verified for X-rays) to all matter.

Combined with Bohr's quantization condition for atomic orbits, requiring an integer number of wavelengths around an orbit

(n\lambda = 2\pi r)

reproduces angular momentum quantization

L = n\hbar

— beautiful internal consistency.

Limiting cases

p \to 0

⟶

\lambda \to \infty

A particle at rest has infinite wavelength — its position is maximally uncertain.

p \to \infty

⟶

\lambda \to 0

Highly energetic particles behave like point particles; wave nature becomes invisible.

macroscopic m, ordinary v⟶

\lambda

≪ atomA 1 kg ball at

1\,\mathrm{m/s} has \lambda \approx 6.6\times 10^{-34} m

— far below any measurable scale, hence no diffraction.

What if…

What if h were a million times larger?

Macroscopic matter waves would have $\lambda \sim 10^{-29} m$ — still tiny, but quantum effects would creep into chemistry and biology in dramatic ways.

What if you slow an electron down to 1 m/s?

$\lambda = h/(m_e\cdot 1) \approx 0.7\,\mathrm{mm}$ — nearly visible! Ultra-cold neutrons and atoms exploit this.

What if you tried to diffract a proton with the same energy as an electron?

Same kinetic energy → larger momentum (because m_p ≫ $m_e) \to$ smaller $\lambda by$ factor $\sqrt{m_p/m_e} \approx 43$ . Proton beams give shorter-wavelength probes.

1

Electron accelerated through 100 V

Given ·

V:: 100

Find ·

\lambda

Steps

Kinetic energy $E_k = eV = 1.6\times 10^{-19} \times 100 = 1.6\times 10^{-17} J$
Momentum $p = \sqrt{2m_e\cdot E_k} = \sqrt{2 \times 9.11\times 10^{-31} \times 1.6\times 10^{-17}} \approx 5.4\times 10^{-24} kg\cdot m/s$
$\lambda = h/p = 6.626\times 10^{-34} / 5.4\times 10^{-24} \approx 1.23\times 10^{-10} m$
Comparable to atomic spacing — perfect for crystal diffraction.

Result ·

\lambda = h/\sqrt{2m_e\cdot eV} \approx 1.23\times 10^{-10} m = 0.123\,\mathrm{nm}

2

1 kg baseball at 40 m/s

Given ·

m:: 1
v:: 40

Find ·

\lambda

Steps

$p = mv = 40 kg\cdot m/s$
$\lambda = 6.626\times 10^{-34} / 40 \approx 1.66\times 10^{-35} m$
20 orders of magnitude below the proton — utterly unmeasurable. That is why we don't see baseballs diffract.

Result ·

\lambda = h/(mv) = 6.626\times 10^{-34} / 40 \approx 1.66\times 10^{-35} m

Photoelectric Effect→Heisenberg Uncertainty Principle→