de Broglie Wavelength (Relativistic)

Equivalent forms

\lambda = \frac{hc}{\sqrt{E^2 - (mc^2)^2}}

One Planck constant links momentum and wavelength across every particle ever discovered.

Symbol	Name	SI unit	Dimension	Range
$λ$	Wavelengthoutput Quantum mechanical wavelength of the particle	m	L	1e-20 – 0.000001
$h$	Planck constant Planck's constant	J·s	M·L²·T⁻¹	6.62607015e-34 – 6.62607015e-34
$p$	Momentum Relativistic momentum of the particle	kg·m/s	M·L·T⁻¹	1e-25 – 1e-15

Unit systems

Symbol	SI	Natural	CGS
$λ$	m	fm	cm
$h$	J·s	MeV·fm/c	erg·s
$p$	kg·m/s	MeV/c	g·cm/s

Where it holds

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

\gamma mv

. Breaks down when the particle cannot be treated as free (strong external potentials).

Dimensional analysis

\lambda \to [L]

h/p \to [M\cdot L^{2}\cdot T^{-1}] / [M\cdot L\cdot T^{-1}] = [L]

Both sides are

[L] =

meters.

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Discovery

Louis de Broglie · 1924

De Broglie proposed that matter has wave-like properties in his PhD thesis, unifying particle and wave descriptions.

Try this

How small can we 'see' with a particle accelerator?

A proton has momentum 1 GeV/c. What is its de Broglie wavelength?

Research status: stable

Real-world applications

Electron microscopy resolution limits
Neutron diffraction for crystal structure determination
Particle accelerator design (resolving power)
Quantum tunneling probability estimation

Common misconceptions

The formula $\lambda = h/p$ 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.

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

\lambda \approx 2.5\,\mathrm{pm}

. Neutron diffraction in crystallography uses thermal neutrons with

\lambda \approx 1 \text{Å}

.

Derivation

De Broglie postulated that matter exhibits wave-particle duality analogous to light.

For a photon,

E = hf

and

p = E/c = h/\lambda

, giving

\lambda = h/p

.

De Broglie extended this to massive particles by using the relativistic momentum

p = \gamma mv

,

so \lambda = h/(\gamma mv)

.

The formula

\lambda = h/p

holds universally regardless of whether p is classical or relativistic.

Limiting cases

p \to \infty

⟶

\lambda \to 0

Extremely high momentum particles have vanishingly small wavelengths, behaving classically.

p \to 0

⟶

\lambda \to \infty

As momentum approaches zero, the wavelength diverges — the particle becomes completely delocalized.

p = mc

(Compton scale)⟶

\lambda = h/(mc)

At the Compton momentum, the wavelength equals the Compton wavelength — the boundary between particle and field descriptions.

What if…

What if momentum is doubled?

Wavelength is halved exactly $(\lambda = h/p$ is inversely proportional). The resolving power of the particle doubles.

What if we use a heavier particle (e.g., muon instead of electron) at the same kinetic energy?

The heavier particle has more momentum at the same KE, so its wavelength is shorter. Muons at the same KE have $\sim 14\times$ shorter wavelength than electrons.

What if the particle approaches the speed of light?

As $v \to c$ , momentum $p = \gamma mv \to \infty$ , $so \lambda \to 0$ . The particle's wave nature becomes negligible and it behaves classically at extremely high energies.

1

Wavelength of a 1 GeV/c proton

Given ·

p:: $1\,\mathrm{GeV/c} = 5.34 \times 10^{-19} kg\cdot m/s$
h:: 6.62607015e-34

Find ·

\lambda

Steps

Step 1: Convert momentum to SI: $p = 1\,\mathrm{GeV/c} = (1\times 10^{9} \times 1.602176634\times 10^{-19}) / 299792458 = 5.34 \times 10^{-19} kg\cdot m/s$ .
Step 2: Apply formula: $\lambda = h/p = 6.62607015\times 10^{-34} / 5.34\times 10^{-19}$ .
Step 3: Calculate: $\lambda \approx 1.24 \times 10^{-15} m = 1.24\,\mathrm{fm}$ .
Step 4: This is roughly the size of a proton, so 1 GeV/c probes can resolve nuclear structure.

Result ·

\lambda = h/p = 6.626\times 10^{-34} / 5.34\times 10^{-19} \approx 1.24 \times 10^{-15} m = 1.24\,\mathrm{fm}

2

Wavelength of a 100 keV electron

Given ·

KE:: 100 keV
m e:: $9.1093837015 \times 10^{-31} kg$
h:: 6.62607015e-34

Find ·

\lambda

Steps

Step 1: Total energy $E = KE + m_e c^{2} = 100\,\mathrm{keV} + 511\,\mathrm{keV} = 611\,\mathrm{keV}$ .
Step 2: Find momentum: $pc = \surd (E^{2} - (m_e c^{2})^{2}) = \sqrt{611^{2} - 511^{2}} = \sqrt{373321 - 261121} = \sqrt{112200} \approx 335\,\mathrm{keV}$ , so $p = 335\,\mathrm{keV/c}$ .
Step 3: Convert: $p = 335\times 10^{3} \times 1.602176634\times 10^{-19} / 299792458 \approx 1.79 \times 10^{-22} kg\cdot m/s$ .
Step 4: $\lambda = h/p = 6.626\times 10^{-34} / 1.79\times 10^{-22} \approx 3.70 \times 10^{-12} m = 3.70\,\mathrm{pm}$ .

Result ·

\lambda = h/p \approx 3.70 \times 10^{-12} m = 3.70\,\mathrm{pm}

Relativistic Energy–Momentum Relation→Mass–Energy Equivalence→Klein–Nishina Cross Section→