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De Broglie Wavelength Calculator – Formula, Equation & Electron Wavelength

Quantum Physics Tool

De Broglie Wavelength Calculator

Calculate the matter wavelength of any particle using ? = h/mv. Includes electron-specific tools, particle comparison, and relativistic calculations. Also written as debroglie wavelength calculator.

De Broglie Wavelength 3-Way Solver
h = 6.626 × 10-34 J·s (Planck's constant)
? = h / mv
De Broglie wavelength from mass and velocity
?? At this velocity (> 10% of c) relativistic effects become significant. Use the Relativistic tab for accurate results.

What is the De Broglie Wavelength?

The de Broglie wavelength is the wavelength associated with any moving particle — the matter wavelength that makes all particles exhibit wave-like properties. In 1924, French physicist Louis de Broglie proposed that if light (electromagnetic waves) can behave as particles (photons), then all moving particles must also have wave-like properties.

This revolutionary idea extended wave-particle duality from light to all matter. Every moving particle — electrons, protons, neutrons, atoms, and even baseballs — has an associated wavelength called the de Broglie wavelength or matter wavelength. Also commonly written as debroglie wavelength (without space), the concept describes a quantum mechanical property of all matter.

  • Experimental confirmation: The Davisson-Germer experiment (1927) demonstrated electron diffraction — proving that electrons produce interference patterns, exactly as waves do.
  • Nobel Prize: Louis de Broglie received the 1929 Nobel Prize in Physics for this discovery.
  • Large objects: Macroscopic objects (people, cars, tennis balls) have incredibly tiny de Broglie wavelengths — far too small to detect. This is why we don't experience quantum wave behavior in everyday life.
  • Small fast particles: Electrons and neutrons have wavelengths comparable to atomic scales (~0.1–1 nm), making quantum effects dominant and observable.

Why does this matter? The de Broglie wavelength formula ? = h/mv explains why quantum mechanics applies to atoms and subatomic particles but not to everyday objects. It's the dividing line between the quantum world and the classical world.

De Broglie Wavelength Formula

The de Broglie wavelength formula connects particle momentum to its wavelength through Planck's constant:

? = h / mv
Also written as ? = h ÷ p where p = mv (momentum)

Variable Definitions

  • ? (lambda) = de Broglie wavelength in metres (m)
  • h = Planck's constant = 6.62607015 × 10-34 J·s
  • p = momentum of particle = m × v (in kg·m/s)
  • m = mass of particle in kilograms (kg)
  • v = velocity of particle in metres per second (m/s)

All Rearrangements of the De Broglie Equation

? = h/mv  |  v = h/(m?)  |  p = h/?  |  m = h/(?v)

De Broglie Wavelength Units Derivation

Starting with ? = h/mv and checking units:

? = [J·s] ÷ [kg × m/s]

= [kg·m²/s² × s] ÷ [kg·m/s]

= [kg·m²/s] ÷ [kg·m/s]

= [m] ?

De Broglie wavelength units are always metres (m) — typically expressed in nm, pm, or Å for subatomic particles where the values are more convenient.

How to Calculate De Broglie Wavelength

Follow these five steps to calculate the de Broglie wavelength of any particle using the formula ? = h/mv:

  1. Step 1: Find the mass of your particle in kg
    • Electron: m = 9.109 × 10-31 kg
    • Proton: m = 1.673 × 10-27 kg
    • Convert from atomic mass units: 1 u = 1.661 × 10-27 kg
  2. Step 2: Find the velocity in m/s
    • From km/h: divide by 3.6
    • From mph: multiply by 0.4470
  3. Step 3: Calculate momentum: p = m × v
  4. Step 4: Apply de Broglie formula: ? = h ÷ p = 6.626 × 10-34 ÷ p
  5. Step 5: Convert to appropriate units
    • Multiply by 109 to get nm
    • Multiply by 1012 to get pm
    • Multiply by 1010 to get Å

Example 1: Electron at v = 2 × 106 m/s

  1. Mass: m = 9.109 × 10-31 kg
  2. Velocity: v = 2 × 106 m/s
  3. Momentum: p = m × v = 9.109×10-31 × 2×106 = 1.822 × 10-24 kg·m/s
  4. Apply ? = h/mv: ? = 6.626×10-34 ÷ 1.822×10-24
  5. ? = 3.637 × 10-10 m = 0.364 nm = 3.637 Å

Answer: ? = 3.637 × 10-10 m = 0.364 nm (X-ray range)

Example 2: Proton at v = 3 × 105 m/s

  1. Mass: m = 1.673 × 10-27 kg
  2. Velocity: v = 3 × 105 m/s
  3. Momentum: p = 1.673×10-27 × 3×105 = 5.019 × 10-22 kg·m/s
  4. ? = h/p: ? = 6.626×10-34 ÷ 5.019×10-22
  5. ? = 1.320 × 10-12 m = 1.320 pm

Answer: ? = 1.320 pm (gamma ray range)

Example 3: Electron accelerated through V = 150 V

Using the electron diffraction formula: ? = h / v(2meeV)

  1. h = 6.626 × 10-34 J·s
  2. me = 9.109 × 10-31 kg
  3. e = 1.602 × 10-19 C
  4. Calculate: 2 × me × e × V = 2 × 9.109×10-31 × 1.602×10-19 × 150 = 4.382 × 10-47
  5. v(4.382×10-47) = 6.620 × 10-24
  6. ? = 6.626×10-34 ÷ 6.620×10-24 = 1.001 × 10-10 m
  7. ? = 100.1 pm = 1.001 Å

Answer: ? = 100.1 pm (comparable to crystal lattice spacing — useful for electron diffraction)

Example 4: Tennis Ball (57g) at 50 m/s

  1. Mass: m = 0.057 kg
  2. Velocity: v = 50 m/s
  3. Momentum: p = 0.057 × 50 = 2.85 kg·m/s
  4. ? = h/p: ? = 6.626×10-34 ÷ 2.85
  5. ? = 2.325 × 10-34 m

Answer: ? = 2.325 × 10-34 m — This is 1020 times smaller than a proton. Completely undetectable. This is why macroscopic objects do not exhibit quantum wave behavior.

De Broglie Wavelength of an Electron

The electron is the most important application of the de Broglie wavelength formula. Electrons have such small masses that their de Broglie wavelengths are comparable to atomic scales, making quantum effects dominant.

Electron Properties

Mass (me)9.10938 × 10-31 kg
Charge (e)1.60218 × 10-19 C
Rest energy0.511 MeV

Method 1: Electron Wavelength from Velocity (? = h/mv)

Velocity% of cWavelengthNotes
1 × 105 m/s0.033%7.27 nmLow energy (UV range)
1 × 106 m/s0.334%0.727 nmModerate energy
1 × 107 m/s3.34%72.7 pmHigh energy (X-ray range)
1 × 108 m/s33.4%7.27 pmRelativistic (use Tool 4)

Method 2: Electron Wavelength from Accelerating Voltage

When an electron is accelerated through potential difference V volts, it gains kinetic energy KE = eV. The de Broglie wavelength is:

? = h / v(2meeV)
Voltage (V)WavelengthApplication
1 V1.226 nmLow energy electrons
10 V0.388 nm—
100 V0.123 nm = 123 pmElectron diffraction
1,000 V38.8 pmElectron microscopy
10,000 V12.3 pmHigh-res TEM
100,000 V3.70 pmHRTEM (relativistic)

Comparison to Atomic Scales

  • Hydrogen atom radius (Bohr radius): 53 pm — electron at 100V has similar wavelength
  • Carbon atom radius: 77 pm
  • Crystal lattice spacing: 200–300 pm — ideal for electron diffraction
  • DNA helix width: ~2 nm — visible with electron microscope

Wave-Particle Duality and the De Broglie Wavelength

Wave-particle duality is one of the most fundamental and counterintuitive principles of quantum mechanics. It states that quantum objects exhibit both wave-like and particle-like properties depending on how they are observed.

Historical Development

  • Light as waves: Maxwell (1865) showed light is an electromagnetic wave
  • Light as particles: Einstein (1905) explained the photoelectric effect using photons
  • Matter as waves: de Broglie (1924) extended wave-particle duality to all matter — ? = h/mv
  • Experimental proof: Davisson-Germer (1927) observed electron diffraction from crystal lattices

When is Quantum Behavior Observable?

Quantum wave behavior is only significant when the de Broglie wavelength is comparable to the size of the system being studied:

  • Electrons in atoms: wavelengths ~0.1–1 nm — wave behavior dominates, explains atomic orbitals
  • Neutrons in nuclei: wavelengths ~1 pm — relevant in nuclear physics
  • Macroscopic objects: wavelengths < 10-34 m — completely undetectable, classical physics applies

Real Applications of De Broglie Wavelength

  • Transmission Electron Microscopy (TEM): Uses electron waves with pm-scale wavelengths to image individual atoms — impossible with visible light (~500 nm)
  • Electron Diffraction: Determines crystal structures by analyzing how electron waves diffract from atomic lattices (? ˜ lattice spacing)
  • Neutron Diffraction: Used in materials science to probe magnetic structures and molecular arrangements
  • Scanning Tunneling Microscopy (STM): Exploits quantum mechanical tunneling of electron waves through potential barriers — images individual atoms

De Broglie Wavelength vs Photon Wavelength

Both electrons and photons have wavelengths, but the physics behind them is completely different. Understanding the distinction is essential for correctly applying the de Broglie formula ? = h/mv.

Property De Broglie Wavelength Photon Wavelength
What has it Any moving particle with mass Massless photons (light)
Formula ? = h/mv ? = c/f
Depends on Mass and velocity Frequency only
At rest Undefined (v=0 ? ?=8) Always moving at c
Energy formula KE = p²/2m E = hf = hc/?
Example Electron at 106 m/s ? 0.73 nm Green light ? 550 nm

Photons have no rest mass, so the matter wave formula ? = h/mv cannot apply — photons use ? = c/f instead. For photons, momentum is p = h/? = hf/c (massless particle momentum from special relativity). This is the same formula as the de Broglie wavelength for matter — it's mathematically equivalent but derived from different physics.

Key insight: The formula p = h/? applies to both photons AND matter particles — but the wavelength ? arises from completely different physical mechanisms in each case.

Worked Examples

1. Electron at 106 m/s ? de Broglie wavelength

The formula ? = h/mv gives the matter wavelength for an electron traveling at any velocity below ~0.1c.

  1. me = 9.109 × 10-31 kg
  2. v = 1 × 106 m/s
  3. p = m × v = 9.109×10-31 × 1×106 = 9.109 × 10-25 kg·m/s
  4. ? = 6.626×10-34 ÷ 9.109×10-25 = 7.274 × 10-10 m
  5. ? = 0.727 nm = 7.274 Å = 727 pm

Answer: 7.274 × 10-10 m = 0.727 nm

2. Proton at 105 m/s ? de Broglie wavelength

  1. mp = 1.673 × 10-27 kg
  2. v = 1 × 105 m/s
  3. p = 1.673×10-27 × 1×105 = 1.673 × 10-22 kg·m/s
  4. ? = 6.626×10-34 ÷ 1.673×10-22 = 3.961 × 10-12 m
  5. ? = 3.961 pm

Answer: 3.961 pm — comparable to nuclear scale

3. De Broglie wavelength from accelerating voltage (V = 100 V)

  1. Use formula: ? = h / v(2meeV)
  2. Calculate: 2 × 9.109×10-31 × 1.602×10-19 × 100 = 2.921 × 10-47
  3. v(2.921×10-47) = 5.405 × 10-24
  4. ? = 6.626×10-34 ÷ 5.405×10-24 = 1.226 × 10-10 m
  5. ? = 122.6 pm = 1.226 Å

Answer: ? = 122.6 pm = 1.226 Å (100 V electron)

4. De Broglie wavelength of 70 kg person walking at 1.5 m/s

  1. m = 70 kg, v = 1.5 m/s
  2. p = 70 × 1.5 = 105 kg·m/s
  3. ? = 6.626×10-34 ÷ 105 = 6.31 × 10-36 m

Answer: 6.31 × 10-36 m — about 1020 times smaller than a proton. Completely immeasurable. This is why you don't diffract around doorways.

5. Find velocity from de Broglie wavelength (electron, ? = 0.1 nm)

  1. Rearrange ? = h/mv ? v = h/(m?)
  2. ? = 0.1 nm = 1 × 10-10 m
  3. v = 6.626×10-34 ÷ (9.109×10-31 × 1×10-10)
  4. v = 6.626×10-34 ÷ 9.109×10-41 = 7.274 × 106 m/s

Answer: v = 7.274 × 106 m/s = 2.43% of c

6. De Broglie wavelength in nm for electron at 5 × 106 m/s

  1. p = 9.109×10-31 × 5×106 = 4.555 × 10-24 kg·m/s
  2. ? = 6.626×10-34 ÷ 4.555×10-24 = 1.455 × 10-10 m
  3. Convert to nm: 1.455×10-10 × 109 = 0.1455 nm

Answer: 0.1455 nm = 145.5 pm

7. Momentum of particle with de Broglie wavelength 0.1 nm

  1. Using p = h/?
  2. ? = 0.1 nm = 1 × 10-10 m
  3. p = 6.626×10-34 ÷ 1×10-10
  4. p = 6.626 × 10-24 kg·m/s

Answer: p = 6.626 × 10-24 kg·m/s

8. Using de Broglie equation ? = h/mv step by step

General method for any particle:

  1. Identify: particle type ? look up mass in kg
  2. Convert velocity to m/s if needed
  3. Calculate: p = mass (kg) × velocity (m/s)
  4. Apply: ? = 6.626 × 10-34 ÷ p
  5. Convert ? from m to nm (×109), pm (×1012), or Å (×1010)

9. De Broglie wavelength of neutron at thermal energy (0.025 eV)

  1. KE = 0.025 eV = 0.025 × 1.602×10-19 J = 4.005 × 10-21 J
  2. mn = 1.675 × 10-27 kg
  3. KE = ½mv² ? v = v(2KE/m) = v(2 × 4.005×10-21 ÷ 1.675×10-27)
  4. v = v(4.782×106) = 2186 m/s
  5. ? = 6.626×10-34 ÷ (1.675×10-27 × 2186) = 1.811 × 10-10 m = 181.1 pm = 1.81 Å

Answer: 181 pm = 1.81 Å — thermal neutrons have wavelengths ideal for crystal diffraction!

10. How does de Broglie wavelength change when velocity doubles?

Since ? = h/mv, and h and m are constant, wavelength is inversely proportional to velocity.

If v doubles: ?new = h/(m × 2v) = ½ × (h/mv) = ?/2

Doubling the velocity exactly halves the de Broglie wavelength.

Example: Electron at 106 m/s has ? = 0.727 nm. At 2×106 m/s ? ? = 0.364 nm (exactly half).

Frequently Asked Questions

What is the de Broglie wavelength formula?
The de Broglie wavelength formula is ? = h/mv where ? is the wavelength (metres), h = 6.626×10?³4 J·s (Planck's constant), m is the particle mass (kg), and v is the velocity (m/s). It can also be written as ? = h/p where p = mv is the momentum.
What are the units of de Broglie wavelength?
De Broglie wavelength units are metres (m). However, for subatomic particles the values are extremely small, so they're typically expressed in nanometres (nm = 10?? m), picometres (pm = 10?¹² m), femtometres (fm = 10?¹5 m), or angstroms (Å = 10?¹° m).
How do you calculate the de Broglie wavelength of an electron?
Use ? = h/(m_e × v) where m_e = 9.109×10?³¹ kg. For an electron accelerated through voltage V, use ? = h/v(2m_e×e×V). Example: electron at 106 m/s ? ? = 6.626×10?³4 ÷ (9.109×10?³¹ × 106) = 7.27×10?¹° m = 0.727 nm.
What is the de Broglie wavelength of a proton?
The proton has mass 1.673×10?²7 kg — about 1836 times heavier than an electron. At the same velocity, a proton's de Broglie wavelength is 1836 times shorter. At v = 105 m/s: ? = 6.626×10?³4 ÷ (1.673×10?²7 × 105) = 3.96×10?¹² m = 3.96 pm.
Why do macroscopic objects not show de Broglie wave behavior?
Because their de Broglie wavelengths are unimaginably small. A 1 kg ball moving at 1 m/s has ? = 6.626×10?³4 m — about 10¹? times smaller than a proton. Wave effects only appear when wavelength is comparable to system size. For macroscopic objects, this condition is never met.
What is the relationship between de Broglie wavelength and momentum?
They are inversely proportional: ? = h/p. Higher momentum (more mass or faster speed) gives shorter de Broglie wavelength. Lower momentum gives longer wavelength. Planck's constant h is the proportionality constant linking the two.
How does de Broglie wavelength change with velocity?
De Broglie wavelength is inversely proportional to velocity (? = h/mv). If velocity doubles, wavelength halves. If velocity is reduced to one-tenth, wavelength increases tenfold. At v = 0, wavelength would be infinite — but a particle at rest has no defined matter wavelength.
What is the difference between de Broglie wavelength and photon wavelength?
De Broglie wavelength (? = h/mv) applies to massive particles and depends on mass and velocity. Photon wavelength (? = c/f) applies to massless photons and depends on frequency. Both use p = h/? for momentum, but the physics of how that wavelength arises is fundamentally different.
How do you find de Broglie wavelength from kinetic energy?
From KE = ½mv², find velocity: v = v(2KE/m). Then apply ? = h/mv. Alternatively, use ? = h/v(2mKE) directly. Example: electron with KE = 1 eV = 1.602×10?¹? J ? v = v(2×1.602×10?¹?/9.109×10?³¹) = 5.93×105 m/s ? ? = 1.226 nm.
What is wave-particle duality?
Wave-particle duality is the quantum mechanical principle that all matter and energy exhibits both wave-like and particle-like properties. Electrons behave as particles in detectors but create interference patterns like waves in double-slit experiments. De Broglie's formula ? = h/mv quantifies the wave property of matter.

Related Calculators

Key Formula
? = h / mv
h6.626×10-34 J·s
me9.109×10-31 kg
mp1.673×10-27 kg
mn1.675×10-27 kg
1 u1.661×10-27 kg
Unit Conversions
1 nm10-9 m
1 pm10-12 m
1 fm10-15 m
1 Å10-10 m
Scale Reference
Proton radius~0.85 fm
H atom (Bohr)53 pm
Crystal lattice~200 pm
DNA width~2 nm
Visible light380–700 nm
Quick Tips
  • Larger mass ? shorter wavelength
  • Faster speed ? shorter wavelength
  • ? × 2 when v ÷ 2
  • v > 0.1c ? use relativistic
  • Units always = metres

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