De Broglie Wavelength of Matter Waves

Wave-particle duality

Four light particles (photons) of different frequency.

After Albert Einstein explained the photoelectric effect by assuming that light consists of particles (photons), Louis de-Broglie (pronounced: de-broi) came up with a similar idea. He thought: If it was possible to assign a particle character to the wave-like light, then it must also be possible to assign a wave character to a particle-like matter.

And this assumption has actually been experimentally proven (e.g. in the Davisson-Germer experiment) and is nowadays used, for example, in electron diffraction. Just like light, electrons are also able to produce an interference pattern in the double-slit experiment.

This finding that light and matter can behave in a wave-like and particle-like manner depending on the situation is called wave-particle duality.

Matter wavelength of particles

A photon of wavelength $ \lambda $ has the following momentum:

$$ \begin{align} p ~=~ \frac{h}{\lambda} \end{align} $$

Here $h$ is the Planck's constant, a natural constant with the value:

$$ \begin{align} h = 6.6 \cdot 10^{-34} \, \mathrm{Js} \end{align} $$

Analogously, a wavelength can be assigned to a particle with mass, which has a momentum $p$. For this purpose Eq. 1 is rearranged with respect to the wavelength:

$$ \begin{align} \lambda ~=~ \frac{h}{p} \end{align} $$

A particle considered as a wave, with the associated de Broglie wavelength.

The momentum $p$ of a classical particle is defined as the product of the mass $m$ of the particle and its velocity $v$: $ p = m \, v $. We insert the momentum in 2 and interpret the wavelength $\lambda$ as the de Broglie wavelength of a particle:

Formula: De Broglie wavelength

$$ \begin{align} \lambda ~=~ \frac{h}{m \, v} \end{align} $$

From the relation 3 you can read that fast and heavy particles (large momentum) have a shorter de Broglie wavelength than slow and light particles (small momentum).

If the velocity $v$ of the particle and thus the momentum of the particle approaches zero (a particle without momentum is a particle at rest), then the de Broglie wavelength becomes very large. Physically, it again means that the particle, for example in a metal, is "smeared" all over the metal. Interference phenomena between several particle waves in the metal become non-negligible. The particle shows its wave character and behaves more quantum mechanically.

For a large velocity and large mass of the particle, the de Broglie wavelength is negligible. The particle behaves like a real classical particle and can be described with classical mechanics. Quantum effects, such as particle interference, do not play a role here.

What is the de Broglie wavelength good for?

With the help of the de Broglie wavelength you can estimate (e.g. in an experiment) whether an object will behave more wave-like or particle-like.

Example: An electron in a metal

In a metal, there exist free electrons (collectively called electron gas) that have a thermal velocity of $ v \approx 10^6 \, \frac{\mathrm m}{\mathrm s}$ at room temperature. By thermal is meant that the electrons in the metal are moving in a random manner. After a collision with a metal atom, it moves in one direction. After the other collision it moves in the other direction. With a rest mass of $ m_{\text e} = 9.1 \cdot 10^{-31} \, \mathrm{kg} $ the de Broglie wavelength of a free electron becomes:

$$ \begin{align} \lambda &~=~ \frac{ 6.6\cdot 10^{-34}\,\mathrm{Js} }{9.1 \cdot 10^{-31} \, \mathrm{kg} \cdot 10^6 \, \frac{\mathrm m}{\mathrm s} } \\\\
&~=~ 7.2 \cdot 10^{-10}\,\mathrm{m} \end{align} $$

This is $ 0.72 \, \mathrm{nm} $. For comparison, the diameter of the Nucleic acid double helix is in the order of $ 2 \, \mathrm{nm} $.

Classical gas vs. quantum gas

A particle of mass $m$ moving with velocity $v$ has kinetic energy:

$$ \begin{align} W_{\text{kin}} ~=~ \frac{1}{2} \, m \, v^2 \end{align} $$

Consider a gas in a so-called thermodynamic equilibrium, i.e., a collection of unbound particles flying around at a constant temperature $T$. According to thermodynamics, the following mean kinetic energy can be assigned to each gas particle:

$$ \begin{align} W_{\text{kin}} ~=~ \frac{3}{2} \, k_{\text B} \, T \end{align} $$

Here $ k_{\text B} = 1.38 \cdot 10^{-23} \, \frac{\mathrm J}{\mathrm K}$ is the Boltzmann constant and $T$ is the temperature of the gas.

Set 4 and 5 equal and rearrange the result with respect to velocity $v$:

$$ \begin{align} v ~=~ \sqrt{\frac{3\, k_{\text B}\,T}{m}} \end{align} $$

If you know the mass $m$ of the gas particle and the temperature $T$ of the gas, you can substitute Eq. 6 into the de Broglie relation 3 for velocity to express the de Broglie wavelength in terms of temperature:

De Broglie wavelength of a gas particle using temperature

$$ \begin{align} \lambda ~=~ \frac{h}{ \sqrt{3\,m\,k_{\text B}\,T} } \end{align} $$

With the help of the de Broglie wavelength 7 it is possible to estimate whether a gas behaves more classically or quantum mechanically.

A gas that consists of light gas particles and is as cold as possible tends to be a quantum gas. This obeys the laws of quantum mechanics.
A gas that consists of heavy gas particles and is as hot as possible tends to be a classical gas. This obeys the laws of classical mechanics and thermodynamics.

Example: Helium gas

A helium atom has mass $ m = 4u = 6.64 \cdot 10^{-27} \, \mathrm{kg} $. If the helium gas has a temperature of $ 300 \, \mathrm{K}$ (room temperature), then a helium atom in this gas has the following de Broglie wavelength:

$$ \begin{align} \lambda ~=~ \frac{6.6 \cdot 10^{-34} \, \mathrm{Js} }{ \sqrt{ 3 \cdot 6.64\cdot 10^{-27}\,\mathrm{kg} \cdot 1.38 \cdot 10^{-23} \, \frac{\mathrm J}{\mathrm K} \cdot 300 \, \mathrm{K} } } ~=~ 7.3 \cdot 10^{-11} \, \mathrm{m} \end{align} $$

A helium atom has a matter wavelength $ \lambda = 73 \, \mathrm{pm} $ (picometer). For comparison: A helium atom is about $ 30 \, \mathrm{pm} $.

For a very cold helium gas ($ T = 0.1\,\mathrm{K} $), the de Broglie wavelength is larger: $ \lambda = 4 \, \mathrm{nm} $.

Particles in the electric field

In addition to temperature, a electric field can also be used to change the de Broglie wavelength of a particle. Note, however, that only electrically charged particles can be accelerated in the electric field (e.g. electrons, ions). Uncharged particles (neutrons, helium atoms) are unsuitable for this.

Hover the image!

A positive charge is accelerated to the positive plate and thus gains kinetic energy.

Consider a positive charged particle with charge $q$ in an electric field. For example, in the homogeneous field of a plate capacitor. The E-field is generated by applying an electric voltage $U$ between the two plates.

If we now place a resting positive charged particle at the positively charged plate, then the particle will experience a force $F$ towards the negative plate. The force accelerates the particle, increasing its velocity and thus its kinetic energy. Once the positive particle arrives at the negative plate, it has converted its potential energy into kinetic energy. Its kinetic energy in this case is:

$$ \begin{align} W_{\text{kin}} ~=~ q \, U \end{align} $$

Equate the kinetic energy 4 with 8 and rearrange the result with respect to velocity $v$:

$$ \begin{align} v ~=~ \sqrt{ \frac{2\,q\,U}{m} } \end{align} $$

With 9 you can find the velocity of a particle. You can substitute this into the de Broglie relation 3 to determine the de Broglie wavelength of the charged particle:

De Broglie wavelength using voltage

$$ \begin{align} \lambda ~=~ \frac{h}{ \sqrt{2m\,q\,U} } \end{align} $$

Example: Accelerate an electron

A resting electron with elementary charge $ |q| = 1.6 \cdot 10^{-19}\,\mathrm{C} $ and rest mass $ m_{\text e} = 9.1 \cdot 10^{-31} \, \mathrm{kg} $, passes through a voltage of $ U = 1 \, \mathrm{kV} $ (kilovolts). Thus, its de Broglie wavelength becomes:

$$ \begin{align} \lambda ~=~ \frac{6.6\cdot 10^{-34}\,\text{Js} }{ \sqrt{ 2 \cdot 9.1 \cdot 10^{-31} \, \text{kg} \cdot 1.6 \cdot 10^{-19}\,\text{C} \cdot 1000 \, \text{V} } } ~=~ 38.7 \cdot 10^{-12} \, \text{m} \end{align} $$

The matter wavelength is $ 38.7 \, \mathrm{pm} $. If, on the other hand, the electron only passes through a voltage of $ U = 1\,\mathrm{V}$, its de Broglie wavelength is larger: $ 1.2 \, \mathrm{nm} $.

De Broglie wavelength of relativistic particles

If the velocity of a particle becomes large, for example, by passing through a very large voltage or by heating the particle gas, then relativistic effects may no longer be neglected. Note that the de Broglie relation 3 is non-relativistic. That is, it is valid only if the gas particle has a velocity that is very small compared to the speed of light $ c = 3 \cdot 10^8 \, \frac{\mathrm m}{\mathrm s} $.

A thermal velocity in the order of $ v = 10^3 \, \frac{\mathrm m}{\mathrm s}$ is very high from the human point of view. But it is just 0.0003% of the speed of light! So a rather non-relativistic velocity. A hot electron gas with an average velocity of $ v = 10^6 \, \frac{\mathrm m}{\mathrm s}$ is also just 0.3% of the speed of light.

The closer the particle approaches the speed of light, the more inaccurate becomes its de Broglie wavelength calculated with 3. To be able to calculate also the de Broglie wavelength of "light-fast" particles, we modify 3 to make it relativistic.

For $ v \approx c$ the following energy-momentum relation holds (this is a result of special relativity):

$$ \begin{align} W ~=~ \sqrt{ (p \, c)^2 ~+~ (m_0 \, c^2)^2 } \end{align} $$

Here $ W $ is the energy, $p$ the momentum of the particle and $m_0$ its rest mass.

Let us now make an approximation. For particles with a very large momentum $p$, the rest energy $ m_0 \, c^2 $ of the particle becomes negligible compared to its kinetic energy $p \, c$. Thus 11 simplifies to:

$$ \begin{align} W ~=~ p \, c \end{align} $$

Here we just set $ m_0 \, c^2 = 0 $ and take root in 11. Eq. 12 inserted in 2 results:

Formula: Relativistic de Broglie wavelength

$$ \begin{align} \lambda ~=~ \frac{h \, c}{ W } \end{align} $$

If you rearrange 13 for the energy $W$ and by using $ c = \lambda \, f $ you replace the wavelength with the frequency $f$, you get exactly the energy of a photon postulated by Albert Einstein: $ W = h\, f$. This equation is also valid for relativistic particles.

Example: Proton in the Large Hadron Collider

Hover the image!

Two proton packages collide with each other at almost the speed of light.

A proton from the LHC particle accelerator at CERN is accelerated to an energy of $ 7 \, \mathrm{TeV}$ (teraelectron volts). This corresponds to: $W = 1.12 \cdot 10^{-6} \, \mathrm{J}$. Using 13 you can calculate its relativistic de Broglie wavelength:

$$ \begin{align} \lambda ~=~ \frac{6.6\cdot 10^{-34}\,\mathrm{Js} ~\cdot~ 3 \cdot 10^8 \, \frac{\mathrm m}{\mathrm s} }{ 1.12 \cdot 10^{-6} \, \mathrm{J} } ~=~ 1.8 \cdot 10^{-19} \, \mathrm{m} \end{align} $$

This is a very short wavelength. The wave character of the proton in the LHC ring does not matter. It behaves like a particle.