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:

Formula: Momentum of a photon

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

Here \(h\) is the Planck's constant, a natural constant with the value:

Value of Planck constant

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

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:

De Broglie wavelength is Planck constant divided by momentum

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

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

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.

Classical gas vs. quantum gas

A particle of mass \(m\) moving with velocity \(v\) has kinetic energy:

Formula: Classical kinetic energy

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

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:

Formula: Average kinetic energy of a gas

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

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 \(v\):

Mean velocity of the gas particles

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

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

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.

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.

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:

Kinetic energy equals charge times voltage

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

Equate the kinetic energy 4 with 8 and rearrange the result with respect to velocity \(v\):

Velocity of the gas particle using voltage

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

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

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

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):

Relativistic energy-momentum relation

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

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:

Energy equals momentum times speed of light

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

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

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

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.

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