De Broglie Wavelength

Divide Planck constant $h = 6.626\times{}10^{-34} J\cdot{}s$ by the momentum p of the particle: $\lambda{} =\frac{ h}{p}$. For a non-relativistic particle $p = m\cdot{}v$, so $\lambda{} =\frac{ h}{(m\cdot{}v)}$. An electron with $v =\frac{ 10^{6} m}{s}$ has $\lambda{} =\frac{ 6.626\times{}10^{-34}}{(9.11\times{}10^{-31}\cdot{}10^{6})} \approx{} 0.73 nm$, which is on the order of atomic spacings and enables electron diffraction. Insert the mass in kilograms and the speed in metres per second, then the wavelength comes out in metres. Small wavelengths mean large momentum, large wavelengths mean small momentum.

Because their momentum is macroscopically huge and $\lambda{} =\frac{ h}{p}$ therefore gives an unimaginably small wavelength. Planck constant h is extremely tiny at $6.626\times{}10^{-34} J\cdot{}s$. A football of 0.4 kg at $\frac{10 m}{s}$ has a momentum of $\frac{4 kg\cdot{}m}{s}$ and thus a wavelength of about $1.7\times{}10^{-34} m$, billions of times smaller than an atomic nucleus. Such wavelengths cannot be measured in any experiment; there is no object with a suitably small slit. Wave character appears only for very light particles such as electrons, whose small mass leads to measurable wavelengths in the nanometre range.

h is Planck constant at $6.626\times{}10^{-34} J\cdot{}s$ and appears in the de Broglie relation $\lambda{} =\frac{ h}{p}$ as well as in $E = h\cdot{}f$. The reduced Planck constant ℏ is defined as $\hbar{} =\frac{ h}{(2\pi{})} \approx{} 1.055\times{}10^{-34} J\cdot{}s$. It shows up wherever angular frequencies or angular momenta are involved, for example in the Heisenberg uncertainty relation $\Delta{}x\cdot{}\Delta{}p \ge{}\frac{ \hbar{}}{2}$. A common mistake is to confuse h and ℏ; they differ by the factor $2\pi{} \approx{} 6.28$. Check carefully in every formula whether the normal or the reduced constant is required, otherwise your result is off by this factor.

When an electron is accelerated by a voltage U, it gains the kinetic energy $E_{\mathrm{kin}} = e\cdot{}U$. Through $E_{\mathrm{kin}} =\frac{ p^{2}}{(2m)}$ the momentum can be found: $p = \sqrt{(2\cdot{}m\cdot{}e\cdot{}U)}$. Inserting this into $\lambda{} =\frac{ h}{p}$ gives $\lambda{} =\frac{ h}{\sqrt{(2\cdot{}m\cdot{}e\cdot{}U)}}$. The wavelength therefore decreases with increasing voltage, because a faster electron has more momentum. This relationship is the basis of the electron microscope: high voltages produce very small wavelengths and thus a much higher resolution than visible light. For very high voltages you also have to apply a relativistic correction.

In 1924 de Broglie transferred the relation $p =\frac{ h}{\lambda{}}$, known for photons, to matter particles and postulated that electrons too have a wavelength. Shortly afterwards Davisson and Germer confirmed this experimentally through diffraction of electrons at a crystal lattice, a typical wave phenomenon. This made it clear that particles such as electrons show both particle and wave properties, depending on the experiment. The wavelength $\lambda{} =\frac{ h}{p}$ links the particle quantity momentum with the wave quantity wavelength through the natural constant h. This duality is a cornerstone of quantum mechanics and explains phenomena such as interference and diffraction of matter.