De Broglie Wavelength Formula

Matter waves are the central part of the theory of quantum mechanics. All matter can exhibit wave-like behaviour. The concept that matter behaves like a wave this concept was proposed by a French physicist named Louis de Broglie in the year 1924. It is also known as the de Broglie hypothesis. Matter waves are also known as de Broglie waves. The de Broglie wavelength is represented by , it is associated with a massive particle and it is related to its momentum that is represented by p, through the Planck constant that is denoted as h:

λ = \[\frac{h}{p}\] = \[\frac{h}{mv}\], this is the De Broglie wavelength formula. 

At the end of the 19th century, it was thought that light consists of waves of electromagnetic fields that are propagated in accordance with Maxwell's equations. The matter was thought to consist of localized particles. In the year 1900, when investigating the theory of black-body radiation this division was exposed to doubt. Max Planck proposed that light is emitted in discrete quanta of energy. By extending Planck's investigation in several ways, Albert Einstein proposed that light can also be propagated and absorbed in the particles called quanta now these are known as photons. 

The wave-like behaviour of matter was first demonstrated by George Paget Thomson by using a thin metal diffraction experiment. Independently it was demonstrated in the Davisson–Germer experiment, by using electrons and elementary particles such as neutral atoms and even molecules. Let us calculate the De Broglie wavelength of an electron.

Formula for De Broglie Wavelength Experimental Confirmation

Electrons: At Bell Labs in the year 1927, Clinton Davisson and Lester Germer observed slow-moving electrons at a crystalline nickel target. The diffracted electron intensity was measured and determined to have the same diffraction pattern as the ones that are predicted by Bragg for x-rays. The diffraction is considered as a property that can be exhibited only by waves but it happened before the acceptance of the de Broglie hypothesis. The presence of any diffraction hence effects by the matter demonstrated the wave-like nature of matter. De Broglie’s hypothesis was confirmed experimentally by adding the de Broglie wavelength into the Bragg condition, thus the predicted diffraction pattern was observed.

Neutral Atoms: Experiments with Fresnel diffraction and an atomic mirror for the specular reflection of the neutral atoms confirms the application of the de Broglie hypothesis to atoms. The existence of the atomic waves that undergo diffraction and interference thus allows the quantum reflection by the tails of the attractive potential. The thermal De Broglie wavelength came into the micrometre range. By using the Bragg diffraction of atoms and a Ramsey interferometry technique, the cold sodium atoms De Broglie wavelength was explicitly measured and it is found to be consistent with the temperature that is measured by a different method.

Molecules: Recent experiments even confirm the relations for molecules and even macromolecules that otherwise might be supposed too large to undergo quantum mechanical effects. The researchers calculated a De Broglie wavelength of the most probable velocity. Still one step further than Louis de Broglie go theories which in quantum mechanics eliminate the concept of a pointlike classical particle and explain the observed facts by means of wavepackets of matter waves alone.

Calculate the De Broglie Wavelength of an Electron

By the De Broglie wavelength formula the nature of the particle can be determined. Einstein explained the photon momentum and the energy of the photon is given by the formula,

E = mc\[^{2}\] , -------- (1)

We also know that E = hv, by equating equation one we get that,

mc\[^{2}\] = hv

H is the Planck’s constant and the value is 6.627 x 10\[^{-34}\] Js 

De Broglie considered the velocity of the particle as v instead of c, hence we can replace c with v in equation (1).

mv\[^{2}\] = hv

⇒ v = \[\frac{mv^{2}}{h}\]

We know the relation between wavelength, frequency, and velocity that is v = \[\frac{v}{\lambda}\], by replacing the value of in the above equation, we get

\[\frac{v}{\lambda}\] = \[\frac{mv^{2}}{h}\]

⇒ \[\frac{1}{\lambda}\] = \[\frac{mv}{h}\]

⇒ \[\lambda\] = \[\frac{h}{mv}\]

From the definition of the momentum, we can write as, p = mv, by substituting this equation in the above formula we get,

\[\lambda\] = \[\frac{h}{p}\]

The above equation is the De Broglie equation where represents the wavelength.

Solved Examples

1. Calculate the Wavelength of the Electron that is Moving at the Speed of Light.

Ans: The De Broglie wavelength equation is as follows,

\[\lambda\] = \[\frac{h}{mv}\]

\[\lambda\] is the wavelength

h is the Planck’s constant and the value is 6.6260 x 10\[^{-34}\] Js  

v is the velocity, here it is considered as the speed of light, 3 x 10\[^{8}\] ms\[^{-1}\] 

m is the mass of the electron, 9.1 x 10\[^{-31}\] Kg

Substituting all these values we can get,

\[\lambda\] = \[\frac{6.6260 \times 10^{-34} Js}{9.1 \times 10^{-31}Kg \times 3 \times 10^{8}ms^{-1}}\]

⇒ \[\lambda\] = 0.2424 x 10\[^{-11}\]m

⇒ \[\lambda\] = 2.424 nm