Physics Lesson 19.4.1 - De Broglie Wave and De Broglie Relation

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Welcome to our Physics lesson on De Broglie Wave and De Broglie Relation, this is the first lesson of our suite of physics lessons covering the topic of De Broglie Wave, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

De Broglie Wave and De Broglie Relation

In 1924, when working for his PhD thesis, Louis De Broglie presented his hypothesis on the wave nature of electron. By generalizing the particle-wave dualism, he supported the idea that not only EM radiation has a particle nature but the matter particles (electrons) have a wave nature as well.

By "wave nature of electron", De Broglie did not intend the radiation of any kind of special EM wave produced by electrons in motion or the transformation of electrons in any kind of sustainable matter wave. Rather, he supported the idea that "moving electrons have a corresponding kind of wave associated, whose wavelength is determined only by the electron momentum."

De Broglie was not able to explain the nature of this "electron wave" which we call "De Broglie wave", but he postulated the relationship between its wavelength and the impulse of electron. It is

λ = h/p

Remark! Sometimes in quantum physics we used another constant instead of Planck constant h, known as the reduced Planck constant, ℏ. Its relationship with the Planck constant is

ℏ = h/2π

De Broglie applied the analogy with the corresponding equation used for photons. As we know, the two equations used for light waves are

E_ph = h ∙ f and p = E_ph/c

Combining these two equation, we obtain

p = h ∙ f/c
= h/λ

Thus,

λ = h/p

From the last equation, it is clear that the De Broglie relation represents a generalization of the particle-wave dualism.

Example 1

Calculate the wavelength of De Broglie wave of an electron accelerated in a 100 V potential difference. Take the mass of electron m = 9.1 × 10^-31 kg.

Solution 1

Initially, we find the momentum p of electron using its relationship with the kinetic energy. We have

KE = m ∙ v²/2
= p²/2m

On the other hand, it is obvious that the kinetic energy of electron (assumed that initially it was at rest), is numerically equal to the work done by electric forces, i.e.

KE = W
p²/2m = e ∙ ∆V

where e is the elementary charge of electron (e = 1.6 × 10^-19 C). Hence, we obtain for the momentum of electron:

p = √2 ∙ m ∙ e ∙ ∆V
= √2 ∙ (9.1 × 10^-31 kg) ∙ (1.6 × 10^-19 C) ∙ (10² V)
= √29.12 × 10^-48
= 5.396 × 10^-24 kg ∙ m/s

Therefore, the wavelength of De Broglie wave is

λ = h/p
= h/√2 ∙ m ∙ e ∙ ∆V
= 6.626 × 10^-34 J ∙ s/5.396 × 10^-24 kg ∙ m/s
= 1.23 × 10^-10 m
= 0.123 nm

From this example, it is clear that the electron's wavelength is comparable to wavelengths of X-radiation of EM waves. This is why electronic beams incident on crystals gives diffract visibly and produce patterns similar to those produced by X-rays. In this sense, expressions like "electron has a wave nature" or "electron manifests wave behavior" are very meaningful.

Example 2

An electronic beam diffracts after passing through an a = 12 μm narrow gap and the diffraction pattern is observed on a fluorescent screen placed at a distance of L = 4 m from the gap, as shown in the figure.

Physics Tutorials: This image provides visual information for the physics tutorial De Broglie Wave

Calculate the wavelength of De Broglie wave for electrons, given that measurement gave the value of Δy = 0.2 mm for the width of central maximum. Refer to tutorial> 12.4 for the formula of diffraction.

Solution 2

Clues:

a = 12 μm = 12 × 10^-6 m = 1.2 × 10^-5 m
L = 4 m
Δy = 0.2 mm = 2 × 10^-4 m
λ = ?

The diffraction pattern indicates the density of electrons distribution on the thin film acting as a screen. It is obvious that most electrons are incident on the central maximum as the curve is higher at this section. In other words, the x-coordinate of diffraction pattern is proportional to the number of electrons incident on that specific section of the screen.

We know from tutorial>tutorial 12.4 that the condition for a maximum to occur is

sinθ = N ∙ λ/a

where a is the thickness of the gap (slit), λ is the wavelength of the incident EM radiation and N is the order of diffraction (here N = 1 as we are interested only for the central maximum).

The angle θ is very small (we reach in this conclusion by comparing the vertical displacement Δy, which acts as a vertical component for the angle θ, to the distance L of the gap from the screen which here acts as a horizontal component of the angle θ). Thus, we can write (based on the trigonometric approximation you know from math):

sinθ ≈ tanθ
= ∆y/2/L
= ∆y/2L

Thus, comparing the two above equations for sin θ, we obtain

N ∙ λ/a = ∆y/2L

Hence, we obtain for De Broglie wavelength λ:

λ = a ∙ ∆y/2L
= (1.2 × 10^-5 m) ∙ (2 × 10^-4 m)/2 ∙ (4 m)
= 0.3 × 10^-9 m
= 0.3 nm

You have reached the end of Physics lesson 19.4.1 De Broglie Wave and De Broglie Relation. There are 2 lessons in this physics tutorial covering De Broglie Wave, you can access all the lessons from this tutorial below.

More De Broglie Wave Lessons and Learning Resources

Modern Physics Learning Material
Tutorial ID	Physics Tutorial Title	Revision Notes	Revision Questions
19.4	De Broglie Wave
Lesson ID	Physics Lesson Title	Lesson	Video Lesson
19.4.1	De Broglie Wave and De Broglie Relation
19.4.2	De Broglie Wave is a Probability Wave

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