De Broglie Wave

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19.4De Broglie Wave

In this Physics tutorial, you will learn:

  • Can particles manifest wave behavior?
  • What is the relationship between energy of particle and its wavelength?
  • What is the De Broglie relation?
  • How the wavelength of a particle is related to its momentum?
  • What is the wavelength of electron comparable with?
  • What is the probability approach used for the De Broglie wave?
  • Is the particle-wave nature specific only for EM waves or it extends in other elements of the universe?

Introduction

In previous articles we have seen that a light wave can be thought as a set of microscopic particles in the form of sparks, known as photons. In other words, light waves can be considered as sets of tiny particles. Do you think the reverse can be true as well, i.e. can particles manifest wave properties? Think about this.

If the above assumption is true (i.e. particles can behave as waves), why we are not able to notice this?

Do you think it is 100% sure that an electron will strike a given point of a screen when emitted from a source? Recall the analogy with a gun; is it granted that a bullet will hit the target or maybe it may miss it?

This tutorial is dedicated to the particle-wave analogy. We will close the circle of particle-wave dualism by proving that not only EM waves manifest particle nature but all particles present a wave behavior as well, as postulated by Lois De Broglie about one century ago.

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, . Its relationship with the Planck constant is

ℏ = h/

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

Eph = h ∙ f and p = Eph/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

De Broglie Wave is a Probability Wave

The phenomena of interference and diffraction of electrons are well explained and interpreted through probability pictures, similar to the method used to figure out photons. Given this, we can assert that: "De Broglie wave for electrons (as well as for other matter particles) is a probability wave." This means that in places of the fluorescent screen where the maxima (bright regions) produced by interference or diffraction do appear, the intensity of De Broglie's wave has the highest intensity while in the positions where there are minima, this intensity is almost zero. In this way, the intensity I of this wave as well as the square of its amplitude a, represent the probability for the electron to strike a specific point of the screen. In other words, it is not taken for granted that an electron will hit a specific point of the screen; everything is just probability (chance for an event to occur).

Summarizing what said above, we can write:

For electrons (and other material particles):

I∝ a2

the probability to detect the electron at a specific point

For photons:

I∝ a2

the probability to detect the photon at a specific point

  1. The particle-wave dualistic nature is a fundamental feature for both EM radiation and matter.
  2. The particle-wave dualism is universal; it includes the entire universe.

Example 3

An electron and a photon have an energy of 1eV each. Compare their De Broglie wavelength. Take the mass of photon as 9.1 × 10-31 kg.

Solution 3

Clues:

Eph = Ee = 1 eV = 1.6 × 10-19 J
me = 9.1 × 10-31 kg
(c = 3 × 108 m/s)
(h = 6.626 × 10-34 J · s)
λph = ?
λe = ?

The (kinetic) energy of electron is

Ee = me ∙ v2/2

and its momentum is

pe = me ∙ v

So, combining these equation we obtain the known expression

Ee = pe2/2me

Thus,

pe = √2 ∙ me ∙ Ee
= √2 ∙ (9.1 × 10-31 kg) ∙ (1.6 × 10-19 J)
= 5.396 × 10-25 kg ∙ m/s

Therefore, the De Broglie wave for this electron is

λe = h/pe
= 6.626 × 10-34 J ∙ s/5.396 × 10-25 kg
= 1.228 × 10-9 m
= 1.228 nm

As for the photon, we have:

Eph = h ∙ f
= h ∙ c/λph

Thus, the wavelength of the 1 eV photon is

λph = h ∙ c/Eph
= (6.626 × 10-34 J ∙ s) ∙ (3 × 108 m/s)/(1.6 × 10-19 J)
= 1.242 × 10-6 m
= 1242 nm

As you see, the wavelength of photon is about 1000 times greater than De Broglie wavelength of an electron with the same energy. That's why we often consider only the energy and wavelength of photons in most events of quantum physics.

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