6.5 De Broglie’s Matter Waves - University Physics Volume 3 | OpenStax (2024)

Learning Objectives

By the end of this section, you will be able to:

  • Describe de Broglie’s hypothesis of matter waves
  • Explain how the de Broglie’s hypothesis gives the rationale for the quantization of angular momentum in Bohr’s quantum theory of the hydrogen atom
  • Describe the Davisson–Germer experiment
  • Interpret de Broglie’s idea of matter waves and how they account for electron diffraction phenomena

Compton’s formula established that an electromagnetic wave can behave like a particle of light when interacting with matter. In 1924, Louis de Broglie proposed a new speculative hypothesis that electrons and other particles of matter can behave like waves. Today, this idea is known as de Broglie’s hypothesis of matter waves. In 1926, De Broglie’s hypothesis, together with Bohr’s early quantum theory, led to the development of a new theory of wave quantum mechanics to describe the physics of atoms and subatomic particles. Quantum mechanics has paved the way for new engineering inventions and technologies, such as the laser and magnetic resonance imaging (MRI). These new technologies drive discoveries in other sciences such as biology and chemistry.

According to de Broglie’s hypothesis, massless photons as well as massive particles must satisfy one common set of relations that connect the energy E with the frequency f, and the linear momentum p with the wavelength λ.λ. We have discussed these relations for photons in the context of Compton’s effect. We are recalling them now in a more general context. Any particle that has energy and momentum is a de Broglie wave of frequency f and wavelength λ:λ:

E=hfE=hf

6.53

λ=hp.λ=hp.

6.54

Here, E and p are, respectively, the relativistic energy and the momentum of a particle. De Broglie’s relations are usually expressed in terms of the wave vector k,k, k=2π/λ,k=2π/λ, and the wave frequency ω=2πf,ω=2πf, as we usually do for waves:

E=ωE=ω

6.55

p=k.p=k.

6.56

Wave theory tells us that a wave carries its energy with the group velocity. For matter waves, this group velocity is the velocity u of the particle. Identifying the energy E and momentum p of a particle with its relativistic energy mc2mc2 and its relativistic momentum mu, respectively, it follows from de Broglie relations that matter waves satisfy the following relation:

λf=ωk=E/p/=Ep=mc2mu=c2u=cβλf=ωk=E/p/=Ep=mc2mu=c2u=cβ

6.57

where β=u/c.β=u/c. When a particle is massless we have u=cu=c and Equation 6.57 becomes λf=c.λf=c.

Example 6.11

How Long Are de Broglie Matter Waves?

Calculate the de Broglie wavelength of: (a) a 0.65-kg basketball thrown at a speed of 10 m/s, (b) a nonrelativistic electron with a kinetic energy of 1.0 eV, and (c) a relativistic electron with a kinetic energy of 108keV.108keV.

Strategy

We use Equation 6.57 to find the de Broglie wavelength. When the problem involves a nonrelativistic object moving with a nonrelativistic speed u, such as in (a) when β=u/c1,β=u/c1, we use nonrelativistic momentum p. When the nonrelativistic approximation cannot be used, such as in (c), we must use the relativistic momentum p=mu=m0γu=E0γβ/c,p=mu=m0γu=E0γβ/c, where the rest mass energy of a particle is E0=mc2E0=mc2 and γγ is the Lorentz factor γ=1/1β2.γ=1/1β2. The total energy E of a particle is given by Equation 6.53 and the kinetic energy is K=EE0=(γ1)E0.K=EE0=(γ1)E0. When the kinetic energy is known, we can invert Equation 6.18 to find the momentum p=(E2E02)/c2=K(K+2E0)/cp=(E2E02)/c2=K(K+2E0)/c and substitute in Equation 6.57 to obtain

λ=hp=hcK(K+2E0).λ=hp=hcK(K+2E0).

6.58

Depending on the problem at hand, in this equation we can use the following values for hc: hc=(6.626×10−34J·s)(2.998×108m/s)=1.986×10−25J·m=1.241eV·μmhc=(6.626×10−34J·s)(2.998×108m/s)=1.986×10−25J·m=1.241eV·μm

Solution

  1. For the basketball, the kinetic energy is

    K=mu2/2=(0.65kg)(10m/s)2/2=32.5JK=mu2/2=(0.65kg)(10m/s)2/2=32.5J

    and the rest mass energy is

    E0=mc2=(0.65kg)(2.998×108m/s)2=5.84×1016J.E0=mc2=(0.65kg)(2.998×108m/s)2=5.84×1016J.

    We see that K/(K+E0)1K/(K+E0)1 and use p=mu=(0.65kg)(10m/s)=6.5J·s/m:p=mu=(0.65kg)(10m/s)=6.5J·s/m:

    λ=hp=6.626×10−34J·s6.5J·s/m=1.02×10−34m.λ=hp=6.626×10−34J·s6.5J·s/m=1.02×10−34m.

  2. For the nonrelativistic electron,

    E0=mc2=(9.109×10−31kg)(2.998×108m/s)2=511keVE0=mc2=(9.109×10−31kg)(2.998×108m/s)2=511keV

    and when K=1.0eV,K=1.0eV, we have K/(K+E0)=(1/512)×10−31,K/(K+E0)=(1/512)×10−31, so we can use the nonrelativistic formula. However, it is simpler here to use Equation 6.58:

    λ=hp=hcK(K+2E0)=1.241eV·μm(1.0eV)[1.0eV+2(511keV)]=1.23nm.λ=hp=hcK(K+2E0)=1.241eV·μm(1.0eV)[1.0eV+2(511keV)]=1.23nm.

    If we use nonrelativistic momentum, we obtain the same result because 1 eV is much smaller than the rest mass of the electron.
  3. For a fast electron with K=108keV,K=108keV, relativistic effects cannot be neglected because its total energy is E=K+E0=108keV+511keV=619keVE=K+E0=108keV+511keV=619keV and K/E=108/619K/E=108/619 is not negligible:

    λ=hp=hcK(K+2E0)=1.241eV·μm108keV[108keV+2(511keV)]=3.55pm.λ=hp=hcK(K+2E0)=1.241eV·μm108keV[108keV+2(511keV)]=3.55pm.

Significance

We see from these estimates that De Broglie’s wavelengths of macroscopic objects such as a ball are immeasurably small. Therefore, even if they exist, they are not detectable and do not affect the motion of macroscopic objects.

Check Your Understanding 6.11

What is de Broglie’s wavelength of a nonrelativistic proton with a kinetic energy of 1.0 eV?

Using the concept of the electron matter wave, de Broglie provided a rationale for the quantization of the electron’s angular momentum in the hydrogen atom, which was postulated in Bohr’s quantum theory. The physical explanation for the first Bohr quantization condition comes naturally when we assume that an electron in a hydrogen atom behaves not like a particle but like a wave. To see it clearly, imagine a stretched guitar string that is clamped at both ends and vibrates in one of its normal modes. If the length of the string is l (Figure 6.18), the wavelengths of these vibrations cannot be arbitrary but must be such that an integer k number of half-wavelengths λ/2λ/2 fit exactly on the distance l between the ends. This is the condition l=kλ/2l=kλ/2 for a standing wave on a string. Now suppose that instead of having the string clamped at the walls, we bend its length into a circle and fasten its ends to each other. This produces a circular string that vibrates in normal modes, satisfying the same standing-wave condition, but the number of half-wavelengths must now be an even number k,k=2n,k,k=2n, and the length l is now connected to the radius rnrn of the circle. This means that the radii are not arbitrary but must satisfy the following standing-wave condition:

2πrn=2nλ2.2πrn=2nλ2.

6.59

If an electron in the nth Bohr orbit moves as a wave, by Equation 6.59 its wavelength must be equal to λ=2πrn/n.λ=2πrn/n. Assuming that Equation 6.58 is valid, the electron wave of this wavelength corresponds to the electron’s linear momentum, p=h/λ=nh/(2πrn)=n/rn.p=h/λ=nh/(2πrn)=n/rn. In a circular orbit, therefore, the electron’s angular momentum must be

Ln=rnp=rnnrn=n.Ln=rnp=rnnrn=n.

6.60

This equation is the first of Bohr’s quantization conditions, given by Equation 6.36. Providing a physical explanation for Bohr’s quantization condition is a convincing theoretical argument for the existence of matter waves.

6.5 De Broglie’s Matter Waves - University Physics Volume 3 | OpenStax (1)

Figure 6.18 Standing-wave pattern: (a) a stretched string clamped at the walls; (b) an electron wave trapped in the third Bohr orbit in the hydrogen atom.

Example 6.12

The Electron Wave in the Ground State of Hydrogen

Find the de Broglie wavelength of an electron in the ground state of hydrogen.

Strategy

We combine the first quantization condition in Equation 6.60 with Equation 6.36 and use Equation 6.38 for the first Bohr radius with n=1.n=1.

Solution

When n=1n=1 and rn=a0=0.529Å,rn=a0=0.529Å, the Bohr quantization condition gives a0p=1·p=/a0.a0p=1·p=/a0. The electron wavelength is:

λ=h/p=h//a0=2πa0=2π(0.529Å)=3.324Å.λ=h/p=h//a0=2πa0=2π(0.529Å)=3.324Å.

Significance

We obtain the same result when we use Equation 6.58 directly.

Check Your Understanding 6.12

Find the de Broglie wavelength of an electron in the third excited state of hydrogen.

Experimental confirmation of matter waves came in 1927 when C. Davisson and L. Germer performed a series of electron-scattering experiments that clearly showed that electrons do behave like waves. Davisson and Germer did not set up their experiment to confirm de Broglie’s hypothesis: The confirmation came as a byproduct of their routine experimental studies of metal surfaces under electron bombardment.

In the particular experiment that provided the very first evidence of electron waves (known today as the Davisson–Germer experiment), they studied a surface of nickel. Their nickel sample was specially prepared in a high-temperature oven to change its usual polycrystalline structure to a form in which large single-crystal domains occupy the volume. Figure 6.19 shows the experimental setup. Thermal electrons are released from a heated element (usually made of tungsten) in the electron gun and accelerated through a potential difference ΔV,ΔV, becoming a well-collimated beam of electrons produced by an electron gun. The kinetic energy K of the electrons is adjusted by selecting a value of the potential difference in the electron gun. This produces a beam of electrons with a set value of linear momentum, in accordance with the conservation of energy:

eΔV=K=p22mp=2meΔV.eΔV=K=p22mp=2meΔV.

6.61

The electron beam is incident on the nickel sample in the direction normal to its surface. At the surface, it scatters in various directions. The intensity of the beam scattered in a selected direction φφ is measured by a highly sensitive detector. The detector’s angular position with respect to the direction of the incident beam can be varied from φ=0°φ=0° to φ=90°.φ=90°. The entire setup is enclosed in a vacuum chamber to prevent electron collisions with air molecules, as such thermal collisions would change the electrons’ kinetic energy and are not desirable.

6.5 De Broglie’s Matter Waves - University Physics Volume 3 | OpenStax (2)

Figure 6.19 Schematics of the experimental setup of the Davisson–Germer diffraction experiment. A well-collimated beam of electrons is scattered off the nickel target. The kinetic energy of electrons in the incident beam is selected by adjusting a variable potential, ΔV,ΔV, in the electron gun. Intensity of the scattered electron beam is measured for a range of scattering angles φ,φ, whereas the distance between the detector and the target does not change.

When the nickel target has a polycrystalline form with many randomly oriented microscopic crystals, the incident electrons scatter off its surface in various random directions. As a result, the intensity of the scattered electron beam is much the same in any direction, resembling a diffuse reflection of light from a porous surface. However, when the nickel target has a regular crystalline structure, the intensity of the scattered electron beam shows a clear maximum at a specific angle and the results show a clear diffraction pattern (see Figure 6.20). Similar diffraction patterns formed by X-rays scattered by various crystalline solids were studied in 1912 by father-and-son physicists William H. Bragg and William L. Bragg. The Bragg law in X-ray crystallography provides a connection between the wavelength λλ of the radiation incident on a crystalline lattice, the lattice spacing, and the position of the interference maximum in the diffracted radiation (see Diffraction).

The lattice spacing of the Davisson–Germer target, determined with X-ray crystallography, was measured to be a=2.15Å.a=2.15Å. Unlike X-ray crystallography in which X-rays penetrate the sample, in the original Davisson–Germer experiment, only the surface atoms interact with the incident electron beam. For the surface diffraction, the maximum intensity of the reflected electron beam is observed for scattering angles that satisfy the condition nλ=asinφnλ=asinφ (see Figure 6.21). The first-order maximum (for n=1n=1) is measured at a scattering angle of φ50°φ50° at ΔV54V,ΔV54V, which gives the wavelength of the incident radiation as λ=(2.15Å)sin50°=1.64Å.λ=(2.15Å)sin50°=1.64Å. On the other hand, a 54-V potential accelerates the incident electrons to kinetic energies of K=54eV.K=54eV. Their momentum, calculated from Equation 6.61, is p=2.478×10−5eV·s/m.p=2.478×10−5eV·s/m. When we substitute this result in Equation 6.58, the de Broglie wavelength is obtained as

λ=hp=4.136×10−15eV·s2.478×10−5eV·s/m=1.67Å.λ=hp=4.136×10−15eV·s2.478×10−5eV·s/m=1.67Å.

6.62

The same result is obtained when we use K=54eVK=54eV in Equation 6.61. The proximity of this theoretical result to the Davisson–Germer experimental value of λ=1.64Åλ=1.64Å is a convincing argument for the existence of de Broglie matter waves.

6.5 De Broglie’s Matter Waves - University Physics Volume 3 | OpenStax (3)

Figure 6.20 The experimental results of electron diffraction on a nickel target for the accelerating potential in the electron gun of about ΔV=54V:ΔV=54V: The intensity maximum is registered at the scattering angle of about φ=50°.φ=50°.

6.5 De Broglie’s Matter Waves - University Physics Volume 3 | OpenStax (4)

Figure 6.21 In the surface diffraction of a monochromatic electromagnetic wave on a crystalline lattice structure, the in-phase incident beams are reflected from atoms on the surface. A ray reflected from the left atom travels an additional distance D=asinφD=asinφ to the detector, where a is the lattice spacing. The reflected beams remain in-phase when D is an integer multiple of their wavelength λ.λ. The intensity of the reflected waves has pronounced maxima for angles φφ satisfying nλ=asinφ.nλ=asinφ.

Diffraction lines measured with low-energy electrons, such as those used in the Davisson–Germer experiment, are quite broad (see Figure 6.20) because the incident electrons are scattered only from the surface. The resolution of diffraction images greatly improves when a higher-energy electron beam passes through a thin metal foil. This occurs because the diffraction image is created by scattering off many crystalline planes inside the volume, and the maxima produced in scattering at Bragg angles are sharp (see Figure 6.22).

6.5 De Broglie’s Matter Waves - University Physics Volume 3 | OpenStax (5)

Figure 6.22 Diffraction patterns obtained in scattering on a crystalline solid: (a) with X-rays, and (b) with electrons. The observed pattern reflects the symmetry of the crystalline structure of the sample.

Since the work of Davisson and Germer, de Broglie’s hypothesis has been extensively tested with various experimental techniques, and the existence of de Broglie waves has been confirmed for numerous elementary particles. Neutrons have been used in scattering experiments to determine crystalline structures of solids from interference patterns formed by neutron matter waves. The neutron has zero charge and its mass is comparable with the mass of a positively charged proton. Both neutrons and protons can be seen as matter waves. Therefore, the property of being a matter wave is not specific to electrically charged particles but is true of all particles in motion. Matter waves of molecules as large as carbon C60C60 have been measured. All physical objects, small or large, have an associated matter wave as long as they remain in motion. The universal character of de Broglie matter waves is firmly established.

Example 6.13

Neutron Scattering

Suppose that a neutron beam is used in a diffraction experiment on a typical crystalline solid. Estimate the kinetic energy of a neutron (in eV) in the neutron beam and compare it with kinetic energy of an ideal gas in equilibrium at room temperature.

Strategy

We assume that a typical crystal spacing a is of the order of 1.0 Å. To observe a diffraction pattern on such a lattice, the neutron wavelength λλ must be on the same order of magnitude as the lattice spacing. We use Equation 6.61 to find the momentum p and kinetic energy K. To compare this energy with the energy ETET of ideal gas in equilibrium at room temperature T=300K,T=300K, we use the relation K=32kBT,K=32kBT, where kB=8.62×10−5eV/KkB=8.62×10−5eV/K is the Boltzmann constant.

Solution

We evaluate pc to compare it with the neutron’s rest mass energy E0=940MeV:E0=940MeV:

p=hλpc=hcλ=1.241×10−6eV·m10−10m=12.41keV.p=hλpc=hcλ=1.241×10−6eV·m10−10m=12.41keV.

We see that p2c2E02p2c2E02 so KE0KE0 and we can use the nonrelativistic kinetic energy:

K=p22mn=h22λ2mn=(6.63×10−34J·s)2(2×10−20m2)(1.66×10−27kg)=1.32×10−20J=82.7meV.K=p22mn=h22λ2mn=(6.63×10−34J·s)2(2×10−20m2)(1.66×10−27kg)=1.32×10−20J=82.7meV.

Kinetic energy of ideal gas in equilibrium at 300 K is:

KT=32kBT=32(8.62×10−5eV/K)(300K)=38.8meV.KT=32kBT=32(8.62×10−5eV/K)(300K)=38.8meV.

We see that these energies are of the same order of magnitude.

Significance

Neutrons with energies in this range, which is typical for an ideal gas at room temperature, are called “thermal neutrons.”

Example 6.14

Wavelength of a Relativistic Proton

In a supercollider at CERN, protons can be accelerated to velocities of 0.75c. What are their de Broglie wavelengths at this speed? What are their kinetic energies?

Strategy

The rest mass energy of a proton is E0=m0c2=(1.672×10−27kg)(2.998×108m/s)2=938MeV.E0=m0c2=(1.672×10−27kg)(2.998×108m/s)2=938MeV. When the proton’s velocity is known, we have β=0.75β=0.75 and βγ=0.75/10.752=1.134.βγ=0.75/10.752=1.134. We obtain the wavelength λλ and kinetic energy K from relativistic relations.

Solution

λ=hp=hcpc=hcβγE0=1.241eV·μm1.134(938MeV)=1.16fmλ=hp=hcpc=hcβγE0=1.241eV·μm1.134(938MeV)=1.16fm

K=E0(γ1)=938MeV(1/10.7521)=480.1MeVK=E0(γ1)=938MeV(1/10.7521)=480.1MeV

Significance

Notice that because a proton is 1835 times more massive than an electron, if this experiment were performed with electrons, a simple rescaling of these results would give us the electron’s wavelength of (1835)0.77fm=1.4pm(1835)0.77fm=1.4pm and its kinetic energy of 480.1MeV/1835=261.6keV.480.1MeV/1835=261.6keV.

Check Your Understanding 6.13

Find the de Broglie wavelength and kinetic energy of a free electron that travels at a speed of 0.75c.

6.5 De Broglie’s Matter Waves - University Physics Volume 3 | OpenStax (2024)

FAQs

What is the Compton effect Openstax? ›

The Compton effect is the term used for an unusual result observed when X-rays are scattered on some materials. By classical theory, when an electromagnetic wave is scattered off atoms, the wavelength of the scattered radiation is expected to be the same as the wavelength of the incident radiation.

What observation confirmed de Broglie's theory of matter waves? ›

The validity of de Broglie's proposal was confirmed by electron diffraction experiments of G.P. Thomson in 1926 and of C. Davisson and L. H.

What is the de Broglie theory of matter waves? ›

In 1924, Louis de Broglie hypothesised that all matter is wave-like. It means that particles (like atoms and electrons) have both a particle and wave nature and that the wave nature of particles can only be observed when they are being followed. The wavelength of a particle is inversely proportional to its momentum.

What is the de Broglie wavelength in quantum mechanics? ›

According to wave-particle duality, the De Broglie wavelength is a wavelength manifested in all the objects in quantum mechanics which determines the probability density of finding the object at a given point of the configuration space. The de Broglie wavelength of a particle is inversely proportional to its momentum.

What is the Compton effect for dummies? ›

In the Compton effect, individual photons collide with single electrons that are free or quite loosely bound in the atoms of matter. Colliding photons transfer some of their energy and momentum to the electrons, which in turn recoil.

What does the Compton effect tell us? ›

The Compton effect is defined as the effect that is observed when x-rays or gamma rays are scattered on a material with an increase in wavelength. Arthur Compton studied this effect in the year 1922. During the study, Compton found that wavelength is not dependent on the intensity of incident radiation.

What is the de Broglie formula? ›

De Broglie Wavelength for an Electron

Now, putting these values in the equation λ = h/mv, which yields λ = 3.2 Å. This value is measurable. Therefore, we can say that electrons have wave-particle duality. Thus all the big objects have a wave nature and microscopic objects like electrons have wave-particle nature.

What is the formula for matter waves? ›

The relationship between momentum and wavelength for matter waves is given by p = h/λ, and the relationship energy and frequency is E = hf. The wavelength λ = h/p is called the de Broglie wavelength, and the relations λ = h/p and f = E/h are called the de Broglie relations.

Is matter a wave or particle? ›

Matter waves are a central part of the theory of quantum mechanics, being half of wave–particle duality. At all scales where measurements have been practical, matter exhibits wave-like behavior. For example, a beam of electrons can be diffracted just like a beam of light or a water wave.

What is a de Broglie wave in simple terms? ›

De Broglie waves account for the appearance of subatomic particles at conventionally unexpected sites because their waves penetrate barriers much as sound passes through walls. Thus a heavy atomic nucleus occasionally can eject a piece of itself in a process called alpha decay.

Why is the de Broglie wave important? ›

The importance of the de Broglie relation is that it is more useful for microscopic and fundamental particles like electrons. De Broglie's equation helps us understand the idea of matter having a wavelength. Therefore, if we look at every moving particle, whether microscopic or macroscopic, it will have a wavelength.

What are de Broglie waves called? ›

De-Broglie waves: The waves which are associated with matter are called matter waves or de-Broglie waves. Wave equation, λ=hmv. where, λ - wave length, h - plank's constant, m - mass, v- velocity of wave.

What is the evidence that electrons are waves? ›

Davisson and L. Germer performed a series of electron-scattering experiments that clearly showed that electrons do behave like waves. The experiment was as follows: When electrons pass through a double slit and strike a screen behind the slits, an interference pattern of bright and dark bands is formed on the screen.

What are the four quantum numbers? ›

In atoms, there are a total of four quantum numbers: the principal quantum number (n), the orbital angular momentum quantum number (l), the magnetic quantum number (ml), and the electron spin quantum number (ms).

What does de Broglie wavelength prove? ›

All particles can show wave-like properties. The de Broglie wavelength of a particle indicates the length scale at which wave-like properties are important for that particle.

What is the Doppler effect in Openstax physics? ›

The Doppler Effect of Sound Waves

The Doppler effect is a change in the observed pitch of a sound, due to relative motion between the source and the observer.

Which best describes the Compton effect? ›

The Compton effect is a partial absorption process and as the original photon has lost energy, known as Compton shift (i.e. a shift of wavelength/frequency). The wavelength change of the scattered photon can be determined by 0.024 (1- cos θ), where θ is scattered photon angle.

What is Hooke's law in Openstax? ›

The simplest oscillations occur when the restoring force is directly proportional to displacement. Recall that Hooke's law describes this situation with the equation F = −kx. Therefore, Hooke's law describes and applies to the simplest case of oscillation, known as simple harmonic motion.

What is the reaction rate in Openstax? ›

Likewise, the rate of a chemical reaction is a measure of how much reactant is consumed, or how much product is produced, by the reaction in a given amount of time. The rate of reaction is the change in the amount of a reactant or product per unit time.

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