state the de broglie hypothesis and the corresponding equation

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de Broglie Wavelength

The de Broglie wavelength is a fundamental concept in quantum mechanics that profoundly explains particle behavior at the quantum level. According to de Broglie hypothesis, particles like electrons, atoms, and molecules exhibit wave-like and particle-like properties.

This concept was introduced by French physicist Louis de Broglie in his doctoral thesis in 1924, revolutionizing our understanding of the nature of matter.

de Broglie Equation

A fundamental equation core to de Broglie hypothesis establishes the relationship between a particle’s wavelength and momentum . This equation is the cornerstone of quantum mechanics and sheds light on the wave-particle duality of matter. It revolutionizes our understanding of the behavior of particles at the quantum level. Here are some of the critical components of the de Broglie wavelength equation:

1. Planck’s Constant (h)

Central to this equation is Planck’s constant , denoted as “h.” Planck’s constant is a fundamental constant of nature, representing the smallest discrete unit of energy in quantum physics. Its value is approximately 6.626 x 10 -34 Jˑs. Planck’s constant relates the momentum of a particle to its corresponding wavelength, bridging the gap between classical and quantum physics.

2. Particle Momentum (p)

The second critical component of the equation is the particle’s momentum, denoted as “p”. Momentum is a fundamental property of particles in classical physics, defined as the product of an object’s mass (m) and its velocity (v). In quantum mechanics, however, momentum takes on a slightly different form. It is the product of the particle’s mass and its velocity, adjusted by the de Broglie wavelength.

The mathematical formulation of de Broglie wavelength is

We can replace the momentum by p = mv to obtain

The SI unit of wavelength is meter or m. Another commonly used unit is nanometer or nm.

This equation tells us that the wavelength of a particle is inversely proportional to its mass and velocity. In other words, as the mass of a particle increases or its velocity decreases, its de Broglie wavelength becomes shorter, and it behaves more like a classical particle. Conversely, as the mass decreases or velocity increases, the wavelength becomes longer, and the particle exhibits wave-like behavior. To grasp the significance of this equation, let us consider the example of an electron . 

de Broglie Wavelength of Electron

Electrons are incredibly tiny and possess a minimal mass. As a result, when they are accelerated, such as when they move around the nucleus of an atom , their velocities can become significant fractions of the speed of light, typically ~1%.

Consider an electron moving at 2 x 10 6 m/s. The rest mass of an electron is 9.1 x 10 -31 kg. Therefore,

These short wavelengths are in the range of the sizes of atoms and molecules, which explains why electrons can exhibit wave-like interference patterns when interacting with matter, a phenomenon famously observed in the double-slit experiment.

Thermal de Broglie Wavelength

The thermal de Broglie wavelength is a concept that emerges when considering particles in a thermally agitated environment, typically at finite temperatures. In classical physics, particles in a gas undergo collision like billiard balls. However, particles exhibit wave-like behavior at the quantum level, including wave interference phenomenon. The thermal de Broglie wavelength considers the kinetic energy associated with particles due to their thermal motion.

At finite temperatures, particles within a system possess a range of energies described by the Maxwell-Boltzmann distribution. Some particles have relatively high energies, while others have low energies. The thermal de Broglie wavelength accounts for this distribution of kinetic energies. It helps to understand the statistical behavior of particles within a thermal ensemble.

Mathematical Expression

The thermal de Broglie wavelength (λ th ) is determined by incorporating both the mass (m) of the particle and its thermal kinetic energy (kT) into the de Broglie wavelength equation:

Here, k is the Boltzmann constant, and T is the temperature in Kelvin.

• de Broglie Wave Equation – Chem.libretexts.org  
• de Broglie Wavelength – Spark.iop.org
• de Broglie Matter Waves – Openstax.org
• Wave Nature of Electron – Hyperphysics.phy-astr.gsu.edu

Article was last reviewed on Friday, October 6, 2023

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De Broglie Hypothesis

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Today we know that every particle exhibits both matter and wave nature. This is called wave-particle duality . The concept that matter behaves like wave is called the de Broglie hypothesis , named after Louis de Broglie, who proposed it in 1924.

De Broglie Equation

Explanation of bohr's quantization rule.

De Broglie gave the following equation which can be used to calculate de Broglie wavelength, \(\lambda\), of any massed particle whose momentum is known:

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

where \(h\) is the Plank's constant and \(p\) is the momentum of the particle whose wavelength we need to find.

With some modifications the following equation can also be written for velocity \((v)\) or kinetic energy \((K)\) of the particle (of mass \(m\)):

\[\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mK}}.\]

Notice that for heavy particles, the de Broglie wavelength is very small, in fact negligible. Hence, we can conclude that though heavy particles do exhibit wave nature, it can be neglected as it's insignificant in all practical terms of use.

Calculate the de Broglie wavelength of a golf ball whose mass is 40 grams and whose velocity is 6 m/s. We have \[\lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{40 \times 10^{-3} \times 6} \text{ m}=2.76 \times 10^{-33} \text{ m}.\ _\square\]

One of the main limitations of Bohr's atomic theory was that no justification was given for the principle of quantization of angular momentum. It does not explain the assumption that why an electron can rotate only in those orbits in which the angular momentum of the electron, \(mvr,\) is a whole number multiple of \( \frac{h}{2\pi} \).

De Broglie successfully provided the explanation to Bohr's assumption by his hypothesis.

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6.5 De Broglie’s Matter Waves

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 :

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 = 2 π / λ , k = 2 π / λ , and the wave frequency ω = 2 π f , ω = 2 π f , as we usually do for waves:

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 m c 2 m c 2 and its relativistic momentum mu , respectively, it follows from de Broglie relations that matter waves satisfy the following relation:

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

Example 6.11

How long are de broglie matter waves.

Depending on the problem at hand, in this equation we can use the following values for hc : h c = ( 6.626 × 10 −34 J · s ) ( 2.998 × 10 8 m/s ) = 1.986 × 10 −25 J · m = 1.241 eV · μ m h c = ( 6.626 × 10 −34 J · s ) ( 2.998 × 10 8 m/s ) = 1.986 × 10 −25 J · m = 1.241 eV · μ m

• For the basketball, the kinetic energy is K = m u 2 / 2 = ( 0.65 kg ) ( 10 m/s ) 2 / 2 = 32.5 J K = m u 2 / 2 = ( 0.65 kg ) ( 10 m/s ) 2 / 2 = 32.5 J and the rest mass energy is E 0 = m c 2 = ( 0.65 kg ) ( 2.998 × 10 8 m/s ) 2 = 5.84 × 10 16 J. E 0 = m c 2 = ( 0.65 kg ) ( 2.998 × 10 8 m/s ) 2 = 5.84 × 10 16 J. We see that K / ( K + E 0 ) ≪ 1 K / ( K + E 0 ) ≪ 1 and use p = m u = ( 0.65 kg ) ( 10 m/s ) = 6.5 J · s/m : p = m u = ( 0.65 kg ) ( 10 m/s ) = 6.5 J · s/m : λ = h p = 6.626 × 10 −34 J · s 6.5 J · s/m = 1.02 × 10 −34 m . λ = h p = 6.626 × 10 −34 J · s 6.5 J · s/m = 1.02 × 10 −34 m .
• For the nonrelativistic electron, E 0 = m c 2 = ( 9.109 × 10 −31 kg ) ( 2.998 × 10 8 m/s ) 2 = 511 keV E 0 = m c 2 = ( 9.109 × 10 −31 kg ) ( 2.998 × 10 8 m/s ) 2 = 511 keV and when K = 1.0 eV , K = 1.0 eV , we have K / ( K + E 0 ) = ( 1 / 512 ) × 10 −3 ≪ 1 , K / ( K + E 0 ) = ( 1 / 512 ) × 10 −3 ≪ 1 , so we can use the nonrelativistic formula. However, it is simpler here to use Equation 6.58 : λ = h p = h c K ( K + 2 E 0 ) = 1.241 eV · μ m ( 1.0 eV ) [ 1.0 eV+ 2 ( 511 keV ) ] = 1.23 nm . λ = h p = h c K ( K + 2 E 0 ) = 1.241 eV · μ m ( 1.0 eV ) [ 1.0 eV+ 2 ( 511 keV ) ] = 1.23 nm . If we use nonrelativistic momentum, we obtain the same result because 1 eV is much smaller than the rest mass of the electron.
• For a fast electron with K = 108 keV, K = 108 keV, relativistic effects cannot be neglected because its total energy is E = K + E 0 = 108 keV + 511 keV = 619 keV E = K + E 0 = 108 keV + 511 keV = 619 keV and K / E = 108 / 619 K / E = 108 / 619 is not negligible: λ = h p = h c K ( K + 2 E 0 ) = 1.241 eV · μm 108 keV [ 108 keV + 2 ( 511 keV ) ] = 3.55 pm . λ = h p = h c K ( K + 2 E 0 ) = 1.241 eV · μm 108 keV [ 108 keV + 2 ( 511 keV ) ] = 3.55 pm .

Significance

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 / 2 λ / 2 fit exactly on the distance l between the ends. This is the condition l = k λ / 2 l = k λ / 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 = 2 n , k , k = 2 n , and the length l is now connected to the radius r n r n of the circle. This means that the radii are not arbitrary but must satisfy the following standing-wave condition:

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

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.

Example 6.12

The electron wave in the ground state of hydrogen, 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 Δ V , Δ V , 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:

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 φ φ 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.

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 λ λ 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 = 2.15 Å . a = 2.15 Å . 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 n λ = a sin φ n λ = a sin φ (see Figure 6.21 ). The first-order maximum (for n = 1 n = 1 ) is measured at a scattering angle of φ ≈ 50 ° φ ≈ 50 ° at Δ V ≈ 54 V , Δ V ≈ 54 V , which gives the wavelength of the incident radiation as λ = ( 2.15 Å ) sin 50 ° = 1.64 Å . λ = ( 2.15 Å ) sin 50 ° = 1.64 Å . On the other hand, a 54-V potential accelerates the incident electrons to kinetic energies of K = 54 eV . K = 54 eV . Their momentum, calculated from Equation 6.61 , is p = 2.478 × 10 −5 eV · s / m . p = 2.478 × 10 −5 eV · s / m . When we substitute this result in Equation 6.58 , the de Broglie wavelength is obtained as

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

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 ).

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 C 60 C 60 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.

We see that p 2 c 2 ≪ E 0 2 p 2 c 2 ≪ E 0 2 so K ≪ E 0 K ≪ E 0 and we can use the nonrelativistic kinetic energy:

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

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

Example 6.14

Wavelength of a relativistic proton, check your understanding 6.13.

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

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The de Broglie Wavelength ( AQA A Level Physics )

Revision note.

The de Broglie Wavelength

• The de Broglie equation relates wavelength (a property of waves) to momentum (a property of matter)
• Increasing the speed of electrons, and hence their momentum , causes the angle of diffraction to decrease (as seen by the decrease in the diameter of the diffraction rings)
• Therefore, the greater the momentum of the particle, the smaller the de Broglie wavelength
• Using ideas based on quantum theory and Einstein’s theory of relativity, de Broglie suggested that the momentum (p) of a particle and its associated wavelength (λ) are related by the equation:
• Combining this with the de Broglie equation gives a form which relates the de Broglie wavelength of a particle to its kinetic energy:
• λ = the de Broglie wavelength (m)
• h = -34 Js" data-title="Planck's Constant" data-toggle="popover">Planck’s constant (J s)
• p = momentum of the particle (kg m s -1 )
• E = kinetic energy of the particle (J)
• m = mass of the particle (kg)
• v = speed of the particle (m s -1 )

Worked example

• Mass of a proton = 1.67 × 10 –27 kg
• Mass of an electron = 9.11 × 10 –31 kg

Step 1: Determine how the proton and electron can be related via their mass

• The only information we are given is the mass of the proton and the electron
• When the proton and electron are accelerated through a potential difference, their kinetic energy will increase
• Therefore, we can use kinetic energy to relate them via their mass

Step 2: Write out the equation for the de Broglie wavelength in terms of the kinetic energy of the particle

• The de Broglie wavelength
• Kinetic energy
• Kinetic energy in terms of momentum
• Substitute the expression for momentum into the de Broglie wavelength equation

Step 3: Find the proportional relationship between the de Broglie wavelength and the mass of the particle

• Since h is constant, and E is equal, then:

Step 4: Calculate the ratio

• This means that the de Broglie wavelength of the proton is 0.0234 times smaller than that of the electron
• Or that the de Broglie wavelength of the electron is about 40 times larger than that of the proton

Particles with a greater mass, such as a proton, have a greater momentum. The greater the momentum, the smaller the de Broglie wavelength. Always perform a logic check on your answer to check that makes sense.

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De Broglie Wavelength Example Problem

Finding the Wavelength of a Moving Particle

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This example problem demonstrates how to find the wavelength of a moving electron using de Broglie's equation .​ While an electron has properties of a particle, the de Broglie equation may be used to describe its wave properties.

What is the wavelength of an electron moving at 5.31 x 10 6 m/sec? Given: mass of electron = 9.11 x 10 -31 kg h = 6.626 x 10 -34 J·s

de Broglie's equation is λ = h/mv λ = 6.626 x 10 -34 J·s/ 9.11 x 10 -31 kg x 5.31 x 10 6 m/sec λ = 6.626 x 10 -34 J·s/4.84 x 10 -24 kg·m/sec λ = 1.37 x 10 -10 m λ = 1.37 Å

The wavelength of an electron moving 5.31 x 10 6 m/sec is 1.37 x 10 -10 m or 1.37 Å.

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• De Broglie Equation

Introduction

The wave nature of light was the only aspect that was considered until Neil Bohr’s model. Later, however, Max Planck in his explanation of quantum theory hypothesized that light is made of very minute pockets of energy which are in turn made of photons or quanta. It was then considered that light has a particle nature and every packet of light always emits a certain fixed amount of energy. 

By this, the energy of photons can be expressed as:

E = hf = h * c/λ

Here, h is Plank’s constant

F refers to the frequency of the waves

Λ implies the wavelength of the pockets

Therefore, this basically insinuates that light has both the properties of particle duality as well as wave. 

Louis de Broglie was a student of Bohr, who then formulated his own hypothesis of wave-particle duality, drawn from this understanding of light. Later on, when this hypothesis was proven true, it became a very important concept in particle physics. 

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What is the De Broglie Equation?

Quantum mechanics assumes matter to be both like a wave as well as a particle at the sub-atomic level. The De Broglie equation states that every particle that moves can sometimes act as a wave, and sometimes as a particle. The wave which is associated with the particles that are moving are known as the matter-wave, and also as the De Broglie wave. The wavelength is known as the de Broglie wavelength. 

For an electron, de Broglie wavelength equation is:       

λ = \[\frac{h}{mv}\]

Here, λ points to the wave of the electron in question

M is the mass of the electron

V is the velocity of the electron

Mv is the momentum that is formed as a result

It was found out that this equation works and applies to every form of matter in the universe, i.e, Everything in this universe, from living beings to inanimate objects, all have wave particle duality. 

Significance of De Broglie Equation

De Broglie says that all the objects that are in motion have a particle nature. However, if we look at a moving ball or a moving car, they don’t seem to have particle nature. To make this clear, De Broglie derived the wavelengths of electrons and a cricket ball. Now, let’s understand how he did this.  

De Broglie Wavelength 

1. De Broglie Wavelength for a Cricket Ball

Let’s say,Mass of the ball  = 150 g (150 x 10⁻³ kg),

Velocity = 35 m/s, 

and  h = 6.626 x 10⁻³⁴ Js

Now, putting these values in the equation 

λ = (6.626 * 10 to power of -34)/ (150 * 10 to power of -3 *35) 

This yields

λBALL = 1.2621 x 10 to the power of -34 m,

Which is 1.2621 x 10 to the power of -24 Å.

We know that Å is a very small unit, and therefore the value is in the power of 10−24−24^{-24}, which is a very small value. From here, we see that the moving cricket ball is a particle.

Now, the question arises if this ball has a wave nature or not. Your answer will be a big no because the value of λBALL is immeasurable. This proves that de Broglie’s theory of wave-particle duality is valid for the moving objects ‘up to’ the size (not equal to the size) of the electrons.

De Broglie Wavelength for an Electron

We know that me  = 9.1 x 10 to power of -31 kg

and ve = 218 x 10 to power of -6 m/s

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.

E  = hν  = \[\frac{hc}{\lambda }\]

The Conclusion of De Broglie Hypothesis

From de Broglie equation for a material particle, i.e.,  

λ = \[\frac{h}{p}\]or \[\frac{h}{mv}\], we conclude the following:

i. If v = 0, then λ = ∞, and

If v = ∞, then λ = 0

It means that waves are associated with the moving material particles only. This implies these waves are independent of their charge. 

FAQs on De Broglie Equation

1.The De Broglie hypothesis was confirmed through which means?

De Broglie had not proved the validity of his hypothesis on his own, it was merely a hypothetical assumption before it was tested out and consequently, it was found that all substances in the universe have wave-particle duality. A number of experiments were conducted with Fresnel diffraction as well as a specular reflection of neutral atoms. These experiments proved the validity of De Broglie’s statements and made his hypothesis come true. These experiments were conducted by some of his students. 

2.What exactly does the De Broglie equation apply to?

In very broad terms, this applies to pretty much everything in the tangible universe. This means that people, non-living things, trees and animals, all of these come under the purview of the hypothesis. Any particle of any substance that has matter and has linear momentum also is a wave. The wavelength will be inversely related to the magnitude of the linear momentum of the particle. Therefore, everything in the universe that has matter, is applicable to fit under the De Broglie equation. 

3.Is it possible that a single photon also has a wavelength?

When De Broglie had proposed his hypothesis, he derived from the work of Planck that light is made up of small pockets that have a certain energy, known as photons. For his own hypothesis, he said that all things in the universe that have to matter have wave-particle duality, and therefore, wavelength. This extends to light as well, since it was proved that light is made up of matter (photons). Hence, it is true that even a single photon has a wavelength. 

4.Are there any practical applications of the De Broglie equation?

It would be wrong to say that people use this equation in their everyday lives, because they do not, not in the literal sense at least. However, practical applications do not only refer to whether they can tangibly be used by everyone. The truth of the De Broglie equation lies in the fact that we, as human beings, also are made of matter and thus we also have wave-particle duality. All the things we work with have wave-particle duality. 

5.Does the De Broglie equation apply to an electron?

Yes, this equation is applicable for every single moving body in the universe, down to the smallest subatomic levels. Just how light particles like photons have their own wavelengths, it is also true for an electron. The equation treats electrons as both waves as well as particles, only then will it have wave-particle duality. For every electron of every atom of every element, this stands true and using the equation mentioned, the wavelength of an electron can also be calculated.  

6.Derive the relation between De Broglie wavelength and temperature.

We know that the average KE of a particle is:

                       K = 3/2 k b T

Where k b is Boltzmann’s constant, and

T   = temperature in Kelvin

The kinetic energy of a particle is  ½ mv²

The momentum of a particle, p = mv = √2mK

= √2m(3/2)KbT = √2mKbT 

de Broglie wavelength, λ = h/p = h√2mkbT 

7.If an electron behaves like a wave, what should determine its wavelength and frequency?

Momentum and energy determine the wavelength and frequency of an electron.

8. Find λ associated with an H 2 of mass 3 a.m.u moving with a velocity of 4 km/s.

Here,  v = 4 x 10³ m/s 

Mass of hydrogen = 3 a.m.u = 3 x 1.67 x 10⁻²⁷kg = 5 x 10⁻²⁷kg    

On putting these values in the equation λ = h/mv we get

λ = (6.626 x 10⁻³⁴)/(4 x 10³ x 5 x 10⁻²⁷) = 3 x 10⁻¹¹ m.

9. If the KE of an electron increases by 21%, find the percentage change in its De Broglie wavelength.

We know that  λ = h/√2mK

So,  λ i = h/√(2m x 100) , and λ f = h/√(2m x 121)

% change in λ is:

Change in wavelength/Original x 100 = (λ fi - λ f )/λ i = ((h/√2m)(1/10 - 1/21))/(h/√2m)(1/10) 

On solving, we get

% change in λ = 5.238 %

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De-Broglie Wavelength Formula: Equations, Solved Examples

De-Broglie Wavelength Formula – Einstein proposed that any electromagnetic radiation, including light which was, till then, considered an electromagnetic wave, in fact, showed particle-like nature. He coined the word “photon” for the quanta or particle of light. Soon, scientists began to wonder if other particles could also have a dual wave-particle nature.

In 1924, French scientist Louis de Broglie derived an equation, known as the De Broglie Wavelength Formula, that described the wave nature of any particle. Thus, establishing the wave-particle duality for the matter. Microscopic particle-like electrons also proved to possess this dual nature property. Let us learn about the equation proposed by de-Broglie in detail in this article.

De Broglie’s Hypothesis and Equation

Louis-de-Broglie explained the concept of de-Broglie waves in the year 1923. In his thesis, he suggested that any moving particle, whether microscopic or macroscopic, will be related to a wave character. This was later experimented with and proved by Davisson and Germer within the year 1927. The waves associated with matter were called ‘Matter Waves’. These waves explain the character of the wave associated with the particle. We know that electromagnetic radiation exhibit the dual nature of a particle (having a momentum) and wave (expressed in frequency, and wavelength).

He further proposed a relation between the speed and momentum with the wavelength if the particle had to behave as a wave. Since, at non-relativistic speeds, the momentum of a particle will be adequate to its mass \(\text {m}\), multiplied by its velocity \(\text {v}\). Thus, according to de-Broglie, the wavelength \(\left( \lambda \right)\) of any moving object is given by:

\(\lambda = \frac{{\rm{h}}}{{\rm{p}}}\)

Where \(\text {h}\) is Planck’s constant and \(\text {p}\) is the momentum of the particle.

Derivation of De Broglie’s Wavelength

From Einstein’s relation of mass-energy equivalence, we know that,

\({\text{E}} = {\text{m}}{{\text{c}}^2} \cdots (1)\)

\(\text {E}=\) energy of the particle

\(\text {m}=\) mass of the particle

\(\text {c}=\) speed of light

According to Planck’s theory, every quantum of a wave has a discrete amount of energy associated with it, and he gave the equation:

\({\text{E}} = {\text{hf}} \cdots (2)\)

\(\text {E}:\) energy of the particle

\(\text {h}=6.62607 \times 10^{-34} \mathrm{Js}:\) Planck’s constant

\(\text {f}=\) frequency

De-Broglie’s hypothesis suggested that particles and waves behave as similar entities. Thus, he equated the energy relation for both particle and wave; equating equations \((1)\) and \((2)\), we get:

\(\text {mc}^{2}=\text {hf}\)

Since the particles generally do not travel at the speed of light, De Broglie substituted the speed of light \(\text {c}\), with the velocity of a real particle \(\text {v}\), and obtained:

\({\text{m}}{{\text{v}}^2} = {\text{hf}} \cdots (3)\)

If \(\lambda \) be the wavelength of the wave, then the frequency will be: \({\rm{f}} = \frac{{\rm{v}}}{\lambda }\)

 Substituting this in equation \((3)\), we get:

\({\rm{m}}{{\rm{v}}^2} = \frac{{{\rm{hv}}}}{\lambda }\)

\(\lambda = \frac{{\rm{h}}}{{{\rm{mv}}}}\)

or, \(\lambda = \frac{{\rm{h}}}{{\rm{p}}}\,\,\, \cdots (4)\)

Where \(\text {p}\) is the momentum of the particle.

De Broglie Wavelength and Kinetic Energy

The kinetic energy of an object of mass \(\text {m}\) moving with velocity \(\text {v}\) is given as:

\({\text{K}} = \frac{1}{2}{\text{m}}{{\text{v}}^2}\)

or, \({\text{K}} = \frac{1}{2}{\text{mv}} \cdot {\text{v}}\)

\({\text{m}} \cdot {\text{K}} = \frac{1}{2}{({\text{mv}})^2}\)

Since, \(\text {p}=\text {mv}\), Thus:

\({\text{m}}.{\text{K}} = \frac{1}{2}{({\text{p}})^2}\)

From equation \((4),\,{\rm{p}} = \frac{{\rm{h}}}{\lambda }\)

\( \Rightarrow {\rm{m}} \cdot {\rm{K}} = \frac{1}{2}{\left( {\frac{{\rm{h}}}{\lambda }} \right)^2}\)

\({\lambda ^2} = \frac{{{{\rm{h}}^2}}}{{2\,{\rm{mK}}}}\)

\(\lambda = \frac{{\rm{h}}}{{\sqrt {2\,{\rm{mK}}} }}\)

De-Broglie Wavelength and Potential

When a charged particle, having a charge \(\text {q}\) is accelerated through an external potential difference \(\text {V}\), the energy of the particle can be given as:

\({\text{E}} = {\text{qV}} \cdots ({\text{i}})\)

According to Planck’s equation,

\(\text {E}=\text {hf}\)

Since, \({\rm{f}} = \frac{{\rm{v}}}{\lambda }\)

Therefore, \({\rm{E}} = {\rm{h}}\frac{{\rm{v}}}{\lambda }\,\,\, \ldots ({\rm{ii}})\)

Equating the equations \(\left({\text{i}} \right)\) and \(\left({\text{ii}} \right)\),

\({\rm{qV}} = {\rm{h}}\frac{{\rm{v}}}{\lambda }\)

or, \(\lambda = \frac{{{\rm{hv}}}}{{{\rm{qV}}}}\)

Thermal De Broglie Wavelength

There exists a relation between the De-Broglie equation and the temperature of the given gas molecules, and the thermal de Broglie wavelength gives it \(\left( {{\lambda _{{\rm{Th}}}}} \right).\) The Thermal de Broglie equation represents the average value of the de Broglie wavelength of the gas particles at the specified temperature in an ideal gas.

The expression gives the thermal de Broglie wavelength at temperature \(\text {T}\):

\({\lambda _{{\rm{Th}}}} = \lambda  = \frac{{\rm{h}}}{{\sqrt {2\,{\rm{m}}{{\rm{k}}_{\rm{B}}}{\rm{T}}} }}\)

\(h=\) Planck constant

\(m=\) mass of a gas particle

\({{\text{k}}_{\text{B}}} = \) Boltzmann constant

De Broglie Wavelength of Day-to-Day Objects

According to the hypothesis, all particles have a wave associated with them. That is true for us humans and the objects around us. To get an idea of the de-Broglie wavelength associated with macroscopic particles:

Let us find the wavelength of a wave associated with a car of mass \(1000 \mathrm{~kg}\) moving with the velocity of \(10 \mathrm{~m} / \mathrm{s}.\)

The wavelength associated with the car will be: \(\lambda = \frac{{\rm{h}}}{{{\rm{mv}}}}\)

\(\lambda=\frac{6.62607 \times 10^{-34} \mathrm{Js}}{1000 \mathrm{~kg} \times 10 \mathrm{~m} / \mathrm{s}}=6.6 \times 10^{-30} \mathrm{~m}=6.6 \times 10^{-21} \mathrm{~nm}\)

Thus, the value of wavelength associated with this car is insignificant.

Similarly, for other macroscopic objects with large mass values, the wavelength associated with them is so small that it can not be detected.

De-Broglie Wavelength of an Electron

As we have seen above, the matter waves associated with real objects is so small that it is of no good use to us. But for sub-atomic particles with negligible masses, the value of de-Broglie wavelength is substantial. To calculate the de-Broglie wavelength associated with a microscopic particle,

Let us take an electron of mass \({\rm{m}} = 9.1 \times {10^{ – 31}}\;{\rm{kg}},\) moving with the speed of light, i.e., \(\text {c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\), then the de-Broglie wavelength associated with it can be given as:

\(\lambda = \frac{{\rm{h}}}{{{\rm{mc}}}}\)

\(\lambda=\frac{6.62607 \times 10^{-34} \mathrm{Js}}{9.1 \times 10^{-31} \mathrm{~kg} \times 3 \times 10^{8} \mathrm{~m} / \mathrm{s}}=0.7318 \times 10^{-11} \mathrm{~m}=0.073 \mathrm{~A}^{\circ}\)

This is a substantial value. Thus, the de-Broglie wavelength associated has a significant value, and it can be detected. 

Relation Between De-Broglie Wavelength and Potential (for an electron)

The expression for the de-Broglie wavelength of an electron,

If the electron having a charge e is moving under an external potential \(\text {V}\), then,

The kinetic energy of the electron, \({\text{K}} = {\text{eV}}\)

Substituting this expression in the above equation,

\(\lambda = \frac{{\rm{h}}}{{\sqrt {2\,{\rm{meV}}} }}\)

Put, \(h=6.62607 \times 10^{-34} \mathrm{Js}\)

\(\text {e}=1.6 \times 10^{-19} \mathrm{C}\)

\(\text {m}=9.1 \times 10^{-31} \mathrm{~kg}\)

\(\lambda = \frac{{12.27}}{{\sqrt {\rm{V}} }}{\rm{A}}^\circ \)

De Broglie Wavelength- Solved Problems

Q.1. A certain photon has a momentum of \(1.50 \times 10^{-27} \mathrm{~kg} \mathrm{~m} / \mathrm{s}\). What will be the photon’s de Broglie wavelength?

Ans: \(\text {p}=1.50 \times 10^{-27} \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) Plank’s constant, \(\text {h}=6.62607 \times 10^{-34} \mathrm{Js}\) The de Broglie wavelength of the photon can be computed using the formula: \(\lambda = \frac{{\rm{h}}}{{\rm{p}}}\) \(=\frac{6.62607 \times 10^{-34} \mathrm{Js}}{1.50 \times 10^{-27} \mathrm{kgm} / \mathrm{s}}\) \(=4.42 \times 10^{-7} \mathrm{~m}\) \(=442 \times 10^{-9} \mathrm{~m}\) \(=442 \mathrm{~nm}\) The de Broglie wavelength of the photon will be \(442 \mathrm{~nm}\), and this wavelength lies in the blue-violet part of the visible light spectrum.

Q.2. What is the de Broglie wavelength of an electron which is accelerated through a potential difference of \(10\, \mathrm{kV}\) ?

Ans:  If the electron having a charge \(\text {e}\) is moving under an external potential \(\text {V}\), then the expression for the de-Broglie wavelength of an electron is: \(\lambda = \frac{{12.27}}{{\sqrt {\rm{V}} }}{\rm{A}}^\circ \) We are given, \(\text {V}=10 \mathrm{kV}=10 \times 10^{3} \mathrm{~V}=10^{4} \mathrm{~V}\) \(\lambda = \frac{{12.27}}{{\sqrt {{{10}^4}} }}{\rm{A}}^\circ \) \(\lambda = \frac{{12.27}}{{100}}{\rm{A}}^\circ = 0.1227{\mkern 1mu} {\rm{A}}^\circ \)

According to de-Broglie, the wavelength \(\left( \lambda \right)\) of any moving object is given by: \(\lambda = \frac{{\rm{h}}}{{\rm{p}}},\) Where \(\text {h}\) is Planck’s constant and \(\text {p}\) is the mass of the particle.

The relation between de-Broglie wavelength and the kinetic energy of an object of mass \(\text {m}\) moving with velocity \(\text {v}\) is given as: \(\lambda = \frac{{\rm{h}}}{{\sqrt {2\,{\rm{mK}}} }}\)

When a charged particle having a charge \(\text {q}\) is accelerated through an external potential difference \(\text {V}\), de-Broglie wavelength, \(\lambda = \frac{{{\rm{hv}}}}{{{\rm{qV}}}}\)

The expression for the de-Broglie wavelength of an electron, \(\lambda = \frac{{\rm{h}}}{{\sqrt {2\,{\rm{mK}}} }}\) or \(\lambda = \frac{{\rm{h}}}{{\sqrt {2\,{\rm{meV}}} }}\)

Frequently Asked Questions on De Broglie Wavelength Formula

The most commonly asked questions on De Broglie Wavelength Formula are answered here:

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What is De Broglie Hypothesis?

De broglie's hypothesis says that matter consists of both the particle nature as well as wave nature. de broglie wavelength λ is given as λ = h p , where p represents the particle momentum and can be written as: λ = h m v where, h is the planck's constant, m is the mass of the particle, and v is the velocity of the particle. from the above relation, it can be said that the wavelength of the matter is inversely proportional to the magnitude of the particle's linear momentum. this relation is applicable to both microscopic and macroscopic particles the de broglie equation is one of the equations that is commonly used to define the wave properties of matter. electromagnetic radiation exhibits the dual nature of a particle (having a momentum) and wave (expressed in frequency, and wavelength)..

Explain De Broglie’s Hypothesis. - Physics

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Explain De Broglie’s Hypothesis.

Solution Show Solution

• De Broglie proposed that a moving material particle of total energy E and momentum p has a wave associated with it (analogous to a photon).
• He suggested a relation between properties of the wave, like frequency and wavelength, with that of a particle, like energy and momentum. p = `"E"/"c" = "hv"/"c" = "h"/lambda`
• Thus, the frequency and wavelength of a wave associated with a material particle, of mass m moving with a velocity v, are given as `"v" = "E"/"h"` and `lambda = "h"/"p" = "h"/"mv"` ….(1)
• De Broglie referred to these waves associated with material particles as matter waves. The wavelength of the matter waves, given by equation (1), is now known as de Broglie wavelength and the equation is known as de Broglie relation. 

RELATED QUESTIONS

An electron, a proton, an α-particle, and a hydrogen atom are moving with the same kinetic energy. The associated de Broglie wavelength will be longest for ______.

State the importance of Davisson and Germer experiment.

What is the speed of a proton having de Broglie wavelength of 0.08 Å?

Explain what you understand by the de Broglie wavelength of an electron. Will an electron at rest have an associated de Broglie wavelength? Justify your answer.

The de Broglie wavelengths associated with an electron and a proton are the same. What will be the ratio of (i) their momenta (ii) their kinetic energies?

Two particles have the same de Broglie wavelength and one is moving four times as fast as the other. If the slower particle is an α-particle, what are the possibilities for the other particle?

Find the ratio of the de Broglie wavelengths of an electron and a proton when both are moving with the (a) same speed, (b) the same kinetic energy, and (c) the same momentum. State which of the two will have a longer wavelength in each case.

According to De-Broglie, the waves are associated with ______ 

An electron is accelerated through a potential of 120 V. Find its de Broglie wavelength.

Calculate De Broglie's wavelength of the bullet moving with speed 90m/sec and having a mass of 5 gm. 

If the radius of the innermost Bohr orbit is 0.53 Å, the radius of the 4 th orbit is ______

According to de-Broglie hypothesis, the wavelength associated with moving electron of mass 'm' is 'λ e '· Using mass energy relation and Planck's quantum theory, the wavelength associated with photon is  'λ p '. If the energy (E) of electron and photon is same then relation between 'λ e ' and 'λ p ' is ______.

An electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelengths associated with them is ( c being velocity of light) ______.

If the radius of the circular path and frequency of revolution of a particle of mass m are doubled, then the change in its kinetic energy will be (E i and E f are the initial and final kinetic energies of the particle respectively.)

A particle of charge q, mass m and energy E has de-Broglie wavelength `lambda.` For a particle of charge 2q, mass 2m and energy 2E, the de-Broglie wavelength is ____________.

How much energy is imparted to an electron so that its de-Broglie wavelength reduces from 10 -10 m to 0.5 × 10 -10 m? (E =energy of electron)

If '`lambda_1`' and '`lambda_2`' are de-Broglie wavelengths for electrons in first and second Bohr orbits in hydrogen atom, then the ratio '`lambda_2`' to '`lambda_1`' is (E 1 = -13.6 eV) ____________.

If the kinetic energy of a particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is ____________.

Graph shows the variation of de-Broglie wavelength `(lambda)` versus `1/sqrt"V"`, where 'V' is the accelerating potential for four particles carrying same charge but of masses m 1 , m 2 , m 3 , m 4 . Which particle has a smaller mass?

If the potential difference used to accelerate electrons is doubled, by what factor does the de-Broglie wavelength associated with the electrons change?

Explain de-Broglie wavelength.

Obtain an expression for de-Broglie wavelength of wave associated with material particles. The photoelectric work function for metal is 4.2 eV. Find the threshold wavelength.

Calculate the de Broglie wavelength associated with an electron moving with a speed of `5 xx 10^6` m/s. `(m_e = 9.1 xx 10^(-31)kg)`

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