## Pathways to Chemistry

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## The De Broglie Hypothesis

Louis de Broglie (1892 – 1987) was a French physicist who extended the ideas of radiant energy to matter. He suggested that matter has wavelike properties. To develop his theory, de Broglie looked at the inverse relationship between the wavelength and energy for photons.

$$\displaystyle E\;=\;\frac{hc}{\lambda}\;\;\;and\;\;\;\lambda\;=\;\frac{hc}{E}$$

We can substitute Einstein’s equation, E = mc 2 for E,

$$\displaystyle \lambda\;=\;\frac{hc}{mc^2}\;=\;\frac{h}{mc}$$

and then replace the speed of the particle, ν, for c.

$$\displaystyle \lambda\;=\;\frac{h}{m\nu}\;\;\;\;(de\; Broglie\; Equation)$$

The quantity, mν, is the momentum of an object. The mass, m, has units of kJ and the speed, ν has units of m/s. The term matter wave is used to relate wave properties to matter. An example of a matter wave is an electron beam. Matter waves are different from electromagnetic waves because they are the result of the motion of matter. There are no significant fields associated with matter waves whereas with electromagnetic waves there is an electrical and a magnetic field. Matter waves cannot propagate through a vacuum, therefore, the speeds, ν, are less than the speed of light, c. Matter is composed of particles and de Broglie stated that all matter possesses wave like characteristics.

If we have an electron with a mass of 9.11 x 10 -31 kg moving at 2.3 x 10 6 m/s, the wavelength is:

$$\displaystyle \lambda\;=\;\frac{6.626\times\;10^{-34}\;\frac{kg⋅m^2}{s^2}⋅s}{9.11\times\;10^{-31}\;kg\;\times(2.3\times\;10^6\;m/s)}\;=\;3.2\times\;10^{-10}\;m$$

Recall, 1 J = kg⋅m 2 /s 2 . The wavelength, 3.2 x 10 -10 m, is easily measured in the lab. If we have a golf ball with a mass of 0.0457 kg moving at a speed of 53.6 m/s, the wavelength is 2.7 x 10 -34 m. This is not a measurable wavelength — it is much too small. We have never observed such small wavelengths.

Below is an electron microscope image of SARS-CoV-2 emanating from cultured cells. An electron microscope uses a beam of high speed electrons in order to image very small objects like cells and viruses.

Exercise 1 What is the wavelength, in meters, of an electron (mass = 9.11 x 10 -31 kg) that is accelerated to 4.5% the speed of light, c?

Exercise 2 What is the wavelength of a butterfly with a mass of 0.500 g and a speed of 6.0 m/s? Can we observe this wavelength?

Exercise 3 What speed, ν, would an electron (mass = 9.11 x 10 -28 g) be traveling to have a de Broglie wavelength of 625 nm?

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## de Broglie Equation Definition

Chemistry Glossary Definition of de Broglie Equation

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In 1924, Louis de Broglie presented his research thesis, in which he proposed electrons have properties of both waves and particles, like light. He rearranged the terms of the Plank-Einstein relation to apply to all types of matter.

The de Broglie equation is an equation used to describe the wave properties of matter , specifically, the wave nature of the electron :​ λ = h/mv , where λ is wavelength, h is Planck's constant, m is the mass of a particle, moving at a velocity v. de Broglie suggested that particles can exhibit properties of waves.

The de Broglie hypothesis was verified when matter waves were observed in George Paget Thomson's cathode ray diffraction experiment and the Davisson-Germer experiment, which specifically applied to electrons. Since then, the de Broglie equation has been shown to apply to elementary particles, neutral atoms, and molecules.

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## 6.6: De Broglie’s Matter Waves

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## 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 $$λ$$. 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 $$\lambda$$:

$E = h f \label{6.53}$

$\lambda = \frac{h}{p} \label{6.54}$

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 $$\vec{k}$$, $$k = 2 \pi / \lambda$$, and the wave frequency $$\omega = 2 \pi f$$, as we usually do for waves:

\begin{aligned} &E=\hbar \omega \label{6.55}\\ &\vec{p}=\hbar \vec{k} \label{6.56} \end{aligned}

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

$\lambda f =\frac{\omega}{k}=\frac{E / \hbar}{p / \hbar}=\frac{E}{p} = \frac{m c^{2}}{m u}=\frac{c^{2}}{u}=\frac{c}{\beta} \label{6.57}$

where $$\beta = u/c$$. When a particle is massless we have $$u=c$$ and Equation \ref{6.57} becomes $$\lambda f = c$$.

## Example $$\PageIndex{1}$$: How Long are de Broglie Matter Waves?

Calculate the de Broglie wavelength of:

• a 0.65-kg basketball thrown at a speed of 10 m/s,
• a nonrelativistic electron with a kinetic energy of 1.0 eV, and
• a relativistic electron with a kinetic energy of 108 keV.

We use Equation \ref{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 $$\beta=u / c \ll 1$$, we use nonrelativistic momentum p . When the nonrelativistic approximation cannot be used, such as in (c), we must use the relativistic momentum $$p=m u=m_{0} \gamma u=E_{0} \gamma \beta/c$$, where the rest mass energy of a particle is $$E_0 = m c^2$$ and $$\gamma$$ is the Lorentz factor $$\gamma=1 / \sqrt{1-\beta^{2}}$$. The total energy $$E$$ of a particle is given by Equation \ref{6.53} and the kinetic energy is $$K=E-E_{0}=(\gamma-1) E_{0}$$. When the kinetic energy is known, we can invert Equation 6.4.2 to find the momentum

$p=\sqrt{\left(E^{2}-E_{0}^{2}\right) / c^{2}}=\sqrt{K\left(K+2 E_{0}\right)} / c \nonumber$

and substitute into Equation \ref{6.57} to obtain

$\lambda=\frac{h}{p}=\frac{h c}{\sqrt{K\left(K+2 E_{0}\right)}} \label{6.58}$

Depending on the problem at hand, in this equation we can use the following values for hc :

$h c=\left(6.626 \times 10^{-34} \: \mathrm{J} \cdot \mathrm{s}\right)\left(2.998 \times 10^{8} \: \mathrm{m} / \mathrm{s}\right)=1.986 \times 10^{-25} \: \mathrm{J} \cdot \mathrm{m}=1.241 \: \mathrm{eV} \cdot \mu \mathrm{m} \nonumber$

• For the basketball, the kinetic energy is $K=m u^{2} / 2=(0.65 \: \mathrm{kg})(10 \: \mathrm{m} / \mathrm{s})^{2} / 2=32.5 \: \mathrm{J} \nonumber$ and the rest mass energy is $E_{0}=m c^{2}=(0.65 \: \mathrm{kg})\left(2.998 \times 10^{8} \: \mathrm{m} / \mathrm{s}\right)^{2}=5.84 \times 10^{16} \: \mathrm{J} \nonumber$ We see that $$K /\left(K+E_{0}\right) \ll 1$$ and use $$p=m u=(0.65 \: \mathrm{kg})(10 \: \mathrm{m} / \mathrm{s})=6.5 \: \mathrm{J} \cdot \mathrm{s} / \mathrm{m}$$: $\lambda=\frac{h}{p}=\frac{6.626 \times 10^{-34} \: \mathrm{J} \cdot \mathrm{s}}{6.5 \: \mathrm{J} \cdot \mathrm{s} / \mathrm{m}}=1.02 \times 10^{-34} \: \mathrm{m} \nonumber$
• For the nonrelativistic electron, $E_{0}=mc^{2}=\left(9.109 \times 10^{-31} \mathrm{kg}\right)\left(2.998 \times 10^{8} \mathrm{m} / \mathrm{s}\right)^{2}=511 \mathrm{keV} \nonumber$ and when $$K = 1.0 \: eV$$, we have $$K/(K+E_0) = (1/512) \times 10^{-3} \ll 1$$, so we can use the nonrelativistic formula. However, it is simpler here to use Equation \ref{6.58}: $\lambda=\frac{h}{p}=\frac{h c}{\sqrt{K\left(K+2 E_{0}\right)}}=\frac{1.241 \: \mathrm{eV} \cdot \mu \mathrm{m}}{\sqrt{(1.0 \: \mathrm{eV})[1.0 \: \mathrm{eV}+2(511 \: \mathrm{keV})]}}=1.23 \: \mathrm{nm} \nonumber$ 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 $$K=108 \: keV$$, relativistic effects cannot be neglected because its total energy is $$E = K = E_0 = 108 \: keV + 511 \: keV = 619 \: keV$$ and $$K/E = 108/619$$ is not negligible: $\lambda=\frac{h}{p}=\frac{h c}{\sqrt{K\left(K+2 E_{0}\right)}}=\frac{1.241 \: \mathrm{eV} \cdot \mu \mathrm{m}}{\sqrt{108 \: \mathrm{keV}[108 \: \mathrm{keV}+2(511 \: \mathrm{keV})]}}=3.55 \: \mathrm{pm} \nonumber$.

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.

## Exercise $$\PageIndex{1}$$

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 $$\PageIndex{1}$$), the wavelengths of these vibrations cannot be arbitrary but must be such that an integer k number of half-wavelengths $$\lambda/2$$ fit exactly on the distance l between the ends. This is the condition $$l=k \lambda /2$$ 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 $$k$$, $$k=2n$$, and the length l is now connected to the radius $$r_n$$ of the circle. This means that the radii are not arbitrary but must satisfy the following standing-wave condition:

$2 \pi r_{n}=2 n \frac{\lambda}{2} \label{6.59}.$

If an electron in the n th Bohr orbit moves as a wave, by Equation \ref{6.59} its wavelength must be equal to $$\lambda = 2 \pi r_n / n$$. Assuming that Equation \ref{6.58} is valid, the electron wave of this wavelength corresponds to the electron’s linear momentum, $$p = h/\lambda = nh / (2 \pi r_n) = n \hbar /r_n$$. In a circular orbit, therefore, the electron’s angular momentum must be

$L_{n}=r_{n} p=r_{n} \frac{n \hbar}{r_{n}}=n \hbar \label{6.60} .$

This equation is the first of Bohr’s quantization conditions, given by Equation 6.5.6 . Providing a physical explanation for Bohr’s quantization condition is a convincing theoretical argument for the existence of matter waves.

## Example $$\PageIndex{2}$$: The Electron Wave in the Ground State of Hydrogen

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

We combine the first quantization condition in Equation \ref{6.60} with Equation 6.5.6 and use Equation 6.5.9 for the first Bohr radius with $$n = 1$$.

When $$n=1$$ and $$r_n = a_0 = 0.529 \: Å$$, the Bohr quantization condition gives $$a_{0} p=1 \cdot \hbar \Rightarrow p=\hbar / a_{0}$$. The electron wavelength is:

$\lambda=h / p = h / \hbar / a_{0} = 2 \pi a_{0} = 2 \pi(0.529 \: Å)=3.324 \: Å .\nonumber$

We obtain the same result when we use Equation \ref{6.58} directly.

## Exercise $$\PageIndex{2}$$

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

$$\lambda = 2 \pi n a_0 = 2 (3.324 \: Å) = 6.648 \: Å$$

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 $$\PageIndex{2}$$ 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, 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:

$e \Delta V=K=\frac{p^{2}}{2 m} \Rightarrow p=\sqrt{2 m e \Delta V} \label{6.61}$

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° to 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 $$\PageIndex{3}$$). 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 $$\lambda$$ 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 $$a=2.15 \: Å$$. 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 = a sin φ (see Figure $$\PageIndex{4}$$). The first-order maximum (for n=1) is measured at a scattering angle of φ≈50° at ΔV≈54 V, which gives the wavelength of the incident radiation as 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. Their momentum, calculated from Equation \ref{6.61}, is $$p = 2.478 \times 10^{−5} \: eV \cdot s/m$$. When we substitute this result in Equation \ref{6.58}, the de Broglie wavelength is obtained as

$\lambda=\frac{h}{p}=\frac{4.136 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s}}{2.478 \times 10^{-5} \mathrm{eV} \cdot \mathrm{s} / \mathrm{m}}=1.67 \mathrm{Å} \label{6.62}.$

The same result is obtained when we use K = 54eV in Equation \ref{6.61}. The proximity of this theoretical result to the Davisson–Germer experimental value of 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 (Figure $$\PageIndex{3}$$) 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 (Figure $$\PageIndex{5}$$).

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 $$C_{60}$$ 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 $$\PageIndex{3A}$$: 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.

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 λ must be on the same order of magnitude as the lattice spacing. We use Equation \ref{6.61} to find the momentum p and kinetic energy K . To compare this energy with the energy $$E_T$$ of ideal gas in equilibrium at room temperature $$T = 300 \, K$$, we use the relation $$K = 3/2 k_BT$$, where $$k_B = 8.62 \times 10^{-5}eV/K$$ is the Boltzmann constant.

We evaluate pc to compare it with the neutron’s rest mass energy $$E_0 = 940 \, MeV$$:

$p = \frac{h}{\lambda} \Rightarrow pc = \frac{hc}{\lambda} = \frac{1.241 \times 10^{-6}eV \cdot m}{10^{-10}m} = 12.41 \, keV. \nonumber$

We see that $$p^2c^2 << E_0^2$$ and we can use the nonrelativistic kinetic energy:

$K = \frac{p^2}{2m_n} = \frac{h^2}{2\lambda^2 m_n} = \frac{(6.63\times 10^{−34}J \cdot s)^2}{(2\times 10^{−20}m^2)(1.66 \times 10^{−27} kg)} = 1.32 \times 10^{−20} J = 82.7 \, meV. \nonumber$

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

$K_T = \frac{3}{2}k_BT = \frac{3}{2} (8.62 \times 10^{-5}eV/K)(300 \, K) = 38.8 \, MeV. \nonumber$

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

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

## Example $$\PageIndex{3B}$$: Wavelength of a Relativistic Proton

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

The rest mass energy of a proton is $$E_0 = m_0c^2 = (1.672 \times 10^{−27} kg)(2.998 \times 10^8m/s)^2 = 938 \, MeV$$. When the proton’s velocity is known, we have β = 0.75 and $$\beta \gamma = 0.75 / \sqrt{1 - 0.75^2} = 1.714$$. We obtain the wavelength λλ and kinetic energy K from relativistic relations.

$\lambda = \frac{h}{p} = \frac{hc}{\beta \gamma E_0} = \frac{1.241 \, eV \cdot \mu m}{1.714 (938 \, MeV)} = 0.77 \, fm \nonumber$

$K = E_0(\gamma - 1) = 938 \, MeV (1 /\sqrt{1 - 0.75^2} - 1) = 480.1\, MeV \nonumber$

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)0.77 fm = 1.4 pm and its kinetic energy of 480.1 MeV /1835 = 261.6 keV.

## Exercise $$\PageIndex{3}$$

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

$$\lambda = 1.417 \, pm; \, K = 261.56 \, keV$$

## De Broglie Hypothesis

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.

## de Broglie equation, derivation, and its Significance

• November 10, 2021
• Atomic Structure

In 1905, Albert Einstein proposed that light can behave both as a wave and as a particle, implying that it has a dual character.  The French physicist  Louis de Broglie  suggested (1924) a hypothesis to explain the theory of atomic structure, and is therefore the hypothesis is also referred to as the de Broglie hypothesis. The de Broglie hypothesis states that any moving particle/object is associated with both the wave properties as well as particle properties . Later on, Davisson and Germer proved the hypothesis experimentally while studying the diffraction of electrons.

## de Broglie Equation Definition

According to de Broglie equation/hypothesis, “ a matter particle in motion is also associated with waves .” In other words, any moving microscopic or macroscopic particle will be associated with a wave character. Such waves are also called matter waves or de Broglie waves. de Broglie equation is basically used to define the wave properties of matter/electron. Thus, the matter particle-like electrons show dual character i.e. it behaves like a particle as well as a wave.

According to de-Broglie, the particle with wavelength and mass ‘m’ moving with velocity ‘v’ is represented by the relation:

## de Broglie equation significance

The de-Broglie equation is significant only for sub-microscopic objects in the range of atoms, molecules, or smaller sub-atomic particles. The wave nature of matter, on the other hand, has no meaning for ordinary-sized objects because the wavelength of the wave associated with them is too small to detect.

## What is h in de broglie equation?

h in de Broglie equation is Planck’s constant having a value of 6.62607 x 10 -34  J s

## What is the significance of de Broglie equation?

The de-Broglie equation is significant only for sub-microscopic particles that are in the range of atoms, molecules, or smaller sub-atomic particles.

## Derive de Broglie equation for microscopic particle.

According to de Broglie equation/hypothesis, “ a matter particle in motion is also associated with waves.”

• Tags: Atomic Structure , de Broglie equation , de Broglie equation definition chemistry , de Broglie equation derivation , de Broglie equation significance , de Broglie hypothesis , de Broglie wave equation , de Broglie wave equation derivation , de Broglie wavelength equation

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## 38 DeBroglie, Intro to Quantum Mechanics, Quantum Numbers 1-3 (M7Q5)

Introduction.

Bohr’s model explained the experimental data for the hydrogen atom and was widely accepted, but it also raised many questions. Why did electrons orbit at only fixed distances defined by a single quantum number  n = 1, 2, 3, and so on, but never in between? Why did the model work so well describing hydrogen and one-electron ions, but could not correctly predict the emission spectrum for helium or any larger atoms? The goal of this section is to answer these questions by introducing electron orbitals, their different energies, and other properties. This section includes worked examples, sample problems, and a glossary.

Learning Objectives for DeBroglie, Intro to Quantum Mechanics, Quantum Numbers 1-3

• Illustrate how the diffraction of electrons reveals the wave properties of matter. | Behavior of the Microscopic World and De Broglie Wavelength | Interference Patterns are a Hallmark of Wavelike Behavior |
• Recognize that quantum theory leads to discrete energy levels and associated wavefunctions and explain their probabilistic interpretation. | The Quantum Mechanical Model of an Atom |
• Describe the energy levels and wave functions for the hydrogen atom using three quantum numbers. | Principal Quantum Number | Azimuthal Quantum Number | Magnetic Quantum Number | Table of Atomic Orbital Quantum Numbers |

| Key Concepts and Summary | Glossary | End of Section Exercises |

## Behavior in the Microscopic World and De Broglie Wavelength

We know how matter behaves in the macroscopic world—objects that are large enough to be seen by the naked eye follow the rules of classical physics. A billiard ball moving on a table will behave like a particle: It will continue in a straight line unless it collides with another ball or the table cushion, or is acted on by some other force (such as friction). The ball has a well-defined position and velocity (or a well-defined momentum, p = mv, defined by mass m and velocity v ) at any given moment. In other words, the ball is moving in a classical trajectory. This is the typical behavior of a classical object.

One of the first people to pay attention to the special behavior of the microscopic world was Louis de Broglie. He asked the question: If electromagnetic radiation can have particle-like character (as we saw in an earlier section ), can electrons and other submicroscopic particles exhibit wavelike character? In his 1925 doctoral dissertation, de Broglie extended the wave–particle duality of light that Einstein used to resolve the photoelectric effect paradox to material particles. He predicted that a particle with mass m and velocity v (that is, with linear momentum p ) should also exhibit the behavior of a wave with a wavelength value λ , given by this expression in which h is the familiar Planck’s constant:

λ  =  $\frac{h}{mv}$  =  $\frac{h}{p}$

This quantity is called the de Broglie wavelength. Unlike the other values of λ discussed in this chapter, the de Broglie wavelength is a characteristic of particles and other bodies, not electromagnetic radiation (note that this equation involves velocity [ v , with units m/s], not frequency [ ν , with units Hz]. Although these two symbols are similar, they mean very different things). Where Bohr had postulated the electron as being a particle orbiting the nucleus in quantized orbits, de Broglie argued that Bohr’s assumption of quantization can be explained if the electron is considered not as a particle, but rather as a circular standing wave such that only an integer number of wavelengths could fit exactly within the orbit ( Figure 1 ).

## Interference Patterns are a Hallmark of Wavelike Behavior

Shortly after de Broglie proposed the wave nature of matter, two scientists at Bell Laboratories, C. J. Davisson and L. H. Germer, demonstrated experimentally that electrons can exhibit wavelike behavior by showing an interference pattern for electrons reflecting off a crystal. The same interference pattern is also observed when electrons travel through a regular atomic pattern in a crystal. The regularly spaced atomic layers served as slits that diffract the electrons, as used in other interference experiments. Since the spacing between the layers serving as slits needs to be similar in size to the wavelength of the tested wave for an interference pattern to form, Davisson and Germer used a crystalline nickel target for their “slits,” since the spacing of the atoms within the lattice was approximately the same as the de Broglie wavelengths of the electrons that they used. Figure 2 shows an interference pattern. The wave–particle duality of matter can be seen in Figure 2 by observing what happens if electron collisions are recorded over a long period of time. Initially, when only a few electrons have been recorded, they show clear particle-like behavior, having arrived in small localized packets that appear to be random. As more and more electrons arrived and were recorded, a clear interference pattern that is the hallmark of wavelike behavior emerged. Thus, it appears that while electrons are small localized particles, their motion does not follow the equations of motion implied by classical mechanics, but instead it is governed by some type of a wave equation that governs a probability distribution even for a single electron’s motion. Thus the wave–particle duality first observed with photons is actually a fundamental behavior intrinsic to all quantum particles.

View the Dr. Quantum – Double Slit Experiment cartoon for an easy-to-understand description of wave–particle duality and the associated experiments.

## Chemistry in Real Life: Dorothy Hodgkin

Because the wavelengths of X-rays (10-10,000 picometers [pm]) are comparable to the size of atoms, X-rays can be used to determine the structure of molecules. When a beam of X-rays is passed through molecules packed together in a crystal, the X-rays collide with the electrons and scatter. Constructive and destructive interference of these scattered X-rays creates a specific diffraction pattern. Calculating backward from this pattern, the positions of each of the atoms in the molecule can be determined very precisely. One of the pioneers who helped create this technology was Dorothy Crowfoot Hodgkin.

She was born in Cairo, Egypt, in 1910, where her British parents were studying archeology. Even as a young girl, she was fascinated with minerals and crystals. When she was a student at Oxford University, she began researching how X-ray crystallography could be used to determine the structure of biomolecules. She invented new techniques that allowed her and her students to determine the structures of vitamin B 12 , penicillin, and many other important molecules. Diabetes, a disease that affects 382 million people worldwide, involves the hormone insulin. Hodgkin began studying the structure of insulin in 1934, but it required several decades of advances in the field before she finally reported the structure in 1969. Understanding the structure has led to better understanding of the disease and treatment options.

Calculating the Wavelength of a Particle If an electron travels at a velocity of 1.000 × 10 7 m/s and has a mass of 9.109 × 10 –28 g, what is its wavelength?

Solution We can use de Broglie’s equation to solve this problem, but we first must do a unit conversion of Planck’s constant. You learned earlier that 1 J = 1 kg m 2 /s 2 . Thus, we can write h = 6.626 × 10 –34 J·s as 6.626 × 10 –34 kg m 2 /s.

=  $\frac{6.626\;\times\; 10^{-34}\;\text{kg m}^{2}\text{/s}}{(9.190\;\times\; 10^{-31} \;\text{kg})(1.000\;\times\; 10^{7}\;\text{m/s})}$

This is a small value, but it is significantly larger than the size of an electron in the classical (particle) view. This size is the same order of magnitude as the size of an atom. This means that electron wavelike behavior is going to be noticeable in an atom.

Check Your Learning Calculate the wavelength of a softball with a mass of 100 g traveling at a velocity of 35 m/s, assuming that it can be modeled as a single particle.

1.9 × 10 –34 m

We never think of a thrown softball having a wavelength, since this wavelength is so small it is impossible for our senses or any known instrument to detect. The de Broglie wavelength is only appreciable for matter that has a very small mass and/or a very high velocity.

## The Quantum–Mechanical Model of an Atom

Shortly after de Broglie published his ideas that the electron in a hydrogen atom could be better thought of as being a circular standing wave instead of a particle moving in quantized circular orbits, as Bohr had argued, Erwin Schrödinger extended de Broglie’s work by incorporating the de Broglie relation into a wave equation, deriving what is today known as the Schrödinger equation. When Schrödinger applied his equation to hydrogen-like atoms, he was able to reproduce Bohr’s expression for the energy and, thus, the Rydberg formula governing hydrogen spectra. He did so without having to invoke Bohr’s assumptions of stationary states and quantized orbits, angular momenta, and energies. Quantization in Schrödinger’s theory was a natural consequence of the underlying mathematics of the wave equation. Like de Broglie, Schrödinger initially viewed the electron in hydrogen as being a physical wave instead of a particle, but where de Broglie thought of the electron in terms of circular stationary waves, Schrödinger properly thought in terms of three-dimensional stationary waves, or wavefunctions , represented by the Greek letter psi, ψ . A few years later, Max Born proposed an interpretation of the wavefunction, ψ, that is still accepted today: Electrons are still particles, and so the waves represented by ψ are not physical waves. However,  when you square them, you obtain the probability density which  describes the probability of the quantum particle being present near a certain location in space. Wavefunctions, therefore, can be used to determine the distribution of the electron’s density with respect to the nucleus in an atom, but cannot be used to pinpoint the exact location of the electron at any given time. In other words, they predict the energy levels available for electrons in an atom and the probability of finding an electron at a particular place in an atom. Schrödinger’s work, as well as that of Heisenberg and many other scientists following in their footsteps, is generally referred to as quantum mechanics .

You may also have heard of Schrödinger because of his famous thought experiment. This story explains the concepts of superposition and entanglement as related to a cat in a box with poison .

## Understanding Quantum Theory of Electrons in Atoms

As was described previously, electrons in atoms can exist only on discrete energy levels but not between them. It is said that the energy of an electron in an atom is quantized, meaning it can be equal only to certain specific values. The electron can jump from one energy level to another, but it cannot transition smoothly because it cannot exist between the levels.

## Principal Quantum Number

The energy levels are labeled with an n value, where n = 1, 2, 3, …∞. Generally speaking, the energy of an electron in an atom is greater for greater values of n . This number, n , is referred to as the principal quantum number. The principal quantum number defines the location of the energy level. It is essentially the same concept as the n in the Bohr atom description. Another name for the principal quantum number is the shell number. The shells of an atom can be thought of as concentric circles radiating out from the nucleus. The electrons that belong to a specific shell are most likely to be found within the corresponding circular area (not traveling along the circular ring like a planet orbiting the sun). The further we proceed from the nucleus, the higher the shell number, and so the higher the energy level ( Figure 4 ). The positively charged protons in the nucleus stabilize the electronic orbitals by electrostatic attraction between the positive charges of the protons and the negative charges of the electrons. So the further away the electron is from the nucleus, the greater the energy it has.

In a major advance over the Bohr theory of the hydrogen atom, in the quantum mechanical model, one can calculate the quantized energies of any isolated atom.  Knowing these energies one can then predict the frequencies and energies of photons that are emitted or absorbed based on the difference of the calculated energy levels using the equation presented in the previous quantum,   | Δ E | = | E f − E i | = hν.  In the case of the hydrogen atom, the expression simplifies to the previously obtained Bohr result.

• ΔE  =  E final – E initial  =  -2.179 × 10 -18  $(\frac{1}{n^2_\text{f}} - \frac{1}{n^2_\text{i}})$ J

The principal quantum number is one of three quantum numbers used to characterize an orbital. An atomic orbital , which is distinct from an orbit , is a general region in an atom within which an electron is most probable to reside. More precisely, the orbital specifies the probability of finding an electron in the three-dimensional space around the nucleus and is based on solutions of the Schrödinger equation. In addition, the principal quantum number defines the energy of an electron in a hydrogen or hydrogen-like atom or an ion (an atom or an ion with only one electron) and the general region in which discrete energy levels of electrons in multi-electron atoms and ions are located.

## Angular Momentum Quantum Number

Another quantum number is l , the angular momentum quantum number (this is sometimes referred to as the azimuthal quantum number). It is an integer that defines the shape of the orbital, and takes on the values, l = 0, 1, 2, …, n – 1. We will learn what these shapes are in the next section . This means that an orbital with n = 1 can have only one value of l , l = 0, whereas n = 2 permits l = 0 and l = 1, and so on. The principal quantum number defines the general size and energy of the orbital. The l value specifies the shape of the orbital. Orbitals with the same value of l form a subshell .

## Magnetic Quantum Number

Angular momentum is a vector.  Electrons with angular momentum can have this momentum oriented in different directions.  In quantum mechanics it is convenient to describe the z component of the angular momentum.  The magnetic quantum number , called m l, specifies the z component of the angular momentum for a particular orbital. For example, if  l = 0, then the only possible value of m l is zero. When  l = 1, m l can be equal to –1, 0, or +1. Generally speaking, m l can be equal to the set of numbers [– l , –( l – 1), …, –1, 0, +1, …, ( l – 1), l ] . The total number of possible orbitals with the same value of l (a subshell) is 2 l + 1. Thus, there is one orbital with  l = 0 ( m l = 0 is the only orbital), there are three orbitals with  l = 1 ( m l = -1, m l = 0, m l = 1 ), five orbitals with  l = 2 ( m l = -2, m l = -1, m l = 0, m l = 1, m l = 2 ), and so on.

Rather than specifying all the values of n  and  l  every time we refer to a subshell or an orbital, chemists use an abbreviated system with lowercase letters to denote the value of  l for a particular subshell or orbital. Orbitals with l = 0 are called s orbitals (or the s subshell). The value l = 1 corresponds to the p orbitals. For a given n , p orbitals constitute a p subshell (e.g., 3 p subshell if n = 3). The orbitals with l = 2 are called the d orbitals , followed by the f, g, and h orbitals for l = 3, 4, 5, and there are higher values we will not consider. When naming a subshell, it is common to write the principal quantum number ( n)  followed by the subshell letter ( s, p, d, f, etc) . For example, when referring to a subshell with n  = 4 and  l = 2, we would call this the 4 d subshell. We can also say that there are five different 4 d orbitals since there are five values for m l .

As a review, the principal quantum number defines the general value of the electronic energy, with lower values of n indicating lower (more negative) energies and electrons that are closer to the nucleus. The azimuthal quantum number determines the shape of the orbital and we can use s , p , d , f , etc. to designate which subshell the electron is in. And the magnetic quantum number specifies orientation of the orbital in space. Table 1 below provides the possible combinations of  n ,  l , and  m l for the first four shells. You’ll notice that every shell does not contain all shapes of orbitals because of the allowable values for the azimuthal quantum number (e.g., there is  not a 1 p orbital—only shells higher than n = 1 contain a  p subshell).

Working with Shells and Subshells Indicate the number of subshells, the number of orbitals in each subshell, and the values of l and m l for the orbitals in the n = 4 shell of an atom.

Solution For n = 4, l can have values of 0, 1, 2, and 3. Thus, four subshells are found in the n = 4 shell of an atom. For l = 0, m l can only be 0. Thus, there is only one orbital with n = 4 and l = 0. For l = 1,  m l can have values of –1, 0, +1, so we find three orbitals. For l = 2, m l can have values of –2, –1, 0, +1, +2, so we have five orbitals. When l = 3, m l can have values of –3, –2, –1, 0, +1, +2, +3, and we can have seven orbitals. Thus, we find a total of 16 orbitals in the n = 4 shell of an atom.

Check Your Learning How many orbitals are in the  n = 5 shell?

25 orbitals

## Key Concepts and Summary

Macroscopic objects act as particles. Microscopic objects (such as electrons) have properties of both a particle and a wave. Their exact trajectories cannot be determined. The quantum mechanical model of atoms describes the three-dimensional position of the electron in a probabilistic manner according to a mathematical function called a wavefunction, often denoted as ψ . Atomic wavefunctions are also called orbitals and describe the areas in an atom where electrons are most likely to be found.

An atomic orbital is characterized by three quantum numbers. The principal quantum number, n , can be any positive integer. The relative energy of an orbital and the average distance of an electron from the nucleus are related to n . Orbitals having the same value of n are said to be in the same shell. The azimuthal quantum number, l , can have any integer value from 0 to n – 1. This quantum number describes the shape or type of the orbital. Orbitals with the same principal quantum number and the same l value belong to the same subshell. The magnetic quantum number, m l , with 2 l + 1 values ranging from – l to + l , describes the orientation of the orbital in space.

## Chemistry End of Section Exercises

• How are the Bohr model and the quantum mechanical model of the hydrogen atom similar? How are they different?
• Without using quantum numbers, describe the differences between the shells, subshells, and orbitals of an atom.
• How do the quantum numbers of the shells, subshells, and orbitals of an atom differ?
• Describe the wavefunction and in what ways Schrödinger built upon de Broglie’s previous work.
• $c = \lambda\nu$
• $E = h\nu$
• $\lambda = \frac{h}{m\nu}$

## Answers to Chemistry End of Section Exercises

• Both models have a central positively charged nucleus with electrons moving about the nucleus in accordance with the Coulomb electrostatic potential. The Bohr model assumes that the electrons move in circular orbits that have quantized energies, angular momentum, and radii that are specified by a single quantum number, n = 1, 2, 3, …, but this quantization is an ad hoc assumption made by Bohr to incorporate quantization into an essentially classical mechanics description of the atom. Bohr also assumed that electrons orbiting the nucleus normally do not emit or absorb electromagnetic radiation, but do so when the electron switches to a different orbit. In the quantum mechanical model, the electrons do not move in precise orbits. There are inherent limitations in determining simultaneously both the position and energy of a quantum particle like an electron, an outcome of the Heisenberg uncertainty principle, so precise orbits are not possible. Instead, a probabilistic interpretation of the electron’s position at any given instant is used, with a mathematical function ψ called a wavefunction that can be used to determine the electron’s spatial probability distribution. These wavefunctions, or orbitals, are three-dimensional stationary waves that can be specified by three quantum numbers that arise naturally from their underlying mathematics (no ad hoc assumptions required): the principal quantum number, n (the same one used by Bohr), which specifies shells such that orbitals having the same n all have the same energy and approximately the same spatial extent; the angular momentum quantum number l , which is a measure of the orbital’s angular momentum and corresponds to the orbitals’ general shapes, as well as specifying subshells such that orbitals having the same l (and n ) all have the same energy; and the orientation quantum number m , which is a measure of the z component of the angular momentum and corresponds to the orientations of the orbitals. The Bohr model gives the same expression for the energy as the quantum mechanical expression and, hence, both properly account for hydrogen’s discrete spectrum (an example of getting the right answers for the wrong reasons, something that many chemistry students can sympathize with), but gives the wrong expression for the angular momentum (Bohr orbits necessarily all have non-zero angular momentum, but some quantum orbitals [ s orbitals] can have zero angular momentum).
• (a) Shells describe the general size of an orbital (or distance from the nucleus), subshells describe the shape of an orbital, and the actual orbitals include additional information about the orientation of the orbital; (b) the quantum numbers for shells are integers starting from n = 1, for subshells they are integers with values of 0 … ( n – 1), for orbitals they are integers with values of – l … + l .
• Wavefunctions are mathematical functions describing energy and position of electrons in an atom. Building on deBroglie’s work, Schrodinger described the electrons as three-dimensional stationary waves. This was also further extended by Born to show that the square of wavefunction is the probability of finding a quantum particle (electron) in a certain location.
• (a) wavelike behavior because it is describing the relationship between wavelength and frequency of a wave; (b) particle-like behavior because it is describing the energy of a particle (photon) with frequency ν; (c) both because it is describing that a particle with mass m can have a wavelength λ.

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## 7 De Broglie Waves: Are Electrons Waves or Particles?

• Published: May 2020
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In 1924, de Broglie postulated that particles can behave like waves, thus complementing the observation by Einstein in 1905 that light can behave like particles. This wave–particle duality aspect for both particles and waves had a deep impact on the subsequent development of quantum mechanics. Some highly counterintuitive results, like the Heisenberg uncertainty relation and the Bose–Einstein condensation, that were motivated by wave–particle duality are discussed in this chapter. Following de Broglie’s hypothesis, a wave packet description for a particle is described. An analysis of the Heisenberg microscope is presented, thus motivating the Heisenberg uncertainty relation. The Davisson–Germer experiment that showed that electrons can behave like waves and the Compton effect that provided early conclusive evidence that light can behave like particles are also discussed.

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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)..

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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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## 7.3: The Wave-Particle Duality of Matter and Energy

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## Learning Objectives

• To understand the wave–particle duality of matter.

Einstein’s photons of light were individual packets of energy having many of the characteristics of particles. Recall that the collision of an electron (a particle) with a sufficiently energetic photon can eject a photoelectron from the surface of a metal. Any excess energy is transferred to the electron and is converted to the kinetic energy of the ejected electron. Einstein’s hypothesis that energy is concentrated in localized bundles, however, was in sharp contrast to the classical notion that energy is spread out uniformly in a wave. We now describe Einstein’s theory of the relationship between energy and mass, a theory that others built on to develop our current model of the atom.

## The Wave Character of Matter

Einstein initially assumed that photons had zero mass, which made them a peculiar sort of particle indeed. In 1905, however, he published his special theory of relativity, which related energy and mass according to the famous equation:

$E=h u=h\dfrac{c}{\lambda }=mc^{2} \label{6.4.1}$

According to this theory, a photon of wavelength $$λ$$ and frequency $$u$$ has a nonzero mass, which is given as follows:

$m=\dfrac{E}{c^{2}}=\dfrac{h u }{c^{2}}=\dfrac{h}{\lambda c} \label{6.4.2}$

That is, light, which had always been regarded as a wave, also has properties typical of particles, a condition known as wave–particle duality (a principle that matter and energy have properties typical of both waves and particles). Depending on conditions, light could be viewed as either a wave or a particle.

In 1922, the American physicist Arthur Compton (1892–1962) reported the results of experiments involving the collision of x-rays and electrons that supported the particle nature of light. At about the same time, a young French physics student, Louis de Broglie (1892–1972), began to wonder whether the converse was true: Could particles exhibit the properties of waves? In his PhD dissertation submitted to the Sorbonne in 1924, de Broglie proposed that a particle such as an electron could be described by a wave whose wavelength is given by

$\lambda =\dfrac{h}{mv} \label{6.4.3}$

• $$h$$ is Planck’s constant,
• $$m$$ is the mass of the particle, and
• $$v$$ is the velocity of the particle.

This revolutionary idea was quickly confirmed by American physicists Clinton Davisson (1881–1958) and Lester Germer (1896–1971), who showed that beams of electrons, regarded as particles, were diffracted by a sodium chloride crystal in the same manner as x-rays, which were regarded as waves. It was proven experimentally that electrons do exhibit the properties of waves. For his work, de Broglie received the Nobel Prize in Physics in 1929.

If particles exhibit the properties of waves, why had no one observed them before? The answer lies in the numerator of de Broglie’s equation, which is an extremely small number. As you will calculate in Example $$\PageIndex{1}$$, Planck’s constant (6.63 × 10 −34 J•s) is so small that the wavelength of a particle with a large mass is too short (less than the diameter of an atomic nucleus) to be noticeable.

The de Broglie Equation: The de Broglie Equation, YouTube(opens in new window) [youtu.be]

## Example $$\PageIndex{1}$$: Wavelength of a Baseball in Motion

Calculate the wavelength of a baseball, which has a mass of 149 g and a speed of 100 mi/h.

Given: mass and speed of object

• Convert the speed of the baseball to the appropriate SI units: meters per second.
• Substitute values into Equation $$\ref{6.4.3}$$ and solve for the wavelength.

The wavelength of a particle is given by $$λ = h/mv$$. We know that m = 0.149 kg, so all we need to find is the speed of the baseball:

$$v=\left ( \dfrac{100\; \cancel{mi}}{\cancel{h}} \right )\left ( \dfrac{1\; \cancel{h}}{60\; \cancel{min}} \right )\left ( \dfrac{1.609\; \cancel{km}}{\cancel{mi}} \right )\left ( \dfrac{1000\; m}{\cancel{km}} \right )$$

B Recall that the joule is a derived unit, whose units are (kg•m 2 )/s 2 . Thus the wavelength of the baseball is

$\lambda =\dfrac{6.626\times 10^{-34}\; J\cdot s}{\left ( 0.149\; kg \right )\left ( 44.69\; m\cdot s \right )}= \dfrac{6.626\times 10^{-34}\; \cancel{kg}\cdot m{^\cancel{2}\cdot \cancel{s}{\cancel{^{-2}}\cdot \cancel{s}}}}{\left ( 0.149\; \cancel{kg} \right )\left ( 44.69\; \cancel{m}\cdot \cancel{s^{-1}} \right )}=9.95\times 10^{-35}\; m \nonumber$

(You should verify that the units cancel to give the wavelength in meters.) Given that the diameter of the nucleus of an atom is approximately 10 −14 m, the wavelength of the baseball is almost unimaginably small.

## Exercise $$\PageIndex{1}$$: Wavelength of a Neutron in Motion

Calculate the wavelength of a neutron that is moving at 3.00 × 10 3 m/s.

1.32 Å, or 132 pm

As you calculated in Example $$\PageIndex{1}$$, objects such as a baseball or a neutron have such short wavelengths that they are best regarded primarily as particles. In contrast, objects with very small masses (such as photons) have large wavelengths and can be viewed primarily as waves. Objects with intermediate masses, however, such as electrons, exhibit the properties of both particles and waves. Although we still usually think of electrons as particles, the wave nature of electrons is employed in an electron microscope , which has revealed most of what we know about the microscopic structure of living organisms and materials. Because the wavelength of an electron beam is much shorter than the wavelength of a beam of visible light, this instrument can resolve smaller details than a light microscope can (Figure $$\PageIndex{1}$$).

## An Important Wave Property: Phase And Interference

A wave is a disturbance that travels in space. The magnitude of the wave at any point in space and time varies sinusoidally. While the absolute value of the magnitude of one wave at any point is not very important, the relative displacement of two waves, called the phase difference, is vitally important because it determines whether the waves reinforce or interfere with each other. Figure $$\PageIndex{2A}$$ shows an arbitrary phase difference between two wave and Figure $$\PageIndex{2B}$$ shows what happens when the two waves are 180 degrees out of phase. The green line is their sum. Figure $$\PageIndex{2C}$$ shows what happens when the two lines are in phase, exactly superimposed on each other. Again, the green line is the sum of the intensities. A pattern of constructive and destructive interference is obtained when two (or more) diffracting waves interact with each other. This principle of diffraction and interference was used to prove the wave properties of electrons and is the basis for how electron microscopes work.

Photograph of an interference pattern produced by circular water waves in a ripple tank.

For a mathematical analysis of phase aspects in sinusoids, check the math Libretexts library.

## Standing Waves

De Broglie also investigated why only certain orbits were allowed in Bohr’s model of the hydrogen atom. He hypothesized that the electron behaves like a standing wave (a wave that does not travel in space). An example of a standing wave is the motion of a string of a violin or guitar. When the string is plucked, it vibrates at certain fixed frequencies because it is fastened at both ends (Figure $$\PageIndex{3}$$). If the length of the string is $$L$$, then the lowest-energy vibration (the fundamental) has wavelength

\begin{align} \dfrac{\lambda }{2} & =L \nonumber \\ \lambda &= 2L \nonumber \end{align} \label{6.4.4}

Higher-energy vibrations are called overtones (the vibration of a standing wave that is higher in energy than the fundamental vibration) and are produced when the string is plucked more strongly; they have wavelengths given by

$\lambda=\dfrac{2L}{n} \label{6.4.5}$

where n has any integral value. When plucked, all other frequencies die out immediately. Only the resonant frequencies survive and are heard. Thus, we can think of the resonant frequencies of the string as being quantized. Notice in Figure $$\PageIndex{3}$$ that all overtones have one or more nodes, points where the string does not move. The amplitude of the wave at a node is zero.

Quantized vibrations and overtones containing nodes are not restricted to one-dimensional systems, such as strings. A two-dimensional surface, such as a drumhead, also has quantized vibrations. Similarly, when the ends of a string are joined to form a circle, the only allowed vibrations are those with wavelength

$2πr = nλ \label{6.4.6}$

where $$r$$ is the radius of the circle. De Broglie argued that Bohr’s allowed orbits could be understood if the electron behaved like a standing circular wave (Figure $$\PageIndex{4}$$). The standing wave could exist only if the circumference of the circle was an integral multiple of the wavelength such that the propagated waves were all in phase, thereby increasing the net amplitudes and causing constructive interference . Otherwise, the propagated waves would be out of phase, resulting in a net decrease in amplitude and causing destructive interference. The nonresonant waves interfere with themselves! De Broglie’s idea explained Bohr’s allowed orbits and energy levels nicely: in the lowest energy level, corresponding to $$n = 1$$ in Equation $$\ref{6.4.6}$$, one complete wavelength would close the circle. Higher energy levels would have successively higher values of n with a corresponding number of nodes.

Like all analogies, although the standing wave model helps us understand much about why Bohr's theory worked, it also, if pushed too far, can mislead. As you will see, some of de Broglie’s ideas are retained in the modern theory of the electronic structure of the atom: the wave behavior of the electron and the presence of nodes that increase in number as the energy level increases. Unfortunately, his (and Bohr's) explanation also contains one major feature that we now know to be incorrect: in the currently accepted model, the electron in a given orbit is not always at the same distance from the nucleus.

## The Heisenberg Uncertainty Principle

Because a wave is a disturbance that travels in space, it has no fixed position. One might therefore expect that it would also be hard to specify the exact position of a particle that exhibits wavelike behavior. A characteristic of light is that is can be bent or spread out by passing through a narrow slit. You can literally see this by half closing your eyes and looking through your eye lashes. This reduces the brightness of what you are seeing and somewhat fuzzes out the image, but the light bends around your lashes to provide a complete image rather than a bunch of bars across the image. This is called diffraction.

This behavior of waves is captured in Maxwell's equations (1870 or so) for electromagnetic waves and was and is well understood. An "uncertainty principle" for light is, if you will, merely a conclusion about the nature of electromagnetic waves and nothing new. De Broglie's idea of wave-particle duality means that particles such as electrons which exhibit wavelike characteristics will also undergo diffraction from slits whose size is on the order of the electron wavelength.

This situation was described mathematically by the German physicist Werner Heisenberg (1901–1976; Nobel Prize in Physics, 1932), who related the position of a particle to its momentum. Referring to the electron, Heisenberg stated that “at every moment the electron has only an inaccurate position and an inaccurate velocity, and between these two inaccuracies there is this uncertainty relation.” Mathematically, the Heisenberg uncertainty principle states that the uncertainty in the position of a particle (Δ x ) multiplied by the uncertainty in its momentum [Δ( mv )] is greater than or equal to Planck’s constant divided by 4π:

$\left ( \Delta x \right )\left ( \Delta \left [ mv \right ] \right )\ge \dfrac{h}{4\pi } \label{6.4.7}$

Because Planck’s constant is a very small number, the Heisenberg uncertainty principle is important only for particles such as electrons that have very low masses. These are the same particles predicted by de Broglie’s equation to have measurable wavelengths.

If the precise position $$x$$ of a particle is known absolutely (Δ x = 0), then the uncertainty in its momentum must be infinite:

$\left ( \Delta \left [ mv \right ] \right )= \dfrac{h}{4\pi \left ( \Delta x \right ) }=\dfrac{h}{4\pi \left ( 0 \right ) }=\infty \label{6.4.8}$

Because the mass of the electron at rest ($$m$$) is both constant and accurately known, the uncertainty in $$Δ(mv)$$ must be due to the $$Δv$$ term, which would have to be infinitely large for $$Δ(mv)$$ to equal infinity. That is, according to Equation $$\ref{6.4.8}$$, the more accurately we know the exact position of the electron (as $$Δx → 0$$), the less accurately we know the speed and the kinetic energy of the electron (1/2 mv 2 ) because $$Δ(mv) → ∞$$. Conversely, the more accurately we know the precise momentum (and the energy) of the electron [as $$Δ(mv) → 0$$], then $$Δx → ∞$$ and we have no idea where the electron is.

Bohr’s model of the hydrogen atom violated the Heisenberg uncertainty principle by trying to specify simultaneously both the position (an orbit of a particular radius) and the energy (a quantity related to the momentum) of the electron. Moreover, given its mass and wavelike nature, the electron in the hydrogen atom could not possibly orbit the nucleus in a well-defined circular path as predicted by Bohr. You will see, however, that the most probable radius of the electron in the hydrogen atom is exactly the one predicted by Bohr’s model.

## Example $$\PageIndex{1}$$: Quantum Nature of Baseballs

Calculate the minimum uncertainty in the position of the pitched baseball from Example $$\ref{6.4.1}$$ that has a mass of exactly 149 g and a speed of 100 ± 1 mi/h.

Asked for: minimum uncertainty in its position

• Rearrange the inequality that describes the Heisenberg uncertainty principle (Equation $$\ref{6.4.7}$$) to solve for the minimum uncertainty in the position of an object (Δ x ).
• Find Δ v by converting the velocity of the baseball to the appropriate SI units: meters per second.
• Substitute the appropriate values into the expression for the inequality and solve for Δ x .

A The Heisenberg uncertainty principle (Equation \ref{6.4.7}) tells us that $(Δx)(Δ(mv)) = h/4π \nonumber$. Rearranging the inequality gives

$$\Delta x \ge \left( {\dfrac{h}{4\pi }} \right)\left( {\dfrac{1}{\Delta (mv)}} \right)$$

B We know that h = 6.626 × 10 −34 J•s and m = 0.149 kg. Because there is no uncertainty in the mass of the baseball, Δ( mv ) = m Δ v and Δ v = ±1 mi/h. We have

$\Delta u =\left ( \dfrac{1\; \cancel{mi}}{\cancel{h}} \right )\left ( \dfrac{1\; \cancel{h}}{60\; \cancel{min}} \right )\left ( \dfrac{1\; \cancel{min}}{60\; s} \right )\left ( \dfrac{1.609\; \cancel{km}}{\cancel{mi}} \right )\left ( \dfrac{1000\; m}{\cancel{km}} \right )=0.4469\; m/s \nonumber$

## C Therefore,

$\Delta x \ge \left ( \dfrac{6.626\times 10^{-34}\; J\cdot s}{4\left ( 3.1416 \right )} \right ) \left ( \dfrac{1}{\left ( 0.149\; kg \right )\left ( 0.4469\; m\cdot s^{-1} \right )} \right ) \nonumber$

Inserting the definition of a joule (1 J = 1 kg•m 2 /s 2 ) gives

$\Delta x \ge \left ( \dfrac{6.626\times 10^{-34}\; \cancel{kg} \cdot m^{\cancel{2}} \cdot s}{4\left ( 3.1416 \right )\left ( \cancel{s^{2}} \right )} \right ) \left ( \dfrac{1\; \cancel{s}}{\left ( 0.149\; \cancel{kg} \right )\left ( 0.4469\; \cancel{m} \right )} \right ) \nonumber$

$\Delta x \ge 7.92 \pm \times 10^{-34}\; m \nonumber$

This is equal to $$3.12 \times 10^{−32}$$ inches. We can safely say that if a batter misjudges the speed of a fastball by 1 mi/h (about 1%), he will not be able to blame Heisenberg’s uncertainty principle for striking out.

## Exercise $$\PageIndex{2}$$

Calculate the minimum uncertainty in the position of an electron traveling at one-third the speed of light, if the uncertainty in its speed is ±0.1%. Assume its mass to be equal to its mass at rest.

6 × 10 −10 m, or 0.6 nm (about the diameter of a benzene molecule)

An electron possesses both particle and wave properties. The modern model for the electronic structure of the atom is based on recognizing that an electron possesses particle and wave properties, the so-called wave–particle duality . Louis de Broglie showed that the wavelength of a particle is equal to Planck’s constant divided by the mass times the velocity of the particle.

$\lambda =\dfrac{h}{mv} \nonumber$

The electron in Bohr’s circular orbits could thus be described as a standing wave , one that does not move through space. Standing waves are familiar from music: the lowest-energy standing wave is the fundamental vibration, and higher-energy vibrations are overtones and have successively more nodes , points where the amplitude of the wave is always zero. Werner Heisenberg’s uncertainty principle states that it is impossible to precisely describe both the location and the speed of particles that exhibit wavelike behavior.

$\left ( \Delta x \right )\left ( \Delta \left [ mv \right ] \right )\geqslant \dfrac{h}{4\pi } \nonumber$

Modified by Joshua Halpern ( Howard University )

## State de Broglie hypothesis.

Louis de broglie, in 1924, stated that a wave is associated with a moving particle (i.e. matter) and so named these waves as matter waves. he proposed that just like light has dual nature, electrons also have wave like properties. wavelength of a moving particle is given by λ = h p where h is the planck's constant and p is the momentum of moving particle..

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1. Deriving the de Broglie Wavelength

In 1923, Louis de Broglie, a French physicist, proposed a hypothesis to explain the theory of the atomic structure. By using a series of substitution de Broglie hypothesizes particles to hold properties of waves. Within a few years, de Broglie's hypothesis was tested by scientists shooting electrons and rays of lights through slits. What scientists discovered was the electron stream acted the ...

2. de Broglie Equation

The de Broglie equation is one of the equations that is commonly used to define the wave properties of matter. It basically describes the wave nature of the electron. Electromagnetic radiation exhibits the dual nature of a particle (having a momentum) and wave (expressed in frequency and wavelength).

3. The De Broglie Hypothesis

The quantity, mν, is the momentum of an object. The mass, m, has units of kJ and the speed, ν has units of m/s. The term matter wave is used to relate wave properties to matter. An example of a matter wave is an electron beam. Matter waves are different from electromagnetic waves because they are the result of the motion of matter. There are no significant fields associated with matter waves ...

4. What Is the De Broglie Equation?

The de Broglie equation is an equation used to describe the wave properties of matter, specifically, the wave nature of the electron : . λ = h/mv, where λ is wavelength, h is Planck's constant, m is the mass of a particle, moving at a velocity v. de Broglie suggested that particles can exhibit properties of waves.

5. 6.6: De Broglie's Matter Waves

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 ...

6. De Broglie Hypothesis

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

7. De Broglie Relationship

The above equation is known as de Broglie relationship and the wavelength, λ is known as de Broglie wavelength. Diffraction of electron beams explains the de Broglie relationship as diffraction is the property of waves. An electron microscope is a common instrument illustrating this fact. Thus, every object in motion has a wavelike character. Due to a large mass, the wavelengths associated ...

8. de Broglie equation, derivation, and its Significance

According to de Broglie equation/hypothesis, " a matter particle in motion is also associated with waves ." In other words, any moving microscopic or macroscopic particle will be associated with a wave character. Such waves are also called matter waves or de Broglie waves. de Broglie equation is basically used to define the wave properties of matter/electron. Thus, the matter particle-like ...

9. DeBroglie, Intro to Quantum Mechanics, Quantum Numbers 1-3 (M7Q5)

This quantity is called the de Broglie wavelength. Unlike the other values of λ discussed in this chapter, the de Broglie wavelength is a characteristic of particles and other bodies, not electromagnetic radiation (note that this equation involves velocity [ v, with units m/s], not frequency [ ν, with units Hz]. Although these two symbols are similar, they mean very different things). Where ...

10. 1.7: de Broglie Waves can be Experimentally Observed

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. Germer in 1927. In these experiments it was found that electrons were scattered from atoms in a crystal and that these scattered electrons produced an interference pattern. The interference pattern was just like that produced when water waves pass ...

11. De Broglie Equation

De Broglie Relation. To derive the de Broglie equation, he started with Einstein's now famous equation that related matter to energy for light: {eq}E=mc^2 {/eq} De Broglie combined it with Planck ...

12. De Broglie Hypothesis

This lecture is about de Broglie hypothesis and de Broglie wavelength. I will teach you the super easy concept of de Broglie hypothesis with real life examples. At the end of this lecture, you ...

13. De Broglie wavelength (video)

Instead what we do is to use electrons. Since electrons have a rest mass, unlike photons, they have a de Broglie wavelength which is really short, around 0.01 nanometers for easily achievable speeds. This means that a microscope using electron "matter waves" instead of photon light waves can see much smaller things. ( 111 votes) Upvote.

14. 7 De Broglie Waves: Are Electrons Waves or Particles?

where E is the energy, ℏ is Planck's constant ( ⁠ 1. 055 × 10 − 34 J ⋅ s ⁠ ), and ν is the frequency. In 1924, Louis de Broglie argued in his Ph.D. thesis that if light can behave both like waves, as in interference and diffraction, and like particles, as in the photoelectric effect, then particles should also behave like both particles and waves. The de Broglie hypothesis ...

15. De-Broglie Principle and Hypothesis

The de Broglie hypothesis demonstrated that wave-particle duality was a fundamental principle shared by both radiation and matter, not just an aberrant behaviour of light. As a result, if the de Broglie wavelength is properly applied, wave equations can be used to describe material behaviour. This would be critical for quantum mechanics to advance. It is now an accepted part of atomic ...

16. What is De Broglie Hypothesis?

What is de broglie theory. Dual nature of matter was proposed by de Broglie in 1923, it was experimentally verified by Davisson and Germer by diffraction experiment. Wave character of matter has significance only for microscopic particles. de Broglie wavelength or wavelength of matter wave can be calculated using the following relation: λ=hmv ...

17. 2.5: The de Broglie-Bohr Model for the Hydrogen Atom

The de Broglie‐Bohr model of the hydrogen atom presented here treats the electron as a particle on a ring with wave‐like properties. λ = h mev λ = h m e v. de Broglie's hypothesis that matter has wave-like properties. nλ = 2πr n λ = 2 π r. The consequence of de Broglieʹs hypothesis; an integral number of wavelengths must fit within ...

18. How does de Broglie actually prove Bohr's postulates?

10. de Brogile explains why orbitals are quantised. Strictly speaking de Brogile doesn't prove Bohr's postulates which are mostly wrong. But he did provide an explanation for the most important of Bohr's ideas: electron orbitals are quantised. Bohr's whole model starts with the classical idea that electrons "orbit" a nucleus.

19. De Broglie Equation

The wavelength is known as the de Broglie wavelength. For an electron, de Broglie wavelength equation is: λ =. h mv h m v. 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.

20. 7.3: The Wave-Particle Duality of Matter and Energy

If particles exhibit the properties of waves, why had no one observed them before? The answer lies in the numerator of de Broglie's equation, which is an extremely small number. As you will calculate in Example 7.3.1 7.3. 1, Planck's constant (6.63 × 10 −34 J•s) is so small that the wavelength of a particle with a large mass is too short (less than the diameter of an atomic nucleus ...

21. State de Broglie hypothesis.

State de Broglie hypothesis. Louis de Broglie, in 1924, stated that a wave is associated with a moving particle (i.e. matter) and so named these waves as matter waves. He proposed that just like light has dual nature, electrons also have wave like properties. where h is the Planck's constant and p is the momentum of moving particle.