# When You Throw A Baseball, Its De Broglie Wavelength Is

## When you throw a baseball, its de Broglie wavelength is(a)

Problem 8MCQ (Multiple Choice Questionnaire) Problem8MCQ As soon as you toss a baseball, its de Broglie wavelength equals the size of the ball (a). (b) It is approximately the same size as an atom. (c) approximately the same size as the nucleus of an atom. (d) much smaller in size than the nucleus of an atom. Step-by-Step Solution: Step one of three Somethingpersomethingelse=index o something/somethingelse=speed/time=accelerationTofindacceleration o something/somethingelse=index o something/somethingelse=accelerationTofindacceleration.

DataTableWay is a way to organize data in a table.

4 seconds at 45 mph 2 20 10 is a two-digit hexadecimal number that represents the number two in ten.

##### Textbook:Physics: Principles with Applications

Its de Broglie wavelength is equal in size to that of a baseball when you toss it, which is (a) correct. (b) It is approximately the same size as an atom. (c) approximately the same size as the nucleus of an atom. (d) much smaller in size than the nucleus of an atom.” is divided down into a number of simple to follow phases, as well as forty words. This textbook survival guide was written specifically for the textbook Physics: Principles and Applications, 7th edition, which can be found here.

Textbook survival guide with 33 chapters and 9632 solutions.

Since the solution to the 8MCQ from Chapter 27 was posted, more than 447 students have watched the whole step-by-step solution to the question.

## de Broglie Wave Equation

• Explain the de Broglie wave equation and how to use it to calculate the wavelength of a moving object.

### Beyond Bohr’s Model

The model of the atom developed by Bohr was useful in explaining how electrons were capable of absorbing and releasing energy, as well as how atomic emission spectra were generated by combining electrons. However, the model failed to adequately explain why electrons should only be able to live in fixed circular orbits rather than being able to exist in an infinite number of orbits, each with a distinct energy, as suggested by quantum mechanics. In order to understand why atomic energy levels are quantized, physicists had to reconsider their assumptions about the nature of the electron and how it moves across space and time.

It seemed acceptable to speculate about the possibility that electrons may have a dual wave-particle nature as well.

In particular, the wavelength () of any moving object may be calculated as follows_displaystylelambda=frac It is important to note that in this equation,his Planck’s constant is,is the mass of the particle in kilograms, andvis the velocity of the particle in meters per second.

To demonstrate how to compute the wavelength of an electron, consider the example below.

### Sample Problem: de Broglie Wave Equation

An electron with a mass of 9.11 1031 kg travels at a speed that is close to that of light. Calculate the wavelength of the electron with a velocity of 3.00 x 10 8 meters per second. Step 1: Create a list of the known quantities and a plan for solving the problem. Known

• Its mass (m) is 9.11 10 31 kilograms
• Its speed (v) is 3.00 10 8 meters per second
• And its Planck’s constant is 6.62621034 joules per second.

To find the wavelength of the moving electron, use the de Broglie wave equation (displaystyle, lambda=frac) and the displaystyle, lambda=frac (lambda=frac). Step 2: Calculate displaystylelambda=frac=frac text cdottext cdottext cdottext cdottext cdottext cdottext cdottext cdottext cdottext cdottext cdottext cdottext cdottext cdottext cdottext c Step 3: Make a list of all of the things you want to do. Consider the outcome of your decision. This extremely short wavelength is around one-hundredth the diameter of a hydrogen atom.

• When dealing with ordinary massive items with significantly greater masses, the wavelengths should be extremely tiny.
• Even with the most sophisticated scientific equipment, it is difficult to detect this wavelength.
• Particles with detectable wavelengths, on the other hand, are all extremely minuscule in size.
• It is only possible for an electron to circulate about the nucleus of an atom when it is restricted to a certain region around that nucleus in a way that the electron wave produced by that electron “fits” the size of the atom appropriately.
• According to the E=hvequation, quantized frequencies imply that electrons may only exist in an atom at particular energies, which is in accordance with Bohr’s prior theorization.

This is a permitted orbital path. Due to the fact that the electron wave in (B) does not fit correctly into the orbit, this orbit is not permitted.

### Summary

• Because of the deBroglie wave equations, it is possible to calculate the wavelength of any moving object. As the electron’s speed slows, the wavelength of the electron rises.

### Practice

Calculations involving the deBroglie wave equation can be practiced by clicking on the following link:

### Review

1. What was it that the Bohr model couldn’t explain? Give an expression for the deBroglie wave equation. What occurs when the electron’s speed slows down is described below.

## Glossary

• The De Broglie wave equation is_displaystylelambda=lambda=lambda=lambda=lambda=lambda=lambda=lambda=lambda=lambda=lambda=lambda=lambda=lambda=lambda=lambda=lambda frac
• Quantize: Quantum mechanical principles are used to limit the range of potential values of (a magnitude or quantity) to a discrete set of values

## De Broglie’s Matter Waves – University Physics Volume 3

Photons and Matter Waves are two types of particles.

### Learning Objectives

You will be able to do the following by the conclusion of this section:

• Describe de Broglie’s theory of matter waves in more detail. Demonstrate how de Broglie’s hypothesis provides justification for the quantization of angular momentum in Bohr’s quantum theory of the hydrogen atom. Explain the Davisson–Germer experiment in detail. Determine whether or not de Broglie’s theory of matter waves is correct and how it accounts for electron diffraction events.

When an electromagnetic wave interacts with matter, according to Compton’s formula, it can behave like a particle of light, which was previously unknown. A new speculative idea was suggested by Louisde Broglie in 1924: electrons and other bits of matter can act like waves, just as waves can behave like waves. De Broglie’s theory of matter waves is the name given to this concept today. As a result of De Broglie’s hypothesis, which was combined with Niels Bohr’s early quantum theory in 1926, a new theory of wave quantum mechanics was developed, which was used to describe the physics of atoms and subatomic particles.

1. These new technologies serve as a catalyst for advancements in other fields of study, such as biology and chemistry.
2. We have already studied these photon-photon relationships in the context of the Compton effect.
3. Any particle with energy and momentum is a de Broglie wave with a frequency and wavelength of the same value.
4. According to De Broglie’s relations, the wave vector and the wave frequency are commonly written in the same way we normally describe wave relationships: According to wave theory, a wave’s energy is carried along with the group velocity of the wave.

Inferring from de Broglie relations that matter waves fulfill the following relationship by identifying the energyEand momentumpof a particle with its relativistic energyand its relativistic momentummu, respectively, it follows that matter waves satisfy the following relation: where It is possible to have and (Figure)when a particle has no mass.

Determining the wavelength of a de Broglie wave for the following objects: (a) a 0.65-kg basketball thrown at the speed of 10 meters per second, (b) a nonrelativistic electron with a kinetic energy of one electron electron-volt, and (c) a relativistic electron with a kinetic energy of one electron-volt Strategy To determine the de Broglie wavelength, we employ (Figure).

The relativistic momentum must be employed when the nonrelativistic approximation cannot be utilized, as in (c), since the rest mass energy of a particle is and is the Lorentz factor.

Depending on the nature of the situation at hand, we can choose any of the following values forhc in this equation: Solution

1. The kinetic energy of the basketball is and the rest mass energy is for the basketball. That is something we can see
2. We may apply the nonrelativistic electron formula to calculate the nonrelativistic electron, if we have one. However, it is more straightforward to apply in this case (Figure): We get the same conclusion if we use nonrelativistic momentum instead of relativistic momentum since 1 eV is a lot lower than the rest mass of the electron. Because its total energy is and is not negligible, it is impossible to ignore the impacts of a fast electron with relativistic effects:

Significance We can see from these calculations that the wavelengths of De Broglie’s waves for macroscopic objects such as a ball are infinitely short. Consequently, even if they exist, they are undetectable and have no effect on the motion of macroscopic objects. Please double-check your understanding. In the case of a nonrelativistic proton with a kinetic energy of one electronvolt, what is the wavelength of de Broglie’s wavefunction? 29:00 p.m. De Broglie, using the idea of the electron matter wave, offered an explanation for the quantization of the electron’s angular momentum in the hydrogen atom, which had been posited in Bohr’s quantum theory but had not been shown.

1. Think of a stretched guitar string that is clamped at both ends and vibrates in one of its typical modes to better understand what I’m talking about.
2. In order for a standing wave on a string to form, the following conditions must be met: Consider the following scenario: instead of having the string attached to the walls, we bend the length of the rope into a circle and tie the ends together.
3. The following standing-wave condition must be satisfied by the radii, which indicates that they are not arbitrarily chosen.
4. Assuming that (Figure)is correct, the electron wave of this wavelength corresponds to the electron’s linear momentum.
5. This equation is the first of Bohr’s quantization conditions, which are provided by the equation (Figure).
6. For example, a stretched rope clamped against the walls, or an electron wave stuck in the third Bohr orbit of hydrogen atom, are both examples of standing-wave patterns.
7. Determine the de Broglie wavelength of an electron in the ground state of hydrogen by using the formula below.

When the Bohr quantization condition is met, the result is The wavelength of an electron is as follows: Significance When we use (Figure) directly, we get the same result as when we use (Figure) indirectly.

The de Broglie wavelength of an electron in the third excited state of hydrogen is to be determined using this method.

This was the first time that this had been demonstrated.

Instead, the confirmation happened as a result of their usual experimental examinations of metal surfaces subjected to electron bombardment.

It was necessary to prepare their nickel sample in a high-temperature oven in order to transform it from its regular polycrystalline structure to a form in which huge single-crystal domains occupied the whole volume of the sample.

Electron guns use thermal electrons that are emitted from a heated element (typically made of tungsten) in order to accelerate them through a potential difference.

In the electron cannon, the kinetic energyKof the electrons may be modified by selecting a different value for the potential difference between the electrons.

It scatters in a variety of directions as it reaches the surface.

The detector’s angular location with regard to the incoming beam’s direction can be adjusted between and The entire arrangement is contained within a vacuum chamber in order to prevent electron collisions with air molecules, as such thermal collisions would alter the kinetic energy of the electrons, which is not desired in this application.

• The nickel target is dispersed by an electron beam that has been precisely collimated.
• It is possible to measure the intensity of the scattered electron beam over a wide range of scattering angles while maintaining a constant distance between the detector and the target.
• This results in an intensity of the scattered electron beam that is almost constant in all directions, reminiscent of diffuse reflection of light from a porous surface.
• Father and son physicists, William H.Bragg and William L.Bragg, investigated the diffraction patterns created by X-rays dispersed by several crystalline substances in 1912 and discovered a similar pattern.

It was discovered that the lattice spacing of the Davisson–Germer target, which was obtained via X-ray crystallography, was At the surface of the sample, as opposed to X-ray crystallography, in which X-rays enter the sample, only surface atoms interact with the incoming electron beam, as demonstrated by the original Davisson–Germer experiment.

1).

A persuasive evidence for the presence of de Broglie matter waves is provided by the close correspondence between this theoretical conclusion and the Davisson–Germer experimental value of The experimental findings of electron diffraction on a nickel target at an accelerating potential in the electron cannon of approximately 1 kV are presented.

• Because the incident electrons are dispersed solely from the surface, diffraction lines recorded with low-energy electrons, such as those employed in the Davisson–Germer experiment, are relatively broad (see (Figure)).
• Due to the fact that the diffraction picture is made by scattering off a large number of crystalline planes inside the volume, and the maxima produced in scattering at Bragg angles are quite sharp (see (Figure)), this happens.
• Several decades have passed since the work of Davisson and Germer, and de Broglie’s theory has been thoroughly examined using a variety of experimental approaches, with the existence of de Broglie waves being proven for a large number of elementary particles.
• As a neutral particle, the neutron has no charge, and its mass is similar to that of a positively charged proton.
• As a result, the property of being a matter wave is not limited to electrically charged particles, but is applicable to every particle in motion.
• All physical things, no matter how little or huge, have a matter wave linked with them as long as they are in motion.
• Neutron Scattering is a type of scattering that occurs when neutrons collide with one another.

Consider estimating the kinetic energy of a neutron in the neutron beam (in eV), then comparing it to the kinetic energy of an ideal gas in equilibrium at ambient temperature to see which is greater.

It is necessary for the neutron wavelength to be on the same order of magnitude as the lattice spacing in order to witness a diffraction pattern on such a lattice.

For the purpose of comparing this energy to the energy of an ideal gas in equilibrium at ambient temperature, we employ a relationship where is the Boltzmann constant.

It appears to be the case, and we may utilize nonrelativistic kinetic energy: We can see that the kinetic energy of an ideal gas in equilibrium at 300 K is: We can see that these energies are on the same scale in terms of size.

They are distinguished from other types of neutrons by their high energy.

What are the de Broglie wavelengths at this speed, and how long are they?

Strategy A proton’s total rest mass energy is equal to Assuming we know the proton’s velocity, we can get the wavelength and kinetic energyK using relativistic laws, which we can then use to calculate the kinetic energyK.

What You Have Acknowledged Determine the de Broglie wavelength and kinetic energy of a free electron traveling at a speed of 0.75 c while traveling in a straight line.

### Summary

• De Broglie’s hypothesis of matter waves proposes that any particle of matter with linear motion is also a wave, contrary to popular belief at the time. According to Einstein’s equations of motion, the length of the wavelength of a matter wave that is connected with a particle is inversely proportional to its linear momentum. It has been demonstrated that the speed of a matter wave is equal to the speed of a particle
• De Broglie’s concept of the electron matter wave serves as a rationale for the quantization of the electron’s angular momentum in Bohr’s model of the hydrogen atom
• And the Davisson–Germer experiment, in which electrons are scattered off a crystalline nickel surface It has been discovered that electron matter waves exhibit diffraction patterns. They provide as evidence for the presence of matter waves in the universe. In diffraction studies with a variety of particles, matter waves have been seen.
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### Conceptual Questions

Which form of radiation is most suited for observing diffraction patterns on crystalline solids: radio waves, visible light, or X-rays? Which type of radiation is best suited for observing diffraction patterns on crystalline solids? Explain. X-rays have the highest resolving power. If X-rays were employed to create the diffraction patterns of a typical crystal, what effect do you think it would have on the patterns? In a situation when an electron and a proton are moving at the same speed, which one has the shorter de Broglie wavelength is determined.

• Because macroscopic things are so little, it’s difficult to perceive the wave-like character of matter every day.
• What is the wavelength of a neutron when it is not in motion?
• In order for the Davisson–Germer experiment to be successful, it must be carried out in a vacuum chamber.
• in order to prevent colliding with air molecule collisions

### Problems

What is the maximum velocity at which an electron will have a wavelength of 1.00 m? The de Broglie wavelength of an electron traveling at a speed of what is the answer. 145.5 p.m. EST The de Broglie wavelength of an electron that has been accelerated from rest by a potential difference of 20 keV is known as the de Broglie length. When the kinetic energy of a proton is 2.0 MeV, what is the de Broglie wavelength of that proton? What is 10.0 MeV? 20 fm and 9 fm are the frequencies used. When a 10-kg football player is moving at an average pace of 8.0 meters per second, what is the de Broglie wavelength?

• What is the energy of an electron whose de Broglie wavelength is the same as that of a photon of yellow light with a wavelength of 590 nm?
• How much energy does an electron have?
• a.
• 0.846 nm; c.
• Do you know what the speed and energy of this neutron are?
• 8:00 p.m.
• Provide your response in units of c.
• Calculate the wavelength of a proton that is traveling at 1.00 percent the speed of light (where the speed of light is 1.00 percent).

### Glossary

Observation of the Davisson–Germer experiment Particles of matter can behave like waves, according to the de Broglie hypothesis, which was the first experiment to demonstrate this. The de Broglie wave was the first experiment to demonstrate this.

Any object with mass and motion will have a matter wave associated with it. The group velocityof a wave is the speed at which energy moves with the group velocity. The theory of quantum mechanics, which describes the physics of atoms and subatomic particles, is defined as follows:

## De Broglie Wavelength: Definition, Equation & How to Calculate

French physicist Louis de Broglie won the Nobel Prize in 1929 for groundbreaking work in quantum mechanics. His work to show mathematically how subatomic particles share some wave properties was later proven correct through experiment.

## Wave-Particle Duality

Duality between wave and particle qualities is characterized by particles that display both wave and particle properties. Initially discovered as electromagnetic radiation (light), which may be characterized as either an electromagnetic wave or a subatomic particle known as the photon, this natural phenomena was first observed in the universe. When behaving as a wave, light obeys the same laws of physics as other waves in the natural world. As an example, in a double-slit experiment, the patterns of wave interference that occur demonstrate the wave nature of light may be seen.

Photons appear to travel in separate packets of kinetic energy in these situations, obeying the same principles of motion that any other particle would (although photons are massless).

## Matter Waves and the de Broglie Hypothesis

The de Broglie hypothesis is the concept that matter (or anything with mass) may show wavelike features as well as other properties. The fact that these matter waves are produced is important because they are essential to a quantum mechanical explanation of the world – without them, scientists would not be able to describe nature on the smallest possible scale. Because of this, it is in quantum theory where the wave aspect of matter is most obvious, for example, when investigating the behavior of electrons.

Assuming that particles and their wave forms have identical energies, we may solve for E = mc2 = hf (where E is energy, m is mass, c is the speed of light in a vacuum, h is the Planck constant, and f is frequency) by setting the first two equations equal to one another: Due to the fact that large particles do not move at the speed of light, by substituting the particle’s velocity v for the constant c, the result is :mv2 = hf.

After that, substituting f with v/ (from the wave speed equation, where v represents wavelength) and simplifying:lambda = frac is the result.

The de Broglie equation is the name given to this situation.

## de Broglie Wavelength Calculations

• The wavelength of a particle with momentum p is determined by the formula: h/p = h/p.

When the wavelength (m) is expressed as a unit of meters, the Planck’s constant (6.631 10 -34J) is expressed as a unit of time (Js), and the momentum (kgm/s) is expressed as a unit of time. In the following example, what is the de Broglie wavelength of 9.1 10 -31 10 6 meters per second? Since: Remember that for extremely big masses – that is, on the size of ordinary things like a baseball or a vehicle – this wavelength shrinks to vanishingly small values until it becomes negligible. As a result, the de Broglie wavelength has no effect on the behavior of things that we can perceive without the use of a microscope; for example, it is not required to calculate where a baseball pitch will land or how much effort is required to drive a car down the road.

A crucial number in explaining what electrons do, on the other hand, is the de Broglie wavelength of an electron, which is relevant since the rest mass of an electron is tiny enough to be on the quantum scale.

## Wave Nature of Matter and De Broglie’s Equation

The wave nature of matter is one of the most puzzling concepts in physics today. Waves, on the other hand, are scattered over space rather of being restricted to a certain area. It has been proved that light may be either a particle or a wave in its nature, respectively. In the photoelectric effect, electrons and photons behave like billiard balls, exhibiting particle properties similar to those of a billiard ball. You will, however, recall the Diffraction experiment as well as the Interference Rings experiment.

The waveform of light may be seen in a number of situations.

It even has an impact on our ability to see!

However, the absorption of light by the rods and cones of the retina correlates to the particle nature of light!

### Wave Nature of Matter

When it comes to classical mechanics, radiation is represented as a wave, whereas particles are represented as hard billiard balls. This research revealed that radiation may behave both as waves and as particles under certain conditions. Radiation and moving particles are both capable of supplying energy and momentum to a wide range of objects. As a result of nature’s fundamental symmetry, De Broglie postulated in 1924 that matter should have two states: solid and liquid. Unlike objects, particles do not have a fixed position in the space in which they exist.

De Broglie’s Equation is a mathematical formula that describes the relationship between two variables.

As a result of the de Broglie connection, the theory of waves was established.

• When a particle has a wavelength of a certain length, its momentum is equal to that wavelength. The Planck’s constant is equal to the wavelength of the particle times its momentum.

Because this relationship demonstrates that matter may behave similarly to a wave, it is critical to comprehend its significance. According to De Broglie’s Equation, a moving particle, no matter how little or huge it may be, has a unique wavelength that cannot be duplicated. It is possible to perceive the wave element of matter in macroscopic things if we look closely enough. With increasing size, the wavelength of an object becomes smaller and smaller until it is no longer observable, which explains why macroscopic objects in the real world do not exhibit waves of any kind.

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In the equation, the Plank’s constant serves as a connection between the wavelength and the momentum.

The symbol c represents the speed of light in a vacuum, and Planck’s notion argues that the energy of a photon is determined by the frequency and wavelength of the photon.

The momentum of a particle with mass m traveling at a speed of v is represented by the equation p=mvis. As a result, it must have a wavelength equal to = h p = h p. (mv)

### Heisenberg’s Uncertainty Principle

The Davisson-Germer experiment, which included diffracting electrons through a crystal, showed without any reasonable question that matter is a wave. He received the Nobel Prize in Physics in 1929 for his theory of matter waves, which paved the way for a whole new field of research known as Quantum Physics to be discovered and explored. Heisenberg’s Uncertainty Principle integrates the matter-wave concept in a straightforward manner. According to the Uncertainty Principle, it is impossible to know both the momentum and the location of an electron at the same time for any other particle, including an electron.

Heisenberg’s Uncertainty Equation (sometimes known as the Heisenberg Uncertainty Equation): A particle’s momentum and position cannot be precisely calculated at the same time, according to the Uncertainty Principle of quantum mechanics.

The unknowns are connected by the expression,x p h 4where

• X denotes the uncertainty in the particle’s location, while p denotes the uncertainty in the particle’s momentum.

When the momentum of a particle is perfectly measured (i.e., when p=0), the uncertainty x in its position becomes infinite. To be in accordance with de Broglie’s equation, an elementary particle with known momentum must likewise have a known wavelength. A certain wavelength can be found throughout all of space, all the way to the infinitude of the universe. The probability interpretation of Born states that this indicates that the particle is not localized in space and that the uncertainty regarding its position is thus unbounded.

In order to represent any particle, localized waves (wave packets), which comprise wavelengths of varied lengths, should be employed to represent it.

### Sample Problems

TheMatter-wave packet is a technical term that means something different to different people. When compared to a progressive wave, a wave packet is a superposition of sinusoidal waves of different wavelengths that is constrained in space. An accurate representation of the position and momentum of a particle may be obtained by employing a wave packet. The group velocity of the packet is used to compute the velocity of the particle. To characterize a wave packet, two different hypotheses are used: The De Broglie hypothesis and the uncertainty principle.

1. Photons, which are massless particles, are responsible for the transmission of radiation.
2. According to Max Planck’s hypothesis, the energy of a photon is proportional to the frequency and wavelength of the photon.
3. If a baseball weighs 0.1 kg and travels at 60 meters per second, what is the de Broglie wavelength of that baseball?
4. The speed of a baseball is measured in meters per second (v = 60 m s).
5. The de Broglie wavelength of an object is provided as: = h (m v)=6.626 10 34 (0.1 60) The de Broglie wavelength of an object is given as: 1 104 10 34 = 1.104 10 34 = 1.104 10 34m As a result, the de Broglie wavelength is 1.104 10 34 meters.
6. Answer:A proton’s de Broglie wavelength is 1800 times smaller than that of an electron because of the proton’s significantly higher mass.
7. Because of its larger wavelength, the electron has a brighter brilliance than the positron.
8. The electron’s p-momentum uncertainty is 2p10 6 that of the electron’s initial momentum, indicating that the electron is in a state of uncertainty.
9. Answer:Given: The electron’s mass, m, is 9.11031 kilograms.
10. The electron’s momentum is given by the equation p = mv= 9.110 31 20 kg m s= 18210 31 kg m/s.
11. Planck’s constant, h = 6.626 10 34J s, is a measure of the speed of light.

Heisenberg Uncertainty The following is the formula: x p h 4x p h (4x p)x h (4x p)x h (6.626 10 34J s) x h (4x p)x h (6.626 10 34J s) x h (4x p)x h (6.626 10 34J s) x h (4x p)x h (6.626 10 34J s) x h (4 An electron’s location is therefore unknown to the extent of 1.44 meters.

## de Broglie’s proposal

PROFESSOR:ThisisLouis-L-O-U-I-Sd-eBroglie. Andthisisnothyphenatednortogether. Theyareseparate. Andthedisnotcapitalizedapparentlytoo. The photonasaparticleisclear, and the photon is also a wave, and it is the year 1924. AndLouisdeBrogliebasicallyhadagreatinsightinwhichhesaidthatifthisissupposedtobeauniversalorarealbasicphysicalpropertythatphotonsarewavesandparticles,weknewthemaswavesandnowweknowthey’reparticles. Butiftheyaredualedwithrespecttoeachother,bothdescriptionsareindifferentregimesandinasense,aparticleattheendofthedayhaswaveattributesandparticleattributes.

• Notjustthephoton,that’soneexample,buteverybodydoes.
• However, this is fairly intriguing because in quantum mechanics, you have the photon, which is a particle, but it is connected with a wave, and if you are a little quick, you say, oh, yes, the electromagnetism wave, but in quantum mechanics, it is the probability amplitude to be some work.
• We didn’t monitor the waves or a single photon; instead, the wave was a wave of probability amplitude, which they didn’t understand at the time.
• Butwhatisleftunsaidhereisyes,youhaveawave,butawaveofwhat?
• Soit’sverystrangethatthefundamentalequationforawavethatrepresentsaparticleisnotanelectricfieldorasoundwaveorthis,it’sforallofthemisaprobabilitywave.
• But that’s exactly what deBroglie’s intention was.

Moreover, if you conceive of it as a wave, you would say it has a frequency.

Whenwehavethis,wehaveaparticlewaveduality.

Universal.

Thesearethematterwavesthatwe’regoingtotrytodiscuss.

Soyoucouldsaywaveofwhat?

Soremember,theComptonwavelengthwasauniversal-foranyparticle,theComptonwavelengthisjustonenumber,butjustforphotons,thewavelengthdependsonthemomentum,soingeneral,itshouldbedependentonthemomentum.

Soit’saprettydaringstatement.

Itwasaverynaturalconjecture-we’lldiscussitalotmorenextlecture-butthereareverylittleevidenceforit.

Theywouldbehave,collidingintolatticeslikewaves,andthoseareratherfamousexperimentsofDavissonandGermer.

Andthen,eventually,they’vebeendoneforbiggerandbiggerparticles,sothatit’snotjustsomethingthatyoudowithelementaryparticlesnow.

People conduct those experiments, and the results are shown in very beautiful movies in which you see a swarm of electrons hitting the screen and then-I’ll give you some links so you can find the ma- and you see one electron fall here and it gets detected and two electrons, three electrons, four electrons, five electrons, six electrons- by the time you get 10,000 electrons, you see lots of electrons here, very well here, lots of electrons here, and the entire interference pattern is created by Soparticles,bigparticlesinterfere,notjustphotonsinterfere.

Sothoseparticleshavesomewaves,somematterwavesdiscoveredbydeBroglie,andnextlecture,we’regoingtotrackthestoryfromdeBroglietotheSchrodingerequationwherethenatureofthewavesuddenlybecomesclear.

## SOLVED:Louis de Broglie \$ bold hypothesis assumes that it is possible to assign wavelength A to every particle possessing some momentum p by the relationship where h is Plancks constant h _ 6.626 10 S). This applies not only to subatomic particles like electrons_ but every particle and object that has momentum To help you develop some number sense for de Broglie wavelengths of commona everyday objects_ try below calculations_ Use Plancks constant h 6.626 10 J . S; other necessary constants will be given below To enter answers in scientific notation below use the exponential notation For example, 3.14 X 10 would be entered as “3.14E.14”, Hint Air molecules (mostly oxygen and nitrogen) move at speeds of about 380 m/s. If mass of air molecules are about 5 < 10 kg, their de Broglie wavelength Consider baseball thrown at speed 50 m/s. If mass of the baseball is 0.14 kg its de Broglie wavelength is The Earth orbits the Sun at speed of 29800 mfs. Given that the mass of the Earth is about 6.0 * 1024 kg, its de Broglid wavelength Yes, many of these numbers are absurdly srall, which is why think you should enter the powers of 10.

In the text (Section 11.6), it was stated that current theories of atomic structure suggest that all matter and all energy exhibit both particle-like and wave-like properties when the appropriate conditions are met, although the wave-like nature of matter is only visible in extremely small and extremely fast-moving objects. A particle’s wave length \$(lambda)\$ is seen to be related to the particle’s mass and velocity, and this relationship is referred to as the de Broglie relationship. It is represented by the equation \$\$lambda=h / m v\$\$, where \$h\$ is Planck’s constant \$left(6.63 x 10mathrm x right), \$mathrm \$ represents the mass of the particle in kilograms, and \$v\$ represents the velocity of the particle in meters per second.

electron traveling at 0.90 times the speed of light, for example b.

/mathrm \$ c.

a \$150-g\$ ball moving at a speed of \$10.