# 6.5 De broglie’s matter waves  (Page 5/10)

 Page 5 / 10

## Neutron scattering

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

## Strategy

We assume that a typical crystal spacing a is of the order of 1.0 Å. To observe a diffraction pattern on such a lattice, the neutron wavelength $\lambda$ must be on the same order of magnitude as the lattice spacing. We use [link] 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 $T=300\text{K},$ we use the relation $K=3}{2}{k}_{B}T,$ where ${k}_{B}=8.62\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\text{eV}\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}\text{K}$ is the Boltzmann constant.

## Solution

We evaluate pc to compare it with the neutron’s rest mass energy ${E}_{0}=940\phantom{\rule{0.2em}{0ex}}\text{MeV}:$

$p=\frac{h}{\lambda }⇒pc=\frac{hc}{\lambda }=\frac{1.241\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\text{eV}·\text{m}}{{10}^{-10}\text{m}}=12.41\phantom{\rule{0.2em}{0ex}}\text{keV}.$

We see that ${p}^{2}{c}^{2}\ll {E}_{0}^{2}$ so $K\ll {E}_{0}$ and we can use the nonrelativistic kinetic energy:

$K=\frac{{p}^{2}}{2{m}_{n}}=\frac{{h}^{2}}{2{\lambda }^{2}{m}_{n}}=\frac{{\left(6.63\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-34}\text{J}·\text{s}\right)}^{2}}{\left(2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-20}{\text{m}}^{2}\right)\left(1.66\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-27}\text{kg)}}=1.32\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-20}\text{J}=82.7\phantom{\rule{0.2em}{0ex}}\text{meV}.$

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

${K}_{T}=\frac{3}{2}{k}_{B}T=\frac{3}{2}\left(8.62\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\text{eV}\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}\text{K}\right)\left(300\text{K}\right)=38.8\phantom{\rule{0.2em}{0ex}}\text{MeV}.$

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

## Significance

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

## 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?

## Strategy

The rest mass energy of a proton is ${E}_{0}={m}_{0}{c}^{2}=\left(1.672\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-27}\text{kg}\right){\left(2.998\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{8}\text{m/s}\right)}^{2}=938\phantom{\rule{0.2em}{0ex}}\text{MeV}.$ When the proton’s velocity is known, we have $\beta =0.75$ and $\beta \gamma =0.75\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}\sqrt{1-{0.75}^{2}}=1.714.$ We obtain the wavelength $\lambda$ and kinetic energy K from relativistic relations.

## Solution

$\lambda =\frac{h}{p}=\frac{hc}{pc}=\frac{hc}{\beta \gamma {E}_{0}}=\frac{1.241\phantom{\rule{0.2em}{0ex}}\text{eV}·\text{μm}}{1.714\left(938\phantom{\rule{0.2em}{0ex}}\text{MeV}\right)}=0.77\phantom{\rule{0.2em}{0ex}}\text{fm}$
$K={E}_{0}\left(\gamma -1\right)=938\phantom{\rule{0.2em}{0ex}}\text{MeV}\left(1\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}\sqrt{1-{0.75}^{2}}-1\right)=480.1\phantom{\rule{0.2em}{0ex}}\text{MeV}$

## Significance

Notice that because a proton is 1835 times more massive than an electron, if this experiment were performed with electrons, a simple rescaling of these results would give us the electron’s wavelength of $\left(1835\right)0.77\text{fm}=1.4\phantom{\rule{0.2em}{0ex}}\text{pm}$ and its kinetic energy of $480.1\phantom{\rule{0.2em}{0ex}}\text{MeV}\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}1835=261.6\phantom{\rule{0.2em}{0ex}}\text{keV}.$

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

$\lambda =1.417\phantom{\rule{0.2em}{0ex}}\text{pm;}$ $K=261.56\phantom{\rule{0.2em}{0ex}}\text{keV}$

## Summary

• De Broglie’s hypothesis of matter waves postulates that any particle of matter that has linear momentum is also a wave. The wavelength of a matter wave associated with a particle is inversely proportional to the magnitude of the particle’s linear momentum. The speed of the matter wave is the speed of the particle.
• De Broglie’s concept of the electron matter wave provides a rationale for the quantization of the electron’s angular momentum in Bohr’s model of the hydrogen atom.
• In the Davisson–Germer experiment, electrons are scattered off a crystalline nickel surface. Diffraction patterns of electron matter waves are observed. They are the evidence for the existence of matter waves. Matter waves are observed in diffraction experiments with various particles.

## Conceptual questions

Which type of radiation is most suitable for the observation of diffraction patterns on crystalline solids; radio waves, visible light, or X-rays? Explain.

X-rays, best resolving power

Speculate as to how the diffraction patterns of a typical crystal would be affected if $\text{γ}\text{-rays}$ were used instead of X-rays.

If an electron and a proton are traveling at the same speed, which one has the shorter de Broglie wavelength?

proton

If a particle is accelerating, how does this affect its de Broglie wavelength?

Why is the wave-like nature of matter not observed every day for macroscopic objects?

negligibly small de Broglie’s wavelengths

What is the wavelength of a neutron at rest? Explain.

Why does the setup of Davisson–Germer experiment need to be enclosed in a vacuum chamber? Discuss what result you expect when the chamber is not evacuated.

to avoid collisions with air molecules

## Problems

At what velocity will an electron have a wavelength of 1.00 m?

What is the de Broglie wavelength of an electron travelling at a speed of $5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\text{m/s}$ ?

145.5 pm

What is the de Broglie wavelength of an electron that is accelerated from rest through a potential difference of 20 keV?

What is the de Broglie wavelength of a proton whose kinetic energy is 2.0 MeV? 10.0 MeV?

20 fm; 9 fm

What is the de Broglie wavelength of a 10-kg football player running at a speed of 8.0 m/s?

(a) What is the energy of an electron whose de Broglie wavelength is that of a photon of yellow light with wavelength 590 nm? (b) What is the de Broglie wavelength of an electron whose energy is that of the photon of yellow light?

a. 2.103 eV; b. 0.846 nm

The de Broglie wavelength of a neutron is 0.01 nm. What is the speed and energy of this neutron?

What is the wavelength of an electron that is moving at a 3% of the speed of light?

80.9 pm

At what velocity does a proton have a 6.0-fm wavelength (about the size of a nucleus)? Give your answer in units of c .

What is the velocity of a 0.400-kg billiard ball if its wavelength is 7.50 fm?

$2.21\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−}20}\text{m/s}$

Find the wavelength of a proton that is moving at 1.00% of the speed of light (when $\beta =0.01\right).$

#### Questions & Answers

what is an atom
Aroyameh Reply
All matter is composed of two sets of three dimensions. The first set (1,2,3) decay with a positive charge. The second set (4,5,6) decay with a negative charge. As they decay, they create space (7 8,9) dimensions.
John
Two sets of (1,2,3,4,5,6) dimensions create a proton, a neutron, and an electron. This is the primordial atom.
John
A 10kg mass lift to a height of 24m and release. what is the total energy of the system
ADEPOJU Reply
mechanics is that branch of physical and mathatics that
ADEPOJU
E=Mgh=10*10*24=2400J
Adamu
what is the difference between a molecule and atom
Natanim Reply
Atoms are single neutral particles. Molecules are neutral particles made of two or more atoms bonded together.
Manfred
what I'd dynamic propulsion
Elias Reply
A body quadruples its momentum when its speed doubles.What was the initial speed in units of c.i.e..what was u/c ?
Lekshmi Reply
what is enthalpy?
prabir Reply
a thermodynamic quantity equivalent to the total heat content of a system
RAMLA
proparty of tharmo dainamic
bloch
What is the meaning of Nuclear Fission?
Benita Reply
what do you mean by dynamics single particles
Peacekamei Reply
عند قذف جسم إلى أعلى بسرعة إبتدائية فإنه سيصل إلى ارتفاع معين (أقصى ارتفاع) ثم يعود هابطاً نحو سطح الأرض .   إذا قُذِفَ جسم إلى أعلى ووجد أن سرعته 18 م / ث عندما قطع 1/4 المسافة التي تمثل أقصى ارتفاع سيصله فالمطلوب إيجاد السرعة التي قُذِف بها بالمتر / ث . إن هذه السرعة هي واحدة من الإجابات التالية
Aml Reply
what is light
Ayebanifesunday Reply
light is a kind of radiation That stimulates sight brightness a source of illumination.
kenneth
Electromagnet radiation creates space 7th, 8th, and 9th dimensions at the rate of c.
John
That is the reason that the speed of light is constant.
John
This creation of new space is "Dark Energy".
John
The first two sets of three dimensions, 1 through 6, are "Dark Matter".
John
As matter decays into luminous matter, a proton, a neutron, and an electron creat deuterium.
John
There are three sets of three protons, 9.
John
There are three sets of three neutrons, 9.
John
A free neutron decays into a proton, an electron, and a neutrino.
John
There are three sets of five neutrinoes, 15.
John
Neutrinoes are two dimensional.
John
A positron is composed of the first three dimensions.
John
An electron is composed of the second three dimensions.
John
What is photoelectric
Hsssan Reply
light energy (photons) through semiconduction of N-P junction into electrical via excitation of silicon purified and cristalized into wafers with partially contaminated silicon to allow this N-P function to operate.
Michael
i.e. Solar pannel.
Michael
Photoelectric emission is the emission of electrons on a metal surface due to incident rays reflected on it
Benita
If you lie on a beach looking at the water with your head tipped slightly sideways, your polarized sunglasses do not work very well.Why not?
Rakhi Reply
it has everything to do with the angle the UV sunlight strikes your sunglasses.
Jallal
this is known as optical physics. it describes how visible light, ultraviolet light and infrared light interact when they come into contact with physical matter. usually the photons or light upon interaction result in either reflection refraction diffraction or interference of the light.
Jallal
I hope I'm clear if I'm not please tell me to clarify further or rephrase
Jallal
what is bohrs model for hydrogen atom
Swagatika Reply
hi
Tr
Hello
Youte
Hi
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hi
Siddiquee
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Propessor Reply
1.79×10_¹⁹ km per hour
Swagatika
3×10^8
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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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 By Madison Christian By Jemekia Weeden By Madison Christian By OpenStax By Rhodes By OpenStax By Laurence Bailen By Brianna Beck By Angelica Lito By Maureen Miller