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

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## 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

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