## PHYS7635 Solid-state physics课程简介

Prerequisite: good undergraduate solid-state physics course (e.g., PHYS 4454), undergraduate statistical mechanics, and familiarity with graduate-level quantum mechanics.
D. Ralph.
Survey of the physics of solids: crystal structures, X-ray diffraction, phonons, and electrons. Selected topics from semiconductors, magnetism, superconductivity, disordered materials, dielectric properties, and mesoscopic physics. The focus is to enable graduate research at the current frontiers of condensed matter physics.

## PREREQUISITES

Prerequisite: good undergraduate solid-state physics course (e.g., PHYS 4454), undergraduate statistical mechanics, and familiarity with graduate-level quantum mechanics.
D. Ralph.
Survey of the physics of solids: crystal structures, X-ray diffraction, phonons, and electrons. Selected topics from semiconductors, magnetism, superconductivity, disordered materials, dielectric properties, and mesoscopic physics. The focus is to enable graduate research at the current frontiers of condensed matter physics.

## PHYS7635 Solid-state physics HELP（EXAM HELP， ONLINE TUTOR）

Name the three most important probes used in diffraction experiments on crystals. What is the one essential condition that they must all satisfy? Describe briefly what each probe is suitable for.

The three most important probes used in diffraction experiments on crystals are:

1. X-rays
2. Neutrons
3. Electrons

The one essential condition that they all must satisfy is that the wavelength of the probe must be similar in size to the interatomic spacing in the crystal. This is known as the Bragg condition and is expressed as nλ = 2d sinθ, where n is an integer, λ is the wavelength of the probe, d is the interatomic spacing, and θ is the angle of incidence.

Brief descriptions of each probe and what they are suitable for:

1. X-rays: X-rays are the most commonly used probe in crystallography. They have a short wavelength and high penetration power, allowing them to diffract off the atoms in the crystal and produce a diffraction pattern. X-ray diffraction is suitable for determining the crystal structure of a wide range of materials, including small molecules, proteins, and minerals.
2. Neutrons: Neutrons have a longer wavelength than X-rays and can penetrate deeper into the sample, making them useful for studying the structure of materials that are difficult to study with X-rays, such as hydrogen-containing compounds. Neutron diffraction is also sensitive to the positions of light atoms, such as hydrogen, which are often invisible to X-rays.
3. Electrons: Electron diffraction is a powerful technique for studying the structure of small crystals and thin films. Electrons have a much shorter wavelength than X-rays and neutrons, allowing them to diffract off smaller structures. Electron diffraction is also highly sensitive to the arrangement of atoms in a crystal, making it useful for studying defects and disorder in crystal structures.

Calculate the geometrical structure factor for a $b c c$ lattice. Name some important planes which will be missing from the X-ray diffraction pattern.

In a $b c c$ lattice, the basis consists of two atoms located at $(0,0,0)$ and $(\frac{1}{2},\frac{1}{2},\frac{1}{2})$ of the unit cell. The structure factor for a Bragg reflection from a plane with Miller indices $(h,k,l)$ is given by:

F_{hkl} = \sum_j f_j e^{2 \pi i (hx_j + ky_j + lz_j)}Fhkl​=∑j​fj​e2πi(hxj​+kyj​+lzj​)

where $f_j$ is the atomic scattering factor for atom $j$, and $x_j$, $y_j$, and $z_j$ are the fractional coordinates of atom $j$ within the unit cell.

For the $b c c$ lattice, we have $f_1 = f_2 = f$, where $f$ is the atomic scattering factor for the atoms in the basis. Using the coordinates for the two atoms in the basis, we have:

F_{hkl} = f\left[ e^{2 \pi i (0\times h + 0 \times k + 0 \times l)} + e^{2 \pi i (\frac{1}{2} \times h + \frac{1}{2} \times k + \frac{1}{2} \times l)}\right]Fhkl​=f[e2πi(0×h+0×k+0×l)+e2πi(21​×h+21​×k+21​×l)]

Simplifying this expression, we get:

F_{hkl} = f\left[ 1 + (-1)^{h+k+l}\right]Fhkl​=f[1+(−1)h+k+l]

Therefore, the geometrical structure factor for a $b c c$ lattice is given by:

|F_{hkl}|^2 = f^2 \left[ 2 + 2(-1)^{h+k+l}\right]∣Fhkl​∣2=f2[2+2(−1)h+k+l]

This expression shows that the intensity of the diffraction peaks will depend on whether $h+k+l$ is even or odd. If $h+k+l$ is odd, the structure factor will be zero, and no diffraction peak will be observed for that plane.

Some important planes which will be missing from the X-ray diffraction pattern for a $b c c$ lattice are those with odd values of $h+k+l$, such as $(1,1,3)$, $(2,3,1)$, etc. These planes are called “systematically absent” planes, and their absence is a characteristic feature of the $b c c$ lattice.

## Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.

Statistics-lab™可以为您提供cornell.edu PHYS7635 Solid-state physics固体物理的代写代考辅导服务！ 请认准Statistics-lab™. Statistics-lab™为您的留学生涯保驾护航。

## PHYS7635 Solid-state physics课程简介

Prerequisite: good undergraduate solid-state physics course (e.g., PHYS 4454), undergraduate statistical mechanics, and familiarity with graduate-level quantum mechanics.
D. Ralph.
Survey of the physics of solids: crystal structures, X-ray diffraction, phonons, and electrons. Selected topics from semiconductors, magnetism, superconductivity, disordered materials, dielectric properties, and mesoscopic physics. The focus is to enable graduate research at the current frontiers of condensed matter physics.

## PREREQUISITES

Prerequisite: good undergraduate solid-state physics course (e.g., PHYS 4454), undergraduate statistical mechanics, and familiarity with graduate-level quantum mechanics.
D. Ralph.
Survey of the physics of solids: crystal structures, X-ray diffraction, phonons, and electrons. Selected topics from semiconductors, magnetism, superconductivity, disordered materials, dielectric properties, and mesoscopic physics. The focus is to enable graduate research at the current frontiers of condensed matter physics.

## PHYS7635 Solid-state physics HELP（EXAM HELP， ONLINE TUTOR）

What are Laue equations for diffraction of X-rays by a crystalline solid? Show that Bragg’s equation is a special case of Laue equation.

The Laue equations describe the diffraction of X-rays by a crystal, and are based on the principle of constructive interference between X-rays that are diffracted from different planes within the crystal lattice. The equations were first derived by Max von Laue in 1912, and can be written as:

2d\sin\theta_{hkl} = \lambda

where d is the distance between the crystal planes with Miller indices h, k, and l, \theta_{hkl} is the angle between the incident X-ray beam and the plane (known as the Bragg angle), \lambda is the wavelength of the X-rays, and 2d\sin\theta_{hkl} is the path difference between X-rays diffracted from adjacent planes.

Bragg’s law is a special case of the Laue equations, and is derived by considering only those reflections in which the path difference between adjacent planes is equal to the wavelength of the X-rays (i.e., 2d\sin\theta_{hkl} = \lambda). This condition corresponds to the so-called “Bragg reflection”, in which X-rays are diffracted from adjacent planes in phase and reinforce each other, resulting in a sharp peak in the diffracted intensity.

In summary, the Laue equations describe the diffraction of X-rays by a crystal in general, while Bragg’s law is a special case of the Laue equations that describes the conditions for constructive interference between X-rays diffracted from adjacent planes in the crystal lattice.

Name the three most important probes used in diffraction experiments on crystals. What is the one essential condition that they must all satisfy? Describe briefly what each probe is suitable for.

The three most important probes used in diffraction experiments on crystals are X-rays, electrons, and neutrons. The essential condition that they all satisfy is that they have a wavelength comparable to the spacing of atoms in the crystal lattice.

1. X-rays: X-ray diffraction is the most widely used method for determining the atomic and molecular structure of crystals. X-rays have a wavelength of about 0.1 nm, which is comparable to the spacing of atoms in crystals. X-rays are suitable for determining the positions of atoms in a crystal lattice, as well as the arrangement of molecules in large biomolecules like proteins.
2. Electrons: Electron diffraction is a powerful technique for studying the structure of materials at the atomic level. Electrons have a shorter wavelength than X-rays and are therefore better suited for studying smaller structures. Electrons are suitable for studying the structures of metals, semiconductors, and other materials with a relatively simple crystal structure.
3. Neutrons: Neutron diffraction is a technique that uses a beam of neutrons to determine the structure of materials. Neutrons have a longer wavelength than X-rays and are therefore better suited for studying larger structures. Neutron diffraction is suitable for studying the structure of complex materials like ceramics, polymers, and biological materials like DNA.

In summary, X-rays are best for determining atomic positions, electrons for studying small structures, and neutrons for larger or complex materials. The essential condition that they all satisfy is a comparable wavelength to the spacing of atoms in the crystal lattice.

## Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.