/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Imagine the primitive cubic latt... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 23

Imagine the primitive cubic lattice. Now imagine grabbing the top of it and stretching it straight up. All angles remain \(90^{\circ} .\) What kind of primitive lattice have you made?

Short Answer

Expert verified
After stretching the primitive cubic lattice vertically while maintaining all angles at \(90^{\circ}\), we form a Primitive Orthorhombic lattice. It has two edges of equal length 'a' (width and depth), and a vertical edge of different length 'c', with angles remaining unchanged at \(90^{\circ}\).

Step by step solution

01

Understand the initial unit cell

Initially, we have a primitive cubic lattice. This means that the unit cell is a cube with all angles equal to \(90^{\circ}\) and all edges have the same length.
02

Stretching the lattice vertically

As per the exercise, we stretch the top of the lattice along the vertical axis. All angles remain the same, so they continue to be at \(90^{\circ}\). However, stretching changes the edge length along this axis. Let's assume that the initial edge length was 'a' and after stretching the vertical edge length is now 'c'.
03

Identify the new lattice

Now we have a unit cell with two edges (width and depth) having the same length 'a' and the vertical edge (height) with a different length 'c'. The angles remain unchanged at \(90^{\circ}\). Thus, we have a lattice system with right angles and different edge lengths along different axes. This system is known as an Orthorhombic lattice system.
04

Identifying the type of orthorhombic lattice

As we started with a primitive cubic lattice and only stretched the vertical edge, there is only one lattice point per unit cell. Therefore, the new lattice we have formed is a Primitive Orthorhombic lattice. So, after stretching the primitive cubic lattice, a Primitive Orthorhombic lattice is formed.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Indicate whether each statement is true or false: $$ \begin{array}{l}{\text { (a) Elastomers are rubbery solids. }} \\ {\text { (b) Thermosets cannot be reshaped. }} \\ {\text { (c) Thermoplastic polymers can be recycled. }}\end{array} $$

Red light-emitting diodes are made from GaAs and GaP solid solutions, GaP \(_{x} A s_{1-x}(\) see Exercise 12.79\() .\) The original red LEDs emitted light with a wavelength of 660 nm. If we assume that the band gap varies linearly with composition between \(x=0\) and \(x=1,\) estimate the composition (the value of \(x\) ) that is used in these LEDs.

Imagine that you have a metal bar sitting half in the sun and half in the dark. On a sunny day, the part of the metal that has been sitting in the sun feels hot. If you touch the part of the metal bar that has been sitting in the dark, will it feel hot or cold? Justify your answer in terms of thermal conductivity.

Indicate the type of solid (molecular, metallic, ionic, or covalent-network) for each compound: (a) InAs, (b) MgO, \((\mathbf{c}) \mathrm{HgS},(\mathbf{d}) \mathrm{In},(\mathbf{e}) \mathrm{HBr}\)

In their study of X-ray diffraction, William and Lawrence Bragg determined that the relationship among the wavelength of the radiation \((\lambda),\) the angle at which the raveation is diffracted \((\theta),\) and the distance between planes of atoms in the crystal that cause the diffraction \((d)\) is given by \(n \lambda=2 d \sin \theta . X\) rays from a copper \(X\) -ray tube that have a wavelength of 1.54\(\hat{\mathrm{A}}\) are diffracted at an angle of 14.22 degrees by crystalline silicon. Using the Bragg equation, calculate the distance between the planes of atoms responsible for diffraction in this crystal, assuming \(n=1\) (first-order diffraction).
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.