/*! 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 28 The distance between two nearest... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The distance between two nearest neighbours in BCC lattice of axial length, \(l\), is (a) \(l\) (b) \(\frac{\sqrt{3}}{2} l\) (c) \(\frac{\sqrt{2}}{2} l\) (d) \(\frac{1}{2} l\)

Short Answer

Expert verified
\(\frac{\sqrt{3}}{2} l\)

Step by step solution

01

Understanding the Body-Centered Cubic (BCC) Structure

A BCC lattice has atoms at each corner of a cube and a single atom at the center of the cube. The nearest neighbours of the center atom are the corner atoms.
02

Identify the Nearest Neighbours

The shortest distance between the center atom and one of the corner atoms is a body diagonal of the cube.
03

Calculating the Body Diagonal Length

The body diagonal can be found using the Pythagorean theorem in three dimensions for a cube of side length l, which is \(d = l\sqrt{3}\).
04

Determine Nearest Neighbour Distance

Since the central atom is at the midpoint of the body diagonal, the distance to the nearest neighbour is half the length of the body diagonal. Therefore, the nearest neighbour distance is \(\frac{d}{2} = \frac{l\sqrt{3}}{2}\).

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Body-Centered Cubic Structure
The Body-Centered Cubic (BCC) structure is fascinating, especially to students learning about materials science. This crystal architecture is like a dance of atoms, each one positioned meticulously to form an entire lattice. In a BCC lattice, you have a cube with atoms located at all eight corners. But that's not all. Another guest, the body-centered atom, finds its place right at the heart of the cube.

It is akin to a family photo where, apart from the individuals standing at the corners, there's also a person in the center. Every corner atom touches this center atom, meaning they are the nearest neighbours, which is crucial when we want to measure distances within the lattice. These connections are what give the BCC structure its strength and unique properties used in various materials like iron and chromium.
Crystal Lattice
Now, let's spread our gaze beyond a single cube and into the realm of the crystal lattice, a grid-like pattern that extends in all three dimensions. Think of it like a cosmic grid, each node of which is occupied by an atom, molecule, or ion. It's this repetitive pattern that gives a crystal its identity, from the salt on your dinner table to the sapphire in jewelry.

In the context of our BCC structure, the crystal lattice is like a city where each cube is a building with a very specific design. The atoms at the corners and the center are like the foundation stones of each building. Understanding the layout of this 'crystal city' helps us predict how materials will behave when exposed to various conditions, like stress or heat.
Pythagorean Theorem in three dimensions
Our journey through the crystal lattice brings us to a simple yet astounding mathematical tool: the Pythagorean theorem. But, did you know that this theorem also works in three dimensions? Like the traditional version, which we use to find the length of the diagonal of a square, the three-dimensional Pythagorean theorem finds the diagonal length of our cubic 'building' in the BCC 'city.'

Picture each corner atom being a point in a 3D space assigned with coordinates. To link them, imagine a ‘space-diagonal’ that slices through the cube connecting opposing corners. Using this theorem, we can calculate this diagonal's length which is crucial for determining distances inside the lattice. The theorem assures us that for a cube of side length 'l', the body diagonal, which is the line from one corner to the opposite corner through the cube's body, will have a length of \(l\sqrt{3}\). This cornerstone of geometry is, in essence, a bridge between elegantly simple math and the complex world of crystal structures.

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

In an ionic solid \(\mathrm{AB}_{2} \mathrm{O}_{4}\), the oxide ions form CCP. 'A' and 'B' are metal ions in which one is bivalent and another is trivalent (not necessarily in given order). If all the bivalent ions occupy octahedral holes and the trivalent ions occupy tetrahedral and octahedral voids in equal numbers, then the fraction of octahedral voids unoccupied is (a) \(\frac{1}{2}\) (b) \(\frac{3}{4}\) (c) \(\frac{1}{4}\) (d) \(\frac{7}{8}\)

The only incorrect effect on density by the given defect in solids is (a) Density must decrease by vacancy defect. (b) Density must increase by interstitial defect. (c) Density must increase by impurity defect. (d) Density does not change by dislocation defect.

Which of the following have the least void space fraction in their structure? (a) \(\mathrm{BCC}\) (b) \(\mathrm{BCC}\) and \(\mathrm{HCP}\) (c) \(\mathrm{HCP}\) (d) \(\mathrm{FCC}\) and \(\mathrm{HCP}\)

An ionic crystalline solid, \(\mathrm{MX}_{3}\), has a cubic unit cell. Which of the following arrangement of the ions is consistent with the stoichiometry of the compound? (a) \(\mathrm{M}^{3+}\) ions at the corners and \(\mathrm{X}^{-}\) ions at the face centres (b) \(\mathrm{M}^{3+}\) ions at the corners and \(\mathrm{X}^{-}\) ions at the body centres. (c) \(\mathrm{X}^{-}\) ions at the corners and \(\mathrm{M}^{3+}\) ions at the face centres. (d) \(\mathrm{X}^{-}\) ions at the corners and \(\mathrm{M}^{3+}\) ions at the body centres.

A close packing consists of a base of spheres, followed by a second layer where each sphere rests in the hollow at the junction of four spheres below it and the third layer then rests on these in an arrangement which corresponds exactly to that in the first layer. This packing is known as (a) \(\mathrm{HCP}\) (b) \(\mathrm{CCP}\) (c) square close packing (d) \(\mathrm{BCC}\) packing

See all solutions

Recommended explanations on Chemistry Textbooks

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.