Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 65

In FCC lattice A, B, C, D atoms are arranged at corners, face centres, octahedral voids and tetrahedral voids respectively, then the body diagonal contains: (1) \(2 \mathrm{~A}, 2 \mathrm{D}\) (2) \(2 \mathrm{~A}, \mathrm{C}, 2 \mathrm{D}\) (3) \(2 \mathrm{~A}, 2 \mathrm{~B}, \mathrm{D}\) (4) \(2 \mathrm{~A}, 2 \mathrm{~B}, 2 \mathrm{C}\)

Short Answer

Expert verified
Option (2) is correct: 2A, C, 2D.

Step by step solution

01

Identify atom positions in FCC lattice

In a face-centered cubic (FCC) lattice: - 'A' atoms are at the corners, - 'B' atoms are at the face centers, - 'C' atoms are in the octahedral voids, - 'D' atoms are in the tetrahedral voids.
02

Understand body diagonal

A body diagonal in an FCC lattice extends from one corner of the cube to the opposite corner, passing through the interior of the cube.
03

Calculate positions on the body diagonal

The body diagonal of an FCC lattice intersects the following atom positions: - Two 'A' atoms at the corners.- No 'B' atoms on body diagonal (all face center).- One 'C' atom at the center of the cube (octahedral void).- Two 'D' atoms in the path since tetrahedral voids are distributed within the cube.
04

Verify and match the options

From the calculation, the body diagonal contains: - 2 'A' atoms- 1 'C' atom- 2 'D' atomsMatched Option: (2) 2A, C, 2D.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Crystal Structure
Understanding the crystal structure is essential for studying material properties. Crystals are composed of atoms arranged in a highly ordered repeating pattern. One common type of crystal structure is the face-centered cubic (FCC) lattice. Here, atoms are positioned at each of the cube's corners and the centers of all the cube faces. This specific arrangement leads to multiple types of voids and unique attributes for the geometry.
The FCC lattice has key features:
  • 'A' atoms at the cube's corners.
  • 'B' atoms at the face centers.
  • 'C' atoms in octahedral voids.
  • 'D' atoms in tetrahedral voids.
Mastering this layout allows you to predict which atoms lie along specific paths, such as the body diagonal of the cube.
Body Diagonal Calculation
The body diagonal is an imaginary line that stretches from one corner of a cube to the opposite corner, crossing through the center. To determine the atoms present along this diagonal in an FCC lattice, let's break it down.
First, consider where the body diagonal starts and ends. It begins at one corner, passes through the cube's center, and exits through the opposite corner.
In an FCC arrangement:
  • The diagonal intersects two corner atoms, both 'A'.
  • It crosses the center of the cube where one 'C' atom resides (octahedral void).
  • The diagonal also passes through positions related to tetrahedral voids, containing two 'D' atoms distributed within the cube.
Therefore, the body diagonal comprises: 2 'A' atoms, 1 'C' atom, and 2 'D' atoms. It's crucial to visualize this path within the lattice to grasp the arrangement fully.
Octahedral and Tetrahedral Voids
Voids in crystal structures are spaces where atoms are not packed. Understanding voids helps in explaining how atoms and ions pack together within materials.
In an FCC lattice, we encounter:
  • Octahedral Voids (C): These are hollow spaces forming at the body center and edge centers of the cube. Every FCC unit cell has one octahedral void located at the cube's center.
  • Tetrahedral Voids (D): Smaller than octahedral voids, these occur when four atoms form a tetrahedron, creating a void at the center. Within an FCC unit cell, there are eight tetrahedral voids positioned strategically within the lattice.
Recognizing the presence and location of these voids aids in identifying how atoms like 'C' (octahedral void atoms) and 'D' (tetrahedral void atoms) align along specific lattice paths like the body diagonal.
This deeper understanding of voids and how they affect lattice structure solidifies your grasp on finding the specific atoms along unique geometric features like the body diagonal.

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

Select the correct choice(s): (1) The gravitational field inside a spherical cavity, within a spherical planet (mass is uniformly distributed) must be non zero and uniform. (2) When a body is projected horizontally at an appreciable large height above the earth, with a velocity less than for a circular orbit, it will fall to the earth along a parabolic path. (3) A body of zero total mechanical energy placed in a gravitational field will escape the field (4) Earth's satellite must be in equatorial plane.

Zn Amalgam is prepared by electrolysis of aqueous \(\mathrm{ZnCl}_{2}\) using \(\mathrm{Hg}\) cathode \((9 \mathrm{gm})\). How much current is to be passed through \(\mathrm{ZnCl}_{2}\) solution for 1000 seconds to prepare a Zn Amalgam with \(25 \% \mathrm{Zn}\) by wt. \((\mathrm{Zn}=65.4)\) (1) \(5.6\) amp (2) \(7.2 \mathrm{amp}\) (3) \(8.85 \mathrm{amp}\) (4) \(11.2 \mathrm{amp}\)

If \(a, b, c\) are non-zero real constant and the number of terms in the expansion of \((a x+b y+c)^{n}\) is 36 , then \(n(n \in N)\) is equal to (1) 7 (2) 8 (3) 9 (4) 10

A certain acid-base indicator is red in acidic solution and blue in basic solution. At \(\mathrm{pH}=5\), \(75 \%\) of the indicator is present in the solution in its blue form. Calculate the \(\mathrm{pH}\) at which the indicator shows \(90 \%\) red form? (Given \(10^{-4.523}=3 \times 10^{-5}\) ) (1) \(3.56\) (2) \(5.47\) (3) \(2.5\) (4) \(7.4\)

\(\int_{\pi}^{5 \pi / 4} \frac{\sin 2 x}{\cos ^{4} x+\sin ^{4} x} d x\) is equal to (1) \(\frac{\pi}{2}\) (2) \(\frac{\pi}{4}\) (3) \(\frac{\pi}{3}\) (4) \(\frac{\pi}{6}\)
See all solutions

Recommended explanations on English Textbooks

Linguistic Terms

Read Explanation

Rhetorical Analysis Essay

Read Explanation

Global English

Read Explanation

5 Paragraph Essay

Read Explanation

Creative Story

Read Explanation

Language Analysis

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.

Sign-up for free