# cha4_6 - PHYSICS 880.06 (Fall 2004) Problem Set 3 Solution...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PHYSICS 880.06 (Fall 2004) Problem Set 3 Solution 3.1 A&M Chapter 4 Problem 1 (a) Base-centered cubic: is a Bravais lattice, with a set of three primitive vectors that can be chosen as a 1 = a 2 ( x- y ) , a 2 = a 2 ( x + y ) , a 3 = a z , where a is the length of the side of the cube. Other choices of the primitive vectors include, e.g., a 1 = a x , a 2 = a 2 ( x + y ) , a 3 = a z . (b) Side-centered cubic: is NOT a Bravais lattice, since it contains a 2 ( x + z ) and a 2 ( y + z ), but not the sum a 2 ( x + y +2 z ). One representation of this structure with the smallest basis is a simple cubic (Bravais) lattice of side a with a three-point basis , a 2 ( x + z ) , a 2 ( y + z ) . (c) Edge-centered cubic: is NOT a Bravais lattice, since it contains a 2 x and a 2 y , but not the sum a 2 ( x + y ). This structure can be described by a simple cubic lattice of side a with a four-point basis , a 2 x , a 2 y , a 2 z . A&M Chapter 4 Problem 3 See Fig. 4.18 in A&M for a picture of the diamond structure. Look at the triangle formed by the three points located at p 1 = , p 2 = a 4 ( x + y + z ), and p 3 = a 2 ( x + y ). The angle between the two vectors ( p 1- p 2 ) and ( p 3- p 2 ) is cos = ( p 1- p 2 ) ( p 3- p 2 ) | (...
View Full Document

## This note was uploaded on 06/07/2010 for the course MECH 1122 taught by Professor Dont during the Spring '10 term at A.T. Still University.

### Page1 / 3

cha4_6 - PHYSICS 880.06 (Fall 2004) Problem Set 3 Solution...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online