# ENG 2002 Lecture Notes - Diamond Cubic, Monochrome, Diffractometer

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CHAPTER 3

THE STRUCTURE OF CRYSTALLINE SOLIDS

PROBLEM SOLUTIONS

Fundamental Concepts

3.1 What is the difference between atomic structure and crystal structure?

Solution

Atomic structure relates to the number of protons and neutrons in the nucleus of an atom, as well as the

number and probability distributions of the constituent electrons. On the other hand, crystal structure pertains to the

arrangement of atoms in the crystalline solid material.

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Unit Cells

Metallic Crystal Structures

3.2 If the atomic radius of aluminum is 0.143 nm, calculate the volume of its unit cell in cubic meters.

Solution

For this problem, we are asked to calculate the volume of a unit cell of aluminum. Aluminum has an FCC

crystal structure (Table 3.1). The FCC unit cell volume may be computed from Equation 3.4 as

V

C

= 16R32 = (16)(0.143 10-9 m)3(2)= 6.62 10-29 m3

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3.3 Show for the body-centered cubic crystal structure that the unit cell edge length a and the atomic

radius R are related through a =4R/3.

Solution

Consider the BCC unit cell shown below

Using the triangle NOP

(NP)2= a2+ a2=2a2

And then for triangle NPQ,

(NQ)2=(

QP)2+(

NP)2

But

NQ = 4R, R being the atomic radius. Also,

QP = a. Therefore,

(4R)2 = a2 + 2a2

or

a = 4R

3

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