Materials Science and Engineering: An Introduction

Published by Wiley
ISBN 10: 1118324579
ISBN 13: 978-1-11832-457-8

Chapter 3 - The Structure of Crystalline Solids - Questions and Problems - Page 97: 3.4



Work Step by Step

APF = volume of atoms in a unit cell $\div$ total unit cell volume For BCC: Each corner atom counts as 1/8 of an atom in a unit cell. There are 8 corner atoms = 1 atom total. The center atom counts as an entire atom. Therefore, a BCC unit cell has 2 atoms. The volume of a sphere is $\frac{4}{3}\pi r^3$. We have two atoms in the BCC unit cell, so our numerator in the APF is $\frac{8}{3}\pi r^3$. For the denominator, this is just the volume of the unit cell, or $a^3$. We know that for a BCC unit cell, $a = \frac{4R}{\sqrt{3}}$. $a^3$ is therefore $= \frac{64R^3}{3\sqrt{3}}$ Dividing the numerator by denominator, we get: $$\frac{\sqrt{3}\pi}{8}$$ Simplifying, this is $\approx$0.68
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