## Materials Science and Engineering: An Introduction

$d_{110} = 0.2862 nm$
Required: Using data for Aluminum in Table 3.1, compute the interplanar spacing for (110) set of planes Solution: From the given Table, Aluminum has an FCC crysta structure and atomic radius= 0.1431 nm. By using Equation 3.1, compute the lattice parameter $a$, $a= 2R\sqrt 2 = (2)(0.1431 nm)(\sqrt 2) = 0.4047 nm$ Using Equation 3.14, determine the $d_{110}$, $d_{110} = \frac{a}{\sqrt (1)^{2} + (1)^{2} + (0)^{2}} = \frac{0.4047 nm}{\sqrt 2} = 0.2862$