Answer
$d_{110} = 0.2862 nm $
Work Step by Step
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 $d_{110}$,
$d_{110} = \frac{a}{\sqrt (1)^{2} + (1)^{2} + (0)^{2}} = \frac{0.4047 nm}{\sqrt 2} = 0.2862$
