Answer
$d_{111} = 0.1655 nm$
$d_{211} = 0.1170 nm$
Work Step by Step
Required:
Using data for $a$-iron in Table 3.1, compute the interplanar spacings for (111) and (211) sets of planes
Solution:
From Table 3.1, $a$-iron has BCC crystal structure and atomic radius = 0.1241 nm. Using Equation 3.3, compute the lattice parameter $a$,
$a = \frac{4R}{\sqrt 3} = \frac{(4)(0.1241)}{\sqrt 3} = 0.2866 nm$
Using Equation 3.14, $d_{111}$ and $d_{211}$ are computed as,
$d_{111} = \frac{a}{\sqrt (1)^{2} + (1)^{2} + (1)^{2}} = \frac{0.2866}{\sqrt 3} = 0.1655 nm$
$d_{211} = \frac{a}{\sqrt (2)^{2} + (1)^{2} + (1)^{2}} = \frac{0.2866}{\sqrt 6} = 0.1170 nm$
