Question

Consider a cubic crystal with the lattice constant a. Complete the parts (a)-(c) below. (a) Sketch the crystallographic planes with Miller indices 1 10 and 2 20 . Explain your answer and mark the origin and axes. (b) Explain why these planes are parallel. (c) Find the spacing d between these two planes.

Answer 1

Answer:

(a) See attachment

(b) The two planes are parallel because the intercepts for plane [220] are X = 0,5 and Y = 0,5 and for plane [110] are X = 1 and Y = 1. When the planes are drawn, they keep the same slope in a 2D plane.

(c) $d = \frac{a}{\sqrt{h^{2} + k^{2} + l^{2}}} = \frac{1}{\sqrt{2}} = 0,707$

Explanation:

(a) To determine the intercepts for an specific set of Miller indices, the reciprocal intercepts are taken as follows:

For [110]

$X = \frac{1}{1} = 1; Y = \frac{1}{1} = 1; Z = \frac{1}{0} = \inf.$

For [220]

$X = \frac{1}{2} = 0,5;Y = \frac{1}{2} = 0,5;Z = \frac{1}{0} = \inf.$

The drawn of the planes is shown in the attachments.

(b) Considering the planes as two sets of 2D straight lines with no intersection to Z axis, then the slope for these two sets are:

For (1,1):

$K_1 = \frac{1}{1} = 1$

For (0.5, 0.5):

$K_2 = \frac{0.5}{0.5} = 1$

As shown above, the slopes are exactly equal, then, the two straight lines are considered parallel and for instance, the two planes are parallel also.

(c) To calculate the d-spacing between these two planes, the distance is calculated as follows:

The Miller indices are already given in the statement. Then, the distance is:

$\frac{1}{d^{2}} = \frac{h^{2} + k^{2} + l^{2}}{a^{2}}$

$d = \frac{a}{\sqrt{h^{2} + k^{2} + l^{2}}} = \frac{1}{\sqrt{2}} = 0,707$