Answer:
A) The sketches for the required planes were drawn in the first attachment [1 2 1] and the second attachment [1 2 -4].
B) The closest distance between planes are d₁₂₁=a/√6 and d₁₂₋₄=a/√21 with lattice constant a.
C) Five posible directions that electrons can move on the surface of a [1 0 0] silicon crystal are: |0 0 1|, |0 1 3|, |0 1 1|, |0 3 1| and |0 0 1|.
Compleated question:
1. Miller Indices:
a. Sketch (on separate plots) the (121) and (12-4) planes for a face centered cubic crystal structure.
b. What are the closest distances between planes (called d₁₂₁ and d₁₂₋₄)?
c. List five possible directions (using the Miller Indices) the electron can move on the surface of a (100) silicon crystal.
Explanation:
A)To draw a plane in a face centered cubic lattice, you have to follow these instructions:
1- the cube has 3 main directions called "a", "b" and "c" (as shown in the first attachment) and the planes has 3 main coeficients shown as [l m n]
2- The coordinates of that plane are written as: π:[1/a₀ 1/b₀ 1/c₀] (if one of the coordinates is 0, for example [1 1 0], c₀ is ∞, therefore that plane never cross the direction c).
3- Identify the points a₀, b₀, and c₀ at the plane that crosses this main directions and point them in the cubic cell.
4- Join the points.
<u>In this case, for [1 2 1]:</u>
<u>for </u><u>:</u>
B) The closest distance between planes with the same Miller indices can be calculated as:
With ,the distance is
with lattice constant a.
<u>In this case, for [1 2 1]:</u>
<u /><u />
<u>for </u><u>:</u>
C) The possible directions that electrons can move on a surface of a crystallographic plane are the directions contain in that plane that point in the direction between nuclei. In a silicon crystal, an fcc structure, in the plane [1 0 0], we can point in the directions between the nuclei in the vertex (0 0 0) and e nuclei in each other vertex. Also, we can point in the direction between the nuclei in the vertex (0 0 0) and e nuclei in the center of the face of the adjacent crystals above and sideways. Therefore:
dir₁=|0 0 1|
dir₂=|0 0.5 1.5|≡|0 1 3|
dir₃=|0 1 1|
dir₄=|0 1.5 0.5|≡|0 3 1|
dir₅=|0 0 1|
