Question

Miller Indices:

a.Sketch (on separate plots) the (121) and () planes for a face centered cubic crystal structure.
b.What are the closest distances between planes (called and )?
c.List five possible directions (using the Miller Indices) the electron can move on thesurface of a (100) siliconcrystal.

Answer 1

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.

Step-by-step 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.

In this case, for [1 2 1]:

`l=1=1/a_0 \rightarrow a_0=1`

`m=2=2/b_0 \rightarrow b_0=0.5`

`n=1=1/c_0 \rightarrow c_0=1`

for
`[1 2 \overline{4}]`:

`l=1=1/a_0 \rightarrow a_0=1`

`m=2=2/b_0 \rightarrow b_0=0.5`

`n=\overline{4}=-4/c_0 \rightarrow c_0=-0.25`

B) The closest distance between planes with the same Miller indices can be calculated as:

With
`\pi:[l m n]` ,the distance is
`d_(lmn)= \displaystyle (a)/(√(l^2+m^2+n^2))` with lattice constant a.

In this case, for [1 2 1]:

`d_(121)= \displaystyle (a)/(√(1^2+2^2+1^2))=(a)/(√(6))=0.41a`

for
`[1 2 \overline{4}]`:

`d_{12\overline{4}}= \displaystyle (a)/(√(1^2+2^2+(-4)^2))=(a)/(√(21))=0.22a`

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|