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| | + | [[Image:Bcc02-unit cell.png|thumb|right|300px|Example of the [[Lattice:BCC|BCC]] unit cell.]] |
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| | The '''unit cell''' is the basic building block of a crystal [[lattice]] (whether an atomic crystal or a nanoscale [[superlattice]]). Crystalline materials have a periodic structure, with the unit cell being the minimal volume necessary to fully describe the repeating structure. There are a finite number of possible symmetries for the repeating unit cell. | | The '''unit cell''' is the basic building block of a crystal [[lattice]] (whether an atomic crystal or a nanoscale [[superlattice]]). Crystalline materials have a periodic structure, with the unit cell being the minimal volume necessary to fully describe the repeating structure. There are a finite number of possible symmetries for the repeating unit cell. |
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Revision as of 09:52, 18 June 2014
Example of the
BCC unit cell.
The unit cell is the basic building block of a crystal lattice (whether an atomic crystal or a nanoscale superlattice). Crystalline materials have a periodic structure, with the unit cell being the minimal volume necessary to fully describe the repeating structure. There are a finite number of possible symmetries for the repeating unit cell.
A unit cell can be defined by three vectors that lie along the edges of the enclosing parallelepped. We denote the vectors as
,
, and
; alternately the unit cell can be described by the lengths of these vectors (
,
,
), and the angles between them:
, the angle between
and 
, the angle between
and 
, the angle between
and 
Mathematical description
Vectors

Relations



Volume

If a, b, and c are the parallelepiped edge lengths, and α, β, and γ are the internal angles between the edges, the volume is

The volume of a unit cell with all edge-length equal to unity is:

Angles
is the angle between
and 
is the angle between
and 
is the angle between
and 
Unit cell definition using parallelepiped with lengths
a,
b,
c and angles between the sides given by α,β,γ (from Wikipedia
fractional coordinates).
Reciprocal vectors
The repeating structure of a unit cell creates peaks in reciprocal space. In particular, we observe maxima (constructive interference) when:

Where
,
, and
are integers. We define reciprocal-space vectors:

And we can then express the momentum transfer (
) in terms of these reciprocal vectors:

Combining with the three Laue equations yields:

Where
is a vector that defines the position of Bragg reflection
for the reciprocal-lattice.
Examples
Cubic
Since
,
, and:

And in reciprocal-space:

So:

Hexagonal
Since
and
, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V=\frac{\sqrt{3}}{2}abc}
, and:

And in reciprocal-space:
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} \mathbf{u} & =\frac{1}{V} \mathbf{b}\times\mathbf{c} & =\frac{1}{V} \begin{bmatrix} \frac{\sqrt{3}}{2} b c \\ -\frac{1}{2} b c \\ 0 \end{bmatrix} & = \begin{bmatrix} \frac{1}{a} \\ \frac{1}{\sqrt{3}a} \\ 0 \end{bmatrix}\\ \mathbf{v} & =\frac{1}{V} \mathbf{c}\times\mathbf{a} & =\frac{1}{V} \begin{bmatrix} 0 \\ a c \\ 0 \end{bmatrix} & = \begin{bmatrix} 0 \\ \frac{2}{\sqrt{3}b} \\ 0 \end{bmatrix}\\ \mathbf{w} & =\frac{1}{V} \mathbf{a}\times\mathbf{b} & =\frac{1}{V} \begin{bmatrix} 0 \\ 0 \\ \frac{\sqrt{3}}{2} a b \end{bmatrix} & = \begin{bmatrix} 0 \\ 0 \\ \frac{1}{c} \end{bmatrix}\\ \end{alignat} }
So:
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} \mathbf{q}_{hkl} & = (2 \pi h)\mathbf{u} + (2 \pi k)\mathbf{v} + (2 \pi l)\mathbf{w} \\ & = (2 \pi h)\begin{bmatrix} \frac{1}{a} \\ \frac{1}{\sqrt{3}a} \\ 0 \end{bmatrix} + (2 \pi k)\begin{bmatrix} 0 \\ \frac{2}{\sqrt{3}b} \\ 0 \end{bmatrix} + (2 \pi l)\begin{bmatrix} 0 \\ 0 \\ \frac{1}{c} \end{bmatrix} \\ & = \begin{bmatrix} \frac{2 \pi h}{a} \\ \frac{2 \pi h}{\sqrt{3}a} + \frac{4 \pi k}{\sqrt{3}b} \\ \frac{2 \pi l}{c} \end{bmatrix} \\ & = \begin{bmatrix} \frac{2 \pi h}{a} \\ \frac{2 \pi (h + 2 k)}{\sqrt{3}a} \\ \frac{2 \pi l}{c} \end{bmatrix} \end{alignat} }