Building up a face centered cubic lattice

2012-06-24T10:50:33Z

188.123.231.96:

{{Jmol_general|Face_centered_cubic_lattice.xyz|A face centered cubic lattice}}
* Consider:
# a cubic simulation box whose sides are of length <math>\left. L \right. </math>
# a number of lattice positions, <math> \left. M \right. </math> given by <math> \left. M = 4 m^3 \right. </math>,
with <math> m </math> being a positive integer

* The <math> \left. M \right. </math> positions are those given by:

:<math>
\left\{ \begin{array}{l}
x_a = i_a \times (\delta l) \\
y_a = j_a \times (\delta l) \\
z_a = k_a \times (\delta l)
\end{array}
\right\}
</math>

where the indices of a given valid site are integer numbers that must fulfill the following criteria

* <math> 0 \le i_a < 2m </math>
* <math> 0 \le j_a < 2m </math>
* <math> 0 \le k_a < 2m </math>,
* the sum of <math> \left. i_a + j_a + k_a \right. </math> must be, for instance, an even number.

with
<math>
\left.
\delta l = L/(2m)
\right.
</math>

== Atomic position(s) on a cubic cell ==

* Number of atoms per cell: 4
* Coordinates:
Atom 1: <math> \left( x_1, y_1, z_1 \right) = \left( 0, 0, 0 \right) </math>

Atom 2: <math> \left( x_2, y_2, z_2 \right) = \left( 0 , \frac{l}{2}, \frac{l}{2}\right) </math>

Atom 3: <math> \left( x_3, y_3, z_2 \right) = \left( \frac{l}{2}, 0, \frac{l}{2} \right) </math>

Atom 4: <math> \left( x_4, y_4, z_2 \right) = \left( \frac{l}{2}, \frac{l}{2}, 0 \right) </math>

Cell dimensions:
*<math> a=b=c = l </math>

*<math> \alpha = \beta = \gamma = 90^0 </math>
[[category: computer simulation techniques]]
[[category: Contains Jmol]]

x=j+k,y=k+i,z=i+j

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Building up a face centered cubic lattice