AFLOW Prototype: A5B_hR12_167_ce_b-001
If you are using this page, please cite:
H. Eckert, S. Divilov, M. J. Mehl, D. Hicks, A. C. Zettel, M. Esters. X. Campilongo and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 4. Submitted to Computational Materials Science.
Links to this page
https://aflow.org/p/584Z
or
https://aflow.org/p/A5B_hR12_167_ce_b-001
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PDF Version
Prototype | Al$_{5}$Mo |
AFLOW prototype label | A5B_hR12_167_ce_b-001 |
ICSD | 105520 |
Pearson symbol | hR12 |
Space group number | 167 |
Space group symbol | $R\overline{3}c$ |
AFLOW prototype command |
aflow --proto=A5B_hR12_167_ce_b-001
--params=$a, \allowbreak c/a, \allowbreak x_{2}, \allowbreak x_{3}$ |
--hex
. Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $0$ | = | $0$ | (2b) | Mo I |
$\mathbf{B_{2}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}c \,\mathbf{\hat{z}}$ | (2b) | Mo I |
$\mathbf{B_{3}}$ | = | $x_{2} \, \mathbf{a}_{1}+x_{2} \, \mathbf{a}_{2}+x_{2} \, \mathbf{a}_{3}$ | = | $c x_{2} \,\mathbf{\hat{z}}$ | (4c) | Al I |
$\mathbf{B_{4}}$ | = | $- \left(x_{2} - \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(x_{2} - \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(x_{2} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- c \left(x_{2} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Al I |
$\mathbf{B_{5}}$ | = | $- x_{2} \, \mathbf{a}_{1}- x_{2} \, \mathbf{a}_{2}- x_{2} \, \mathbf{a}_{3}$ | = | $- c x_{2} \,\mathbf{\hat{z}}$ | (4c) | Al I |
$\mathbf{B_{6}}$ | = | $\left(x_{2} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{2} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(x_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $c \left(x_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4c) | Al I |
$\mathbf{B_{7}}$ | = | $x_{3} \, \mathbf{a}_{1}- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{8}a \left(4 x_{3} - 1\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{8}a \left(4 x_{3} - 1\right) \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (6e) | Al II |
$\mathbf{B_{8}}$ | = | $\frac{1}{4} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{8}a \left(4 x_{3} - 1\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{8}a \left(4 x_{3} - 1\right) \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (6e) | Al II |
$\mathbf{B_{9}}$ | = | $- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+x_{3} \, \mathbf{a}_{3}$ | = | $- a \left(x_{3} - \frac{1}{4}\right) \,\mathbf{\hat{x}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (6e) | Al II |
$\mathbf{B_{10}}$ | = | $- x_{3} \, \mathbf{a}_{1}+\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $- \frac{1}{8}a \left(4 x_{3} + 3\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{24}a \left(12 x_{3} + 1\right) \,\mathbf{\hat{y}}+\frac{5}{12}c \,\mathbf{\hat{z}}$ | (6e) | Al II |
$\mathbf{B_{11}}$ | = | $\frac{3}{4} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- \frac{1}{8}a \left(4 x_{3} - 1\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{24}a \left(12 x_{3} + 5\right) \,\mathbf{\hat{y}}+\frac{5}{12}c \,\mathbf{\hat{z}}$ | (6e) | Al II |
$\mathbf{B_{12}}$ | = | $\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- x_{3} \, \mathbf{a}_{3}$ | = | $a \left(x_{3} + \frac{1}{4}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{5}{12}c \,\mathbf{\hat{z}}$ | (6e) | Al II |