AFLOW Prototype: A17B2_hR19_166_cegh_c-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/3JAE
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https://aflow.org/p/A17B2_hR19_166_cegh_c-001
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PDF Version
Prototype | Be$_{17}$Nb$_{2}$ |
AFLOW prototype label | A17B2_hR19_166_cegh_c-001 |
ICSD | 58724 |
Pearson symbol | hR19 |
Space group number | 166 |
Space group symbol | $R\overline{3}m$ |
AFLOW prototype command |
aflow --proto=A17B2_hR19_166_cegh_c-001
--params=$a, \allowbreak c/a, \allowbreak x_{1}, \allowbreak x_{2}, \allowbreak x_{4}, \allowbreak x_{5}, \allowbreak z_{5}$ |
$\alpha$-Be$_{17}$Hf$_{2}$, Be$_{17}$Ta$_{2}$, $\alpha$-Be$_{17}$Ti$_{2}$, Be$_{17}$Zr$_{2}$
--hex
. Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $x_{1} \, \mathbf{a}_{1}+x_{1} \, \mathbf{a}_{2}+x_{1} \, \mathbf{a}_{3}$ | = | $c x_{1} \,\mathbf{\hat{z}}$ | (2c) | Be I |
$\mathbf{B_{2}}$ | = | $- x_{1} \, \mathbf{a}_{1}- x_{1} \, \mathbf{a}_{2}- x_{1} \, \mathbf{a}_{3}$ | = | $- c x_{1} \,\mathbf{\hat{z}}$ | (2c) | Be I |
$\mathbf{B_{3}}$ | = | $x_{2} \, \mathbf{a}_{1}+x_{2} \, \mathbf{a}_{2}+x_{2} \, \mathbf{a}_{3}$ | = | $c x_{2} \,\mathbf{\hat{z}}$ | (2c) | Nb I |
$\mathbf{B_{4}}$ | = | $- x_{2} \, \mathbf{a}_{1}- x_{2} \, \mathbf{a}_{2}- x_{2} \, \mathbf{a}_{3}$ | = | $- c x_{2} \,\mathbf{\hat{z}}$ | (2c) | Nb I |
$\mathbf{B_{5}}$ | = | $\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- \frac{1}{4}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{12}a \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}$ | (3e) | Be II |
$\mathbf{B_{6}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}$ | (3e) | Be II |
$\mathbf{B_{7}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{12}a \,\mathbf{\hat{y}}+\frac{1}{3}c \,\mathbf{\hat{z}}$ | (3e) | Be II |
$\mathbf{B_{8}}$ | = | $x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \left(2 x_{4} - 1\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{12}a \left(6 x_{4} + 1\right) \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ | (6g) | Be III |
$\mathbf{B_{9}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}- x_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \left(2 x_{4} + 1\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{12}a \left(6 x_{4} - 1\right) \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ | (6g) | Be III |
$\mathbf{B_{10}}$ | = | $- x_{4} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+x_{4} \, \mathbf{a}_{3}$ | = | $- a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ | (6g) | Be III |
$\mathbf{B_{11}}$ | = | $- x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- \frac{1}{4}a \left(2 x_{4} + 1\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{12}a \left(6 x_{4} - 1\right) \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ | (6g) | Be III |
$\mathbf{B_{12}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+x_{4} \, \mathbf{a}_{3}$ | = | $- \frac{1}{4}a \left(2 x_{4} - 1\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{12}a \left(6 x_{4} + 1\right) \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ | (6g) | Be III |
$\mathbf{B_{13}}$ | = | $x_{4} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}- x_{4} \, \mathbf{a}_{3}$ | = | $a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{6}c \,\mathbf{\hat{z}}$ | (6g) | Be III |
$\mathbf{B_{14}}$ | = | $x_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ | (6h) | Be IV |
$\mathbf{B_{15}}$ | = | $z_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+x_{5} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ | (6h) | Be IV |
$\mathbf{B_{16}}$ | = | $x_{5} \, \mathbf{a}_{1}+z_{5} \, \mathbf{a}_{2}+x_{5} \, \mathbf{a}_{3}$ | = | $- \frac{1}{\sqrt{3}}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ | (6h) | Be IV |
$\mathbf{B_{17}}$ | = | $- z_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}- x_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ | (6h) | Be IV |
$\mathbf{B_{18}}$ | = | $- x_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ | (6h) | Be IV |
$\mathbf{B_{19}}$ | = | $- x_{5} \, \mathbf{a}_{1}- z_{5} \, \mathbf{a}_{2}- x_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{\sqrt{3}}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ | (6h) | Be IV |