AFLOW Prototype: A3B_hR12_166_ach_bc-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/3QUZ
or
https://aflow.org/p/A3B_hR12_166_ach_bc-001
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PDF Version
Prototype | Be$_{3}$Nb |
AFLOW prototype label | A3B_hR12_166_ach_bc-001 |
ICSD | 58723 |
Pearson symbol | hR12 |
Space group number | 166 |
Space group symbol | $R\overline{3}m$ |
AFLOW prototype command |
aflow --proto=A3B_hR12_166_ach_bc-001
--params=$a, \allowbreak c/a, \allowbreak x_{3}, \allowbreak x_{4}, \allowbreak x_{5}, \allowbreak z_{5}$ |
DyNi$_{3}$, ErNi$_{3}$, GdNi$_{3}$, HoNi$_{3}$, LaNi$_{3}$, PrNi$_{3}$, PuCo$_{3}$, PuNi$_{3}$, SmNi$_{3}$, TaBe$_{3}$, TbNi$_{3}$, ThFe$_{3}$, TiBe$_{3}$, TmNi$_{3}$, YNi$_{3}$, YbNi$_{3}$
--hex
. Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $0$ | = | $0$ | (1a) | Be I |
$\mathbf{B_{2}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}c \,\mathbf{\hat{z}}$ | (1b) | Nb I |
$\mathbf{B_{3}}$ | = | $x_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+x_{3} \, \mathbf{a}_{3}$ | = | $c x_{3} \,\mathbf{\hat{z}}$ | (2c) | Be II |
$\mathbf{B_{4}}$ | = | $- x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}- x_{3} \, \mathbf{a}_{3}$ | = | $- c x_{3} \,\mathbf{\hat{z}}$ | (2c) | Be II |
$\mathbf{B_{5}}$ | = | $x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+x_{4} \, \mathbf{a}_{3}$ | = | $c x_{4} \,\mathbf{\hat{z}}$ | (2c) | Nb II |
$\mathbf{B_{6}}$ | = | $- x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}- x_{4} \, \mathbf{a}_{3}$ | = | $- c x_{4} \,\mathbf{\hat{z}}$ | (2c) | Nb II |
$\mathbf{B_{7}}$ | = | $x_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ | (6h) | Be III |
$\mathbf{B_{8}}$ | = | $z_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+x_{5} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ | (6h) | Be III |
$\mathbf{B_{9}}$ | = | $x_{5} \, \mathbf{a}_{1}+z_{5} \, \mathbf{a}_{2}+x_{5} \, \mathbf{a}_{3}$ | = | $- \frac{1}{\sqrt{3}}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}+\frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ | (6h) | Be III |
$\mathbf{B_{10}}$ | = | $- z_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}- x_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ | (6h) | Be III |
$\mathbf{B_{11}}$ | = | $- x_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ | (6h) | Be III |
$\mathbf{B_{12}}$ | = | $- x_{5} \, \mathbf{a}_{1}- z_{5} \, \mathbf{a}_{2}- x_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{\sqrt{3}}a \left(x_{5} - z_{5}\right) \,\mathbf{\hat{y}}- \frac{1}{3}c \left(2 x_{5} + z_{5}\right) \,\mathbf{\hat{z}}$ | (6h) | Be III |