AFLOW Prototype: AB3_hP24_185_c_ab2c-001
This structure originally had the label AB3_hP24_185_c_ab2c. Calls to that address will be redirected here.
If you are using this page, please cite:
D. Hicks, M. J. Mehl, E. Gossett, C. Toher, O. Levy, R. M. Hanson, G. L. W. Hart, and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 2, Comp. Mat. Sci. 161, S1-S1011 (2019). (doi=10.1016/j.commatsci.2018.10.043)
Links to this page
https://aflow.org/p/3ERL
or
https://aflow.org/p/AB3_hP24_185_c_ab2c-001
or
PDF Version
Prototype | AsNa$_{3}$ |
AFLOW prototype label | AB3_hP24_185_c_ab2c-001 |
ICSD | 79586 |
Pearson symbol | hP24 |
Space group number | 185 |
Space group symbol | $P6_3cm$ |
AFLOW prototype command |
aflow --proto=AB3_hP24_185_c_ab2c-001
--params=$a, \allowbreak c/a, \allowbreak z_{1}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak z_{5}$ |
Cu$_{3}$P
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $z_{1} \, \mathbf{a}_{3}$ | = | $c z_{1} \,\mathbf{\hat{z}}$ | (2a) | Na I |
$\mathbf{B_{2}}$ | = | $\left(z_{1} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $c \left(z_{1} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2a) | Na I |
$\mathbf{B_{3}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ | (4b) | Na II |
$\mathbf{B_{4}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4b) | Na II |
$\mathbf{B_{5}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4b) | Na II |
$\mathbf{B_{6}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ | (4b) | Na II |
$\mathbf{B_{7}}$ | = | $x_{3} \, \mathbf{a}_{1}+z_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a x_{3} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ | (6c) | As I |
$\mathbf{B_{8}}$ | = | $x_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a x_{3} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ | (6c) | As I |
$\mathbf{B_{9}}$ | = | $- x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ | = | $- a x_{3} \,\mathbf{\hat{x}}+c z_{3} \,\mathbf{\hat{z}}$ | (6c) | As I |
$\mathbf{B_{10}}$ | = | $- x_{3} \, \mathbf{a}_{1}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a x_{3} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (6c) | As I |
$\mathbf{B_{11}}$ | = | $- x_{3} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a x_{3} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (6c) | As I |
$\mathbf{B_{12}}$ | = | $x_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a x_{3} \,\mathbf{\hat{x}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (6c) | As I |
$\mathbf{B_{13}}$ | = | $x_{4} \, \mathbf{a}_{1}+z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (6c) | Na III |
$\mathbf{B_{14}}$ | = | $x_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (6c) | Na III |
$\mathbf{B_{15}}$ | = | $- x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $- a x_{4} \,\mathbf{\hat{x}}+c z_{4} \,\mathbf{\hat{z}}$ | (6c) | Na III |
$\mathbf{B_{16}}$ | = | $- x_{4} \, \mathbf{a}_{1}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (6c) | Na III |
$\mathbf{B_{17}}$ | = | $- x_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (6c) | Na III |
$\mathbf{B_{18}}$ | = | $x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a x_{4} \,\mathbf{\hat{x}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (6c) | Na III |
$\mathbf{B_{19}}$ | = | $x_{5} \, \mathbf{a}_{1}+z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a x_{5} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (6c) | Na IV |
$\mathbf{B_{20}}$ | = | $x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a x_{5} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (6c) | Na IV |
$\mathbf{B_{21}}$ | = | $- x_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $- a x_{5} \,\mathbf{\hat{x}}+c z_{5} \,\mathbf{\hat{z}}$ | (6c) | Na IV |
$\mathbf{B_{22}}$ | = | $- x_{5} \, \mathbf{a}_{1}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a x_{5} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (6c) | Na IV |
$\mathbf{B_{23}}$ | = | $- x_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a x_{5} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (6c) | Na IV |
$\mathbf{B_{24}}$ | = | $x_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a x_{5} \,\mathbf{\hat{x}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (6c) | Na IV |