AFLOW Prototype: AB_hP16_186_2a2b_2a2b-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/W6UW
or
https://aflow.org/p/AB_hP16_186_2a2b_2a2b-001
or
PDF Version
Prototype | GaSe |
AFLOW prototype label | AB_hP16_186_2a2b_2a2b-001 |
ICSD | 2002 |
Pearson symbol | hP16 |
Space group number | 186 |
Space group symbol | $P6_3mc$ |
AFLOW prototype command |
aflow --proto=AB_hP16_186_2a2b_2a2b-001
--params=$a, \allowbreak c/a, \allowbreak z_{1}, \allowbreak z_{2}, \allowbreak z_{3}, \allowbreak z_{4}, \allowbreak z_{5}, \allowbreak z_{6}, \allowbreak z_{7}, \allowbreak z_{8}$ |
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) | Ga 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) | Ga I |
$\mathbf{B_{3}}$ | = | $z_{2} \, \mathbf{a}_{3}$ | = | $c z_{2} \,\mathbf{\hat{z}}$ | (2a) | Ga II |
$\mathbf{B_{4}}$ | = | $\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2a) | Ga II |
$\mathbf{B_{5}}$ | = | $z_{3} \, \mathbf{a}_{3}$ | = | $c z_{3} \,\mathbf{\hat{z}}$ | (2a) | Se I |
$\mathbf{B_{6}}$ | = | $\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2a) | Se I |
$\mathbf{B_{7}}$ | = | $z_{4} \, \mathbf{a}_{3}$ | = | $c z_{4} \,\mathbf{\hat{z}}$ | (2a) | Se II |
$\mathbf{B_{8}}$ | = | $\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2a) | Se II |
$\mathbf{B_{9}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (2b) | Ga III |
$\mathbf{B_{10}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2b) | Ga III |
$\mathbf{B_{11}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ | (2b) | Ga IV |
$\mathbf{B_{12}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2b) | Ga IV |
$\mathbf{B_{13}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{7} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{7} \,\mathbf{\hat{z}}$ | (2b) | Se III |
$\mathbf{B_{14}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{7} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{7} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2b) | Se III |
$\mathbf{B_{15}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{8} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{8} \,\mathbf{\hat{z}}$ | (2b) | Se IV |
$\mathbf{B_{16}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{8} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{8} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (2b) | Se IV |