AFLOW Prototype: A13B5_hP36_194_a2e4f_b2f-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/G1KV
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https://aflow.org/p/A13B5_hP36_194_a2e4f_b2f-001
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PDF Version
Prototype | Na$_{13}$Pb$_{5}$ |
AFLOW prototype label | A13B5_hP36_194_a2e4f_b2f-001 |
ICSD | none |
Pearson symbol | hP36 |
Space group number | 194 |
Space group symbol | $P6_3/mmc$ |
AFLOW prototype command |
aflow --proto=A13B5_hP36_194_a2e4f_b2f-001
--params=$a, \allowbreak c/a, \allowbreak z_{3}, \allowbreak z_{4}, \allowbreak z_{5}, \allowbreak z_{6}, \allowbreak z_{7}, \allowbreak z_{8}, \allowbreak z_{9}, \allowbreak z_{10}$ |
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $0$ | = | $0$ | (2a) | Na I |
$\mathbf{B_{2}}$ | = | $\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}c \,\mathbf{\hat{z}}$ | (2a) | Na I |
$\mathbf{B_{3}}$ | = | $\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}c \,\mathbf{\hat{z}}$ | (2b) | Pb I |
$\mathbf{B_{4}}$ | = | $\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{3}{4}c \,\mathbf{\hat{z}}$ | (2b) | Pb I |
$\mathbf{B_{5}}$ | = | $z_{3} \, \mathbf{a}_{3}$ | = | $c z_{3} \,\mathbf{\hat{z}}$ | (4e) | Na II |
$\mathbf{B_{6}}$ | = | $\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4e) | Na II |
$\mathbf{B_{7}}$ | = | $- z_{3} \, \mathbf{a}_{3}$ | = | $- c z_{3} \,\mathbf{\hat{z}}$ | (4e) | Na II |
$\mathbf{B_{8}}$ | = | $- \left(z_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- c \left(z_{3} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4e) | Na II |
$\mathbf{B_{9}}$ | = | $z_{4} \, \mathbf{a}_{3}$ | = | $c z_{4} \,\mathbf{\hat{z}}$ | (4e) | Na III |
$\mathbf{B_{10}}$ | = | $\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4e) | Na III |
$\mathbf{B_{11}}$ | = | $- z_{4} \, \mathbf{a}_{3}$ | = | $- c z_{4} \,\mathbf{\hat{z}}$ | (4e) | Na III |
$\mathbf{B_{12}}$ | = | $- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4e) | Na III |
$\mathbf{B_{13}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (4f) | Na IV |
$\mathbf{B_{14}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | Na IV |
$\mathbf{B_{15}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ | (4f) | Na IV |
$\mathbf{B_{16}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | Na IV |
$\mathbf{B_{17}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ | (4f) | Na V |
$\mathbf{B_{18}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | Na V |
$\mathbf{B_{19}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ | (4f) | Na V |
$\mathbf{B_{20}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}- \left(z_{6} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c \left(z_{6} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | Na V |
$\mathbf{B_{21}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{7} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{7} \,\mathbf{\hat{z}}$ | (4f) | Na VI |
$\mathbf{B_{22}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{7} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{7} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | Na VI |
$\mathbf{B_{23}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{7} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{7} \,\mathbf{\hat{z}}$ | (4f) | Na VI |
$\mathbf{B_{24}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}- \left(z_{7} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c \left(z_{7} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | Na VI |
$\mathbf{B_{25}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{8} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{8} \,\mathbf{\hat{z}}$ | (4f) | Na VII |
$\mathbf{B_{26}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{8} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{8} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | Na VII |
$\mathbf{B_{27}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{8} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{8} \,\mathbf{\hat{z}}$ | (4f) | Na VII |
$\mathbf{B_{28}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}- \left(z_{8} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c \left(z_{8} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | Na VII |
$\mathbf{B_{29}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{9} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{9} \,\mathbf{\hat{z}}$ | (4f) | Pb II |
$\mathbf{B_{30}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{9} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{9} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | Pb II |
$\mathbf{B_{31}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{9} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{9} \,\mathbf{\hat{z}}$ | (4f) | Pb II |
$\mathbf{B_{32}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}- \left(z_{9} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c \left(z_{9} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | Pb II |
$\mathbf{B_{33}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{10} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{10} \,\mathbf{\hat{z}}$ | (4f) | Pb III |
$\mathbf{B_{34}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{10} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{10} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | Pb III |
$\mathbf{B_{35}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{10} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{10} \,\mathbf{\hat{z}}$ | (4f) | Pb III |
$\mathbf{B_{36}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}- \left(z_{10} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c \left(z_{10} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (4f) | Pb III |