AFLOW Prototype: A3B2_tP10_129_3c_2c-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/T7XE
or
https://aflow.org/p/A3B2_tP10_129_3c_2c-001
or
PDF Version
Prototype | Cu$_{3}$Ti$_{2}$ |
AFLOW prototype label | A3B2_tP10_129_3c_2c-001 |
ICSD | 103133 |
Pearson symbol | tP10 |
Space group number | 129 |
Space group symbol | $P4/nmm$ |
AFLOW prototype command |
aflow --proto=A3B2_tP10_129_3c_2c-001
--params=$a, \allowbreak c/a, \allowbreak z_{1}, \allowbreak z_{2}, \allowbreak z_{3}, \allowbreak z_{4}, \allowbreak z_{5}$ |
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $\frac{1}{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{1} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+c z_{1} \,\mathbf{\hat{z}}$ | (2c) | Cu I |
$\mathbf{B_{2}}$ | = | $\frac{3}{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{1} \, \mathbf{a}_{3}$ | = | $\frac{3}{4}a \,\mathbf{\hat{x}}+\frac{3}{4}a \,\mathbf{\hat{y}}- c z_{1} \,\mathbf{\hat{z}}$ | (2c) | Cu I |
$\mathbf{B_{3}}$ | = | $\frac{1}{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ | (2c) | Cu II |
$\mathbf{B_{4}}$ | = | $\frac{3}{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{2} \, \mathbf{a}_{3}$ | = | $\frac{3}{4}a \,\mathbf{\hat{x}}+\frac{3}{4}a \,\mathbf{\hat{y}}- c z_{2} \,\mathbf{\hat{z}}$ | (2c) | Cu II |
$\mathbf{B_{5}}$ | = | $\frac{1}{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ | (2c) | Cu III |
$\mathbf{B_{6}}$ | = | $\frac{3}{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ | = | $\frac{3}{4}a \,\mathbf{\hat{x}}+\frac{3}{4}a \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ | (2c) | Cu III |
$\mathbf{B_{7}}$ | = | $\frac{1}{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (2c) | Ti I |
$\mathbf{B_{8}}$ | = | $\frac{3}{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ | = | $\frac{3}{4}a \,\mathbf{\hat{x}}+\frac{3}{4}a \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (2c) | Ti I |
$\mathbf{B_{9}}$ | = | $\frac{1}{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (2c) | Ti II |
$\mathbf{B_{10}}$ | = | $\frac{3}{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $\frac{3}{4}a \,\mathbf{\hat{x}}+\frac{3}{4}a \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ | (2c) | Ti II |