AFLOW Prototype: A5B11_mP16_6_2abc_2a3b3c-001
This structure originally had the label A5B11_mP16_6_2abc_2a3b3c. Calls to that address will be redirected here.
If you are using this page, please cite:
D. Hicks, M.J. Mehl, M. Esters, C. Oses, O. Levy, G.L.W. Hart, C. Toher, and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 3, Comp. Mat. Sci. 199, 110450 (2021). (doi=10.1016/j.commatsci.2021.110450)
Links to this page
https://aflow.org/p/EUM5
or
https://aflow.org/p/A5B11_mP16_6_2abc_2a3b3c-001
or
PDF Version
Prototype | Ta$_{5}$Ti$_{11}$ |
AFLOW prototype label | A5B11_mP16_6_2abc_2a3b3c-001 |
ICSD | none |
Pearson symbol | mP16 |
Space group number | 6 |
Space group symbol | $Pm$ |
AFLOW prototype command |
aflow --proto=A5B11_mP16_6_2abc_2a3b3c-001
--params=$a, \allowbreak b/a, \allowbreak c/a, \allowbreak \beta, \allowbreak x_{1}, \allowbreak z_{1}, \allowbreak x_{2}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak z_{5}, \allowbreak x_{6}, \allowbreak z_{6}, \allowbreak x_{7}, \allowbreak z_{7}, \allowbreak x_{8}, \allowbreak z_{8}, \allowbreak x_{9}, \allowbreak y_{9}, \allowbreak z_{9}, \allowbreak x_{10}, \allowbreak y_{10}, \allowbreak z_{10}, \allowbreak x_{11}, \allowbreak y_{11}, \allowbreak z_{11}, \allowbreak x_{12}, \allowbreak y_{12}, \allowbreak z_{12}$ |
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $x_{1} \, \mathbf{a}_{1}+z_{1} \, \mathbf{a}_{3}$ | = | $\left(a x_{1} + c z_{1} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{1} \sin{\beta} \,\mathbf{\hat{z}}$ | (1a) | Ta I |
$\mathbf{B_{2}}$ | = | $x_{2} \, \mathbf{a}_{1}+z_{2} \, \mathbf{a}_{3}$ | = | $\left(a x_{2} + c z_{2} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{2} \sin{\beta} \,\mathbf{\hat{z}}$ | (1a) | Ta II |
$\mathbf{B_{3}}$ | = | $x_{3} \, \mathbf{a}_{1}+z_{3} \, \mathbf{a}_{3}$ | = | $\left(a x_{3} + c z_{3} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{3} \sin{\beta} \,\mathbf{\hat{z}}$ | (1a) | Ti I |
$\mathbf{B_{4}}$ | = | $x_{4} \, \mathbf{a}_{1}+z_{4} \, \mathbf{a}_{3}$ | = | $\left(a x_{4} + c z_{4} \cos{\beta}\right) \,\mathbf{\hat{x}}+c z_{4} \sin{\beta} \,\mathbf{\hat{z}}$ | (1a) | Ti II |
$\mathbf{B_{5}}$ | = | $x_{5} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $\left(a x_{5} + c z_{5} \cos{\beta}\right) \,\mathbf{\hat{x}}+\frac{1}{2}b \,\mathbf{\hat{y}}+c z_{5} \sin{\beta} \,\mathbf{\hat{z}}$ | (1b) | Ta III |
$\mathbf{B_{6}}$ | = | $x_{6} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ | = | $\left(a x_{6} + c z_{6} \cos{\beta}\right) \,\mathbf{\hat{x}}+\frac{1}{2}b \,\mathbf{\hat{y}}+c z_{6} \sin{\beta} \,\mathbf{\hat{z}}$ | (1b) | Ti III |
$\mathbf{B_{7}}$ | = | $x_{7} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+z_{7} \, \mathbf{a}_{3}$ | = | $\left(a x_{7} + c z_{7} \cos{\beta}\right) \,\mathbf{\hat{x}}+\frac{1}{2}b \,\mathbf{\hat{y}}+c z_{7} \sin{\beta} \,\mathbf{\hat{z}}$ | (1b) | Ti IV |
$\mathbf{B_{8}}$ | = | $x_{8} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+z_{8} \, \mathbf{a}_{3}$ | = | $\left(a x_{8} + c z_{8} \cos{\beta}\right) \,\mathbf{\hat{x}}+\frac{1}{2}b \,\mathbf{\hat{y}}+c z_{8} \sin{\beta} \,\mathbf{\hat{z}}$ | (1b) | Ti V |
$\mathbf{B_{9}}$ | = | $x_{9} \, \mathbf{a}_{1}+y_{9} \, \mathbf{a}_{2}+z_{9} \, \mathbf{a}_{3}$ | = | $\left(a x_{9} + c z_{9} \cos{\beta}\right) \,\mathbf{\hat{x}}+b y_{9} \,\mathbf{\hat{y}}+c z_{9} \sin{\beta} \,\mathbf{\hat{z}}$ | (2c) | Ta IV |
$\mathbf{B_{10}}$ | = | $x_{9} \, \mathbf{a}_{1}- y_{9} \, \mathbf{a}_{2}+z_{9} \, \mathbf{a}_{3}$ | = | $\left(a x_{9} + c z_{9} \cos{\beta}\right) \,\mathbf{\hat{x}}- b y_{9} \,\mathbf{\hat{y}}+c z_{9} \sin{\beta} \,\mathbf{\hat{z}}$ | (2c) | Ta IV |
$\mathbf{B_{11}}$ | = | $x_{10} \, \mathbf{a}_{1}+y_{10} \, \mathbf{a}_{2}+z_{10} \, \mathbf{a}_{3}$ | = | $\left(a x_{10} + c z_{10} \cos{\beta}\right) \,\mathbf{\hat{x}}+b y_{10} \,\mathbf{\hat{y}}+c z_{10} \sin{\beta} \,\mathbf{\hat{z}}$ | (2c) | Ti VI |
$\mathbf{B_{12}}$ | = | $x_{10} \, \mathbf{a}_{1}- y_{10} \, \mathbf{a}_{2}+z_{10} \, \mathbf{a}_{3}$ | = | $\left(a x_{10} + c z_{10} \cos{\beta}\right) \,\mathbf{\hat{x}}- b y_{10} \,\mathbf{\hat{y}}+c z_{10} \sin{\beta} \,\mathbf{\hat{z}}$ | (2c) | Ti VI |
$\mathbf{B_{13}}$ | = | $x_{11} \, \mathbf{a}_{1}+y_{11} \, \mathbf{a}_{2}+z_{11} \, \mathbf{a}_{3}$ | = | $\left(a x_{11} + c z_{11} \cos{\beta}\right) \,\mathbf{\hat{x}}+b y_{11} \,\mathbf{\hat{y}}+c z_{11} \sin{\beta} \,\mathbf{\hat{z}}$ | (2c) | Ti VII |
$\mathbf{B_{14}}$ | = | $x_{11} \, \mathbf{a}_{1}- y_{11} \, \mathbf{a}_{2}+z_{11} \, \mathbf{a}_{3}$ | = | $\left(a x_{11} + c z_{11} \cos{\beta}\right) \,\mathbf{\hat{x}}- b y_{11} \,\mathbf{\hat{y}}+c z_{11} \sin{\beta} \,\mathbf{\hat{z}}$ | (2c) | Ti VII |
$\mathbf{B_{15}}$ | = | $x_{12} \, \mathbf{a}_{1}+y_{12} \, \mathbf{a}_{2}+z_{12} \, \mathbf{a}_{3}$ | = | $\left(a x_{12} + c z_{12} \cos{\beta}\right) \,\mathbf{\hat{x}}+b y_{12} \,\mathbf{\hat{y}}+c z_{12} \sin{\beta} \,\mathbf{\hat{z}}$ | (2c) | Ti VIII |
$\mathbf{B_{16}}$ | = | $x_{12} \, \mathbf{a}_{1}- y_{12} \, \mathbf{a}_{2}+z_{12} \, \mathbf{a}_{3}$ | = | $\left(a x_{12} + c z_{12} \cos{\beta}\right) \,\mathbf{\hat{x}}- b y_{12} \,\mathbf{\hat{y}}+c z_{12} \sin{\beta} \,\mathbf{\hat{z}}$ | (2c) | Ti VIII |