AFLOW Prototype: A3B_hP16_194_gh_ac-001
This structure originally had the label A3B_hP16_194_gh_ac. 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/7D2A
or
https://aflow.org/p/A3B_hP16_194_gh_ac-001
or
PDF Version
Prototype | Ni$_{3}$Ti |
AFLOW prototype label | A3B_hP16_194_gh_ac-001 |
Strukturbericht designation | $D0_{24}$ |
ICSD | 30216 |
Pearson symbol | hP16 |
Space group number | 194 |
Space group symbol | $P6_3/mmc$ |
AFLOW prototype command |
aflow --proto=A3B_hP16_194_gh_ac-001
--params=$a, \allowbreak c/a, \allowbreak x_{4}$ |
Al$_{3}$Dy, Al$_{3}$Ru, Co$_{3}$Ti, Ni$_{3}$Ti, Pd$_{3}$Hf, Pd$_{3}$Np, Pd$_{3}$Th, Pd$_{3}$Ti, Pd$_{3}$Zr, Pt$_{3}$Hf, Pt$_{3}$Zr
Basis vectors
Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
---|---|---|---|---|---|---|
$\mathbf{B_{1}}$ | = | $0$ | = | $0$ | (2a) | Ti I |
$\mathbf{B_{2}}$ | = | $\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}c \,\mathbf{\hat{z}}$ | (2a) | Ti I |
$\mathbf{B_{3}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (2c) | Ti II |
$\mathbf{B_{4}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (2c) | Ti II |
$\mathbf{B_{5}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}$ | (6g) | Ni I |
$\mathbf{B_{6}}$ | = | $\frac{1}{2} \, \mathbf{a}_{2}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}$ | (6g) | Ni I |
$\mathbf{B_{7}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}$ | (6g) | Ni I |
$\mathbf{B_{8}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (6g) | Ni I |
$\mathbf{B_{9}}$ | = | $\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{4}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (6g) | Ni I |
$\mathbf{B_{10}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (6g) | Ni I |
$\mathbf{B_{11}}$ | = | $x_{4} \, \mathbf{a}_{1}+2 x_{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{3}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (6h) | Ni II |
$\mathbf{B_{12}}$ | = | $- 2 x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $- \frac{3}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (6h) | Ni II |
$\mathbf{B_{13}}$ | = | $x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $- \sqrt{3}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ | (6h) | Ni II |
$\mathbf{B_{14}}$ | = | $- x_{4} \, \mathbf{a}_{1}- 2 x_{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $- \frac{3}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (6h) | Ni II |
$\mathbf{B_{15}}$ | = | $2 x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{3}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (6h) | Ni II |
$\mathbf{B_{16}}$ | = | $- x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\sqrt{3}a x_{4} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ | (6h) | Ni II |