Numerical recognition of C15 unit in rapid solidification Ni<sub>70</sub>Ag<sub>30 </sub>nanoparticles

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Abstract

Simulation has become an important tool in materials science, it is a prerequisite to study the correlation between the structure and properties of materials, in that the structural characteristics of the system from the atomic coordinates output can be obtained by simulations. For simple (FCC, HCP, and BCC) crystals containing only 2-6 atoms, in the numerical analysis method, what needs to be determined is only the local characteristics of each atom. However, it is extremely computationally intensive to determine the cells containing tens or hundreds of atoms. The combination of numerical analysis and visualization is one of the methods to solve this kind of problem. In this work, Ni ₇₀Ag ₃₀nanoparticles are simulated by molecular dynamics. It is found that the nanoparticles contain FCC crystals and a large number of complex topologically close-packed (TCP) structures. Using the analysis software based on the largest standard cluster analysis (LaSCA), the C15 phase of TCP atoms in nanoparticles is determined by topology configuration analysis and crystallography knowledge. The analytical ideas provide the algorithm logic fordeveloping the numerical recognition software for complex crystal structures in the future.

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Corresponding author:Tian Ze-An,tianzean@hnu.edu.cn

Funds:Project supported by the National Natural Science Foundation of China (Grant Nos. 51661005, U1612442).

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Type	Multiplicity	Wyckoff letter	Site symmetry	Coordinates
Z16	8	b	$\bar{4}3m$	$\left(\dfrac{1}{2, }\dfrac{1}{2, }\dfrac{1}{2}\right)$ $\left(\dfrac{1}{4, }\dfrac{3}{4, }\dfrac{1}{4}\right)$
Z16	8	a	$\bar{4} 3m$	(0, 0, 0) $\left(\dfrac{3}{4, }\dfrac{1}{4, }\dfrac{3}{4}\right)$
Z12	16	d	$.\bar{3}m$	$\left(\dfrac{5}{8, }\dfrac{5}{8, }\dfrac{5}{8}\right)\left(\dfrac{3}{8, }\dfrac{7}{8, }\dfrac{1}{8}\right)$ $\left(\dfrac{7}{8, }\dfrac{1}{8, }\dfrac{3}{8}\right)\left(\dfrac{1}{8, }\dfrac{3}{8, }\dfrac{7}{8}\right)$
Z12	16	c	$.\bar{3}m$	$\left(\dfrac{1}{8, }\dfrac{1}{8, }\dfrac{1}{8}\right)\left(\dfrac{7}{8, }\dfrac{3}{8, }\dfrac{5}{8}\right)$ $\left(\dfrac{3}{8, }\dfrac{5}{8, }\dfrac{7}{8}\right)\left(\dfrac{5}{8, }\dfrac{7}{8, }\dfrac{3}{8}\right)$

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Vol. 71, Issue 17 - 2022-09-05

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