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onexf766

木虫 (正式写手)

[交流] 搭晶胞问题

看到一篇文章用渗杂的方法解决固溶体合金问题计算
这样描述:

The binary alloys were described by supercells consisting of 2 * 2 * 2 elementary cubic or hexagonal unit cells with a total of 16 atoms. A large variety of alloy compositions were studied by systematically replacing Ti atoms by either Nb atoms. The lowest alloy composition was 6.25 at.% (one Nb atom in a 16 atom supercell). For each alloy composition various local arrangements were considered, and in total 48 body-centered cubic (bcc) and 28 hexagonal closed-packed (hcp) configurations were studied.

不太明其中内容:为什么从加入第二种元素(Nb)时,从1分别到加入个16(原本为Ti,共16个位)。
他说对bcc总共有48种排列,而对hcp总共有28种,那么我该分别在原来的结构(bcc或hcp)如何换入第二原子?(加1个时是多少种排列?加2个时又是多少?3个....16个分别是多少?怎么排列?)
下面是我问该文作者这个问题时收到的回复:
the number comes from the different sublattice arrangements to check for short range order effects

看了还是不太明白,望大牛指点。

[ Last edited by onexf766 on 2008-1-8 at 10:51 ]
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onexf766

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帮忙呀

高手帮忙啊~
2楼2008-01-08 10:09:02
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onexf766

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答一个哈):

人呢?
3楼2008-01-10 10:24:49
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onexf766

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??

4楼2008-02-24 11:26:20
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元小雪

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对bcc总共有48种排列,而对hcp总共有28种,指的是什么意思???
无聊的博士
5楼2008-02-24 13:22:11
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