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面心立方晶面计算

創建於：2025年1月9日

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問題

试计算面心立方晶体的(100)，(110)，(111)等晶面的面间距和面致密度，并指出面间距最大的面。帮我复习计算公式

思考

Mapping out plane spacing

OK, let's think through the plane spacing formula for cubic crystals, including FCC: dhkl=ah2+k2+l2d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}. This applies to (100), (110), and (111) planes.

Calculating plane spacing

I’m working through plane spacing: d100=ad_{100} = a, d110=a/2d_{110} = a/\sqrt{2}, and d111=a/3d_{111} = a/\sqrt{3}. The largest spacing is for the (100) plane.

Determining plane density

I’m calculating plane density by looking at the number of atoms per area in the 2D unit cell for the (100), (110), and (111) planes in an FCC crystal.

Assessing plane components

I’m figuring out how to count atoms in the (100) plane: 4 corner atoms, each 1/4 to the plane, and 1 face-centered atom, totaling 1 unit per 2D unit cell.

Laying out the details

I'm mapping out the (110) plane density, considering the 2D repeating area and using the formula Plane Density=AtomsArea\text{Plane Density} = \frac{\text{Atoms}}{\text{Area}}.

Assessing the plane density

OK, let me see. For the (110) plane, the 2D unit cell area is a2/2a^2 / \sqrt{2}, leading to a plane density of 2/a2\sqrt{2} / a^2.

Calculating plane density

I'm working through the (110) plane's plane density, considering the rectangular net area a22a^2 \sqrt{2} and the atoms per unit cell, leading to a plane density of 2/a2\sqrt{2} / a^2.

Mapping the density

I'm evaluating the (111) plane density by repeatedly calculating the area and atom distribution for the (100) and (110) planes, leading to the conclusion that the unit cell parameters are a23a^2 \sqrt{3} and 3/22\sqrt{3}/2 \sqrt{2} respectively.

Determining (110) plane density

I'm analyzing the (110) plane density with a22a^2 \sqrt{2} as the 2D unit cell area, considering atoms at corners and edge centers. This approach helps clarify the structure.

Verifying planar density

I'm checking the planar density of the (110) plane in an FCC crystal. This involves calculating the 2D unit cell area and counting atoms within it to ensure accuracy.

Assessing plane density

OK, let's break this down: for an FCC crystal, plane density depends on atoms per 2D repeat area. The (100) plane density is 2a2\frac{2}{a^2}, with 2 atoms in its 2D repeat area.

Calculating plane density

Taking a closer look at the (110) plane density, I’m examining the 2D repeat area a22a^2 \sqrt{2} to determine the atom count within it.

Revisiting my approach

I’m re-evaluating the vectors and atom positions to accurately determine the area and atom count in the (110) plane, aiming for a precise planar packing fraction.

Identifying key atoms

OK, I’m mapping out the atom positions in the (110) plane: four corners and one interior atom at (1/2,−1/2,0)(1/2, -1/2, 0). This helps define the 2D unit cell.

Checking atom positions

I'm confirming the plane equation x+y=0x + y = 0 to ensure internal atoms are properly accounted for in the (110) plane.

Revisiting standard results

I’m working through standard planar density formulas for (100) and (110) planes, ensuring atom count is accurate per their 2D unit cell areas.

Calculating planar density

I'm thinking through the planar density for the (111) plane in an FCC crystal. The hexagonal unit cell area is 32a2\frac{\sqrt{3}}{2} a^2, with a single atom per 2D cell.

Solving for planar density

I'm calculating the planar density for the (111) plane in an FCC crystal, using the hexagonal unit cell area and distance between nearest neighbors.

Assessing planar density

OK, so the (111) plane emerges with the highest planar density at 43a2≈2.309a2\frac{4}{\sqrt{3} a^2} \approx \frac{2.309}{a^2}, while (110) and (100) follow, matching known rules for each plane type.

Thought about 面心立方晶体面间距计算，持续 42 秒

回答

下面给出在面心立方(fcc)晶体中常见的三组晶面 (100)(100)、(110)(110)、(111)(111) 的面间距及面致密度的复习要点，并说明哪一个面具有最大的面间距。

1. 面间距 (Plane Spacing)

对于所有立方晶系(包括简单立方、体心立方、面心立方)，(hkl)(hkl) 晶面的面间距 dhkld_{hkl} 一般可用统一的公式表示：

dhkl  =  a h2+k2+l2 d_{hkl} \;=\; \frac{a}{\sqrt{\,h^2 + k^2 + l^2\,}}

其中

  • aa 为晶胞常数(立方晶格边长)，
  • (hkl)(hkl) 为密勒指数。

因此，对于面心立方(fcc)晶体的这三个晶面：

  1. (100)(100) 面

    d100  =  a 12+02+02   =  ad_{100} \;=\; \frac{a}{\sqrt{\,1^2 + 0^2 + 0^2\,}} \;=\; a

  2. (110)(110) 面

    d110  =  a 12+12+02   =  a2d_{110} \;=\; \frac{a}{\sqrt{\,1^2 + 1^2 + 0^2\,}} \;=\; \frac{a}{\sqrt{2}}

  3. (111)(111) 面

    d111  =  a 12+12+12   =  a3d_{111} \;=\; \frac{a}{\sqrt{\,1^2 + 1^2 + 1^2\,}} \;=\; \frac{a}{\sqrt{3}}

从数值大小来看

a  >  a2  >  a3a \;>\; \frac{a}{\sqrt{2}} \;>\; \frac{a}{\sqrt{3}}

可见，(100)(100) 面的面间距最大。

2. 面致密度 (Planar Density)

面致密度(又称平面原子密度)指单位面积上所占有的原子数，计算步骤一般分为两步：

  1. 找出该晶面在二维(面内)的最小重复单位(2D晶格)的面积
  2. 数清楚该最小重复单位内“有效占有”的原子个数
  3. 用“有效原子数 / 面积”得到面致密度

在面心立方(fcc)晶体中，不同取向的晶面由于原子在面内排列方式不同，导致面致密度也不同。以下给出常见结论(用aa表示立方晶胞参数)：

2.1 (100)(100) 面

  • 面内最小重复单位面积:
    (100)(100) 面在面内呈正方形网格，其边长为 aa，面积即 a2a^2。

  • 面内“有效占有”原子数:

    • 四个角原子(各占该面单元的 1/4)合计贡献 4×14=14 \times \tfrac{1}{4} = 1
    • 中心(face-center)原子完全位于此面，在 2D 单元中算作 1
    • 合计 =1+1=2= 1 + 1 = 2

  • 面致密度:

    ρ(100)  =  2a2\rho_{(100)} \;=\;\frac{2}{a^2}

2.2 (110)(110) 面

  • 面内最小重复单位面积:
    (110)(110) 面通常可选的 2D 基矢为 T1=[1,−1,0]⋅a\mathbf{T}_1=[1,-1,0]\cdot a 和 T2=[0,0,1]⋅a\mathbf{T}_2=[0,0,1]\cdot a，

    • T1\mathbf{T}_1 长度 =12+(−1)2 a=2 a= \sqrt{1^2 + (-1)^2} \,a = \sqrt{2}\,a
    • T2\mathbf{T}_2 长度 =a= a
    • 二者正交，故该面内二维单元面积为 A(110)  =  (2 a)  ×  (a)  =  a2 2.A_{(110)} \;=\; (\sqrt{2}\,a)\;\times\; (a) \;=\; a^2\,\sqrt{2}.

  • 面内“有效占有”原子数（更细致的数原子方法略繁，这里给出常见结果）：
    该二维单胞中有效占有的原子数通常为 2（可通过数角点、沿棱边或面心等位置的原子综合折算得出）。

  • 面致密度:

    ρ(110)  =  2 a2 2   =  2 a2 .\rho_{(110)} \;=\; \frac{2}{\,a^2\,\sqrt{2}\,} \;=\; \frac{\sqrt{2}}{\,a^2\,}.

2.3 (111)(111) 面

(111)(111) 面是面心立方中最密排的晶面，也被称为密排面。

  • 面内最小重复单位: 它在面内形成了六角密堆(2D)的排列方式。

  • 面内六角最近邻边长
    若最近邻原子间距离为 a/2a/\sqrt{2} (因为 fcc 中最近邻原子间的直线距离是 a/2a/\sqrt{2})，则在 (111)(111) 面形成 2D 六角紧密堆积。

  • 该 2D 六角单胞面积
    六角紧密堆积的单胞(平行四边形或原胞)面积为

    Ahex  =  32 (a′)2,A_{\text{hex}} \;=\; \frac{\sqrt{3}}{2} \, (a')^2,

    其中 a′=a2a' = \tfrac{a}{\sqrt{2}} 为六角网格的边长(即最近邻距离)。
    代入可得

    Ahex=32(a2)2=32⋅a22=3 a24.A_{\text{hex}} = \frac{\sqrt{3}}{2} \left(\frac{a}{\sqrt{2}}\right)^2 = \frac{\sqrt{3}}{2}\cdot \frac{a^2}{2} = \frac{\sqrt{3}\,a^2}{4}.

  • 面内“有效占有”原子数
    在六角最小单胞里有效占有的原子数为 1(六角密排 2D 原胞中每个原胞就是 1 个原子量)。

  • 面致密度:

    ρ(111)  =  1 Ahex   =  1 3 a24   =  4 3 a2 .\rho_{(111)} \;=\; \frac{1}{\,A_{\text{hex}}\,} \;=\; \frac{1}{\,\frac{\sqrt{3}\,a^2}{4}\,} \;=\; \frac{4}{\,\sqrt{3}\,a^2\,}.

    数值上约为 2.309/a22.309/a^2，这在三组面中最大。

3. 小结

  1. 面间距比较

    d100=a,d110=a2,d111=a3.d_{100} = a, \quad d_{110} = \frac{a}{\sqrt{2}}, \quad d_{111} = \frac{a}{\sqrt{3}}.

    因此，(100)面的面间距最大\boxed{(100)\text{面的面间距最大}}。

  2. 面致密度比较(从大到小)

    ρ(111)  >  ρ(100)  >  ρ(110).\rho_{(111)} \;>\; \rho_{(100)} \;>\; \rho_{(110)}.

    其中

    ρ(111)=4 3 a2,ρ(100)=2a2,ρ(110)=2 a2 .\rho_{(111)} = \frac{4}{\,\sqrt{3}\,a^2}, \quad \rho_{(100)} = \frac{2}{a^2}, \quad \rho_{(110)} = \frac{\sqrt{2}}{\,a^2\,}.

    (111)(111) 面是面心立方中最密排的面。

4. 复习提示

  • 面间距通用公式(立方晶系)

    dhkl  =  a h2+k2+l2 .d_{hkl} \;=\; \frac{a}{\sqrt{\,h^2 + k^2 + l^2\,}}.

  • 面致密度计算要点

    1. 在所研究晶面上，先找出面内最小二维晶格(基矢和对应的 2D 单位胞面积)。
    2. 计算单胞内**“有效占有”的原子数**(注意角上、棱上、面心原子在不同晶面单胞中的分摊)。
    3. 用 “有效原子数 / 单胞面积” 即可得到该晶面的原子密度。

通过以上公式和步骤，就可以系统地求解任意 (hkl)(hkl) 面的面间距和面致密度。祝学习顺利!

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