下面是某立方晶系物质的几个晶面,试将它们的面间距从大到小按次序重 新排列:(123 ), (100), (200), ( 311), (121), (111), ( 210), (220), (130),(030), (221), (110)o

步骤 1：理解晶面间距的计算公式
晶面间距 \(d\) 可以通过布拉格方程计算，对于立方晶系，晶面间距 \(d\) 与晶面指数 \(hkl\) 的关系为：
\[ d = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \]
其中，\(a\) 是晶胞的边长，\(h, k, l\) 是晶面指数。

步骤 2：计算每个晶面的面间距
根据上述公式，计算每个晶面的面间距：
- (100): \(d = \frac{a}{\sqrt{1^2 + 0^2 + 0^2}} = a\)
- (110): \(d = \frac{a}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{a}{\sqrt{2}}\)
- (111): \(d = \frac{a}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{a}{\sqrt{3}}\)
- (200): \(d = \frac{a}{\sqrt{2^2 + 0^2 + 0^2}} = \frac{a}{2}\)
- (210): \(d = \frac{a}{\sqrt{2^2 + 1^2 + 0^2}} = \frac{a}{\sqrt{5}}\)
- (220): \(d = \frac{a}{\sqrt{2^2 + 2^2 + 0^2}} = \frac{a}{2\sqrt{2}}\)
- (221): \(d = \frac{a}{\sqrt{2^2 + 2^2 + 1^2}} = \frac{a}{3}\)
- (121): \(d = \frac{a}{\sqrt{1^2 + 2^2 + 1^2}} = \frac{a}{\sqrt{6}}\)
- (130): \(d = \frac{a}{\sqrt{1^2 + 3^2 + 0^2}} = \frac{a}{\sqrt{10}}\)
- (030): \(d = \frac{a}{\sqrt{0^2 + 3^2 + 0^2}} = \frac{a}{3}\)
- (311): \(d = \frac{a}{\sqrt{3^2 + 1^2 + 1^2}} = \frac{a}{\sqrt{11}}\)
- (123): \(d = \frac{a}{\sqrt{1^2 + 2^2 + 3^2}} = \frac{a}{\sqrt{14}}\)

步骤 3：将面间距从大到小排序
根据计算结果，将面间距从大到小排序：
- (100): \(a\)
- (110): \(\frac{a}{\sqrt{2}}\)
- (111): \(\frac{a}{\sqrt{3}}\)
- (200): \(\frac{a}{2}\)
- (210): \(\frac{a}{\sqrt{5}}\)
- (220): \(\frac{a}{2\sqrt{2}}\)
- (221): \(\frac{a}{3}\)
- (121): \(\frac{a}{\sqrt{6}}\)
- (130): \(\frac{a}{\sqrt{10}}\)
- (030): \(\frac{a}{3}\)
- (311): \(\frac{a}{\sqrt{11}}\)
- (123): \(\frac{a}{\sqrt{14}}\)