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PHY.K02UF Molecular and Solid State Physics | ||||
Every point of a Bravais lattice can be reached from the origin by a translation vector of the form,
\[ \begin{equation} \vec{T}_{hkl} = h\vec{a}_1 + k\vec{a}_2 + l\vec{a}_3, \end{equation} \]where $\vec{a}_1$, $\vec{a}_2$, $\vec{a}_3$ are the primitive lattice vectors and $h$, $k$, and $l$ are integers called the Miller indices. The direction that the vector $\vec{T}_{hkl}$ points is denoted with square brackets, $[hkl]$. If there are directions that are equivalent due to symmetry any of the equivalent directions are specified by angular brackets <$hkl$>.
For example, in a cubic system:
<100> = [100], [010], [001], [-100], [0-10], [00-1]
<110> = [110], [-1-10], [1-10], [-110], [101], [-10-1], [-101], [10-1], [011], [0-1-1], [0-11], [01-1].
Often the minus signs are drawn as overlines [-10-1] = $\left[\overline{1}0\overline{1}\right]$. This is pronounced one-bar zero one-bar.
Using Miller indices, planes are denoted with curved brackets $(hkl)$. For cubic crystals, the $(hkl)$ plane is normal to the $[hkl]$ direction. However, this simple relationship does not hold for the other crystal systems. Generally, the $(hkl)$ plane intercepts the axis in the direction of $\vec{a}_1$ at $\frac{|\vec{a}_1|}{h}$, the axis in the direction of $\vec{a}_2$ at $\frac{|\vec{a}_2|}{k}$, and the axis in the direction of $\vec{a}_3$ at $\frac{|\vec{a}_3|}{l}$.
Equivalent planes are denoted with curly brackets $\{hkl\}$. Three points that can be used to define the $(hkl)$ plane are $\left( \frac{a_{1x}}{h},\frac{a_{1y}}{h},\frac{a_{1z}}{h} \right)$, $\left(\frac{a_{2x}}{k},\frac{a_{2y}}{k},\frac{a_{2z}}{k} \right)$, and $\left(\frac{a_{3x}}{l},\frac{a_{3y}}{l},\frac{a_{3z}}{l}\right)$. The normal vector to this plane can be determined by taking the cross product of two vectors in the plane. If $(h,k,l \ne 0)$, two suitable vectors are $\vec{v}_1=\frac{\vec{a}_1}{h}-\frac{\vec{a}_3}{l}$ and $\vec{v}_2=\frac{\vec{a}_2}{k}-\frac{\vec{a}_3}{l}$. The unit vector normal to the $(hkl)$ plane is called the pole of that crystallographic plane,
$$\hat{n}_{hkl}=\frac{\vec{v}_1\times \vec{v}_2}{|\vec{v}_1\times \vec{v}_2|}.$$For cubic crystals, $\vec{T}_{hkl}$ is normal to $(hkl)$ and parallel to $\hat{n}_{hkl}$ but this is not generally true for crystals with other symmetries.
The volume of the primitive unit cell is, $\vec{a}_1\cdot(\vec{a}_2\times\vec{a}_3)=$ 6.9935e-29 [m³]. The translation vector is, $\vec{T}_{hkl}=$ 4.12e-10 $\hat{x}+$ 0 $\hat{y}+$ 0 $\hat{z}$ [m]. The length of the translation vector is, 4.12e-10 [m]. The normal vector to the $(hkl)$ plane is, $\hat{n}_{hkl}=$ 1.00 $\hat{x}+$ 0.00 $\hat{y}+$ 0.00 $\hat{z}$. The unit vector in the $\vec{T}_{hkl}$ direction is, $\hat{T}_{hkl}=$ 1.00 $\hat{x}+$ 0.00 $\hat{y}+$ 0.00 $\hat{z}$. The angle between $\vec{T}_{hkl}$ and $\hat{n}_{hkl}$ is, $\theta=$ 0.00 [rad]. |
JSmol can be used to draw crystalographic planes of different crystal structures. The JSmol models of some common crystal structures are linked below.