Chapter 15, Problem 15A.9P

Interpretation Introduction

Interpretation:

The separation of the $\left\{hkl\right\}$ planes in an orthorhombic crystal with sides a, b and c has to be derived.

Explanation of Solution

As known the trigonometric function as follows,

Figure 1

Consider an orthorhombic crystal with edges a, b and c. Let $\left(hkl\right)$ be the Miller indices of the plane ABC.

Let, $OP=d$ be a normal to the plane passing through the origin O and makes angles ${\alpha }^{\text{'}},{\beta }^{\text{'}}and{\gamma }^{\text{'}}$ with x, y and z-axes respectively.

Figure 2

Then,

  $\begin{array}{c}\cos {\alpha }^{\text{'}}=\frac{OP}{OA}=\frac{d_1}{a/h}=\frac{d_{1}h}{a}\\ \\ \cos {\beta }^{\text{'}}=\frac{OP}{OB}=\frac{d_1}{b/h}=\frac{d_{1}h}{b}\\ \\ \cos {\gamma }^{\text{'}}=\frac{OP}{OC}=\frac{d_1}{c/h}=\frac{d_{1}h}{c}\end{array}$

From the law of directions of cosines, it can be written ${\cos }^2{\alpha }^{\text{'}}+{\cos }^2{\beta }^{\text{'}}+{\cos }^2{\gamma }^{\text{'}}=1$

  $\begin{array}{c}\therefore {\left(\frac{d_{1}h}{a}\right)}^2+{\left(\frac{d_{1}k}{b}\right)}^2+{\left(\frac{d_{1}l}{c}\right)}^2=1\\ \\ \frac{1}{d_1^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}\\ \\ d_1=\frac{1}{\sqrt{\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}}}\end{array}$

Let the next plane parallel to ABC be at a distance OQ from the origin. Then its intercepts are

  $O{A}^{\text{'}}:O{B}^{\text{'}}:O{C}^{\text{'}}=\frac{2a}{h}:\frac{2b}{k}:\frac{2c}{l}$

Figure 3

  $\begin{array}{c}\cos {\alpha }^{\text{'}\text{'}}=\frac{OQ}{O{A}^{\text{'}}}=\frac{d_2}{2a/h}=\frac{d_{2}h}{2a}\\ \\ \cos {\beta }^{\text{'}\text{'}}=\frac{OQ}{O{B}^{\text{'}}}=\frac{d_2}{2b/h}=\frac{d_{2}h}{2b}\\ \\ \cos {\gamma }^{\text{'}\text{'}}=\frac{OQ}{O{C}^{\text{'}}}=\frac{d_2}{2c/h}=\frac{d_{2}h}{2c}\end{array}$

From the law of directions of cosines, it can be written ${\cos }^2{\alpha }^{\text{'}}{}^{\text{'}}+{\cos }^2{\beta }^{\text{'}}{}^{\text{'}}+{\cos }^2{\gamma }^{\text{'}}{}^{\text{'}}=1$

  $\begin{array}{c}\therefore {\left(\frac{d_{2}h}{2a}\right)}^2+{\left(\frac{d_{2}k}{2b}\right)}^2+{\left(\frac{d_{2}l}{2c}\right)}^2=1\\ \\ \frac{1}{d_2^2}=\frac{h^2}{4a^2}+\frac{k^2}{4b^2}+\frac{l^2}{4c^2}\\ \\ d_2=\frac{2}{\sqrt{\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}}}\end{array}$

Interplanar distance, d:

It is the distance between two planes. Here, the two plane are ABC and $A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}$.

Then

  $\begin{array}{c}d=d_2-d_1\\ \\ =\frac{2}{\sqrt{\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}}}-\frac{1}{\sqrt{\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}}}\\ \\ =\frac{1}{\sqrt{\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}}}\end{array}$

  OR

  $\frac{1}{d_{{}_{hkl}}^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}$

This is the expression for the separation of the $\left\{hkl\right\}$ planes in an orthorhombic crystal with sides a, b and c.