Consider a two-dimensional square crystal structure, such as one side of the structure shown in Fig. 36-28a.The largest interplanar spacing of reflecting planes is the unit cell size ${\mathbf{a}}_{\mathbf{o}}$. Calculate and sketch the (a) second largest, (b) third largest, (c) fourth largest, (d) fifth largest, and (e) sixth largest interplanar spacing. (f) Show that your results in (a) through (e) are consistent with the general formula

$\mathbf{d}\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{o}}}{\sqrt{{\mathbf{h}}^{\mathbf{2}}\mathbf{+}{\mathbf{k}}^{\mathbf{2}}}}$

where h and k are relatively prime integers (they have no common factor other than unity).

If the incident direction of the wave, measured from the surfaces of these planes, and the radiation's wavelength$\mathrm{\lambda }$ fulfil Bragg's equation, diffraction maxima (caused by constructive interference) happen

$2\mathrm{dsin\theta }=\mathrm{m\lambda };\mathrm{m}=1,2,\mathrm{3...}$(Bragg’s equation)

Where $d$is the interplanar distance,$\theta$ is the angle of the incident beam,$\lambda$ is the wavelength of the incident beam, and$m$ is the order of the maximum intensity.

The largest interplanar spacing is given is equal to the cell size ${\mathrm{a}}_{\mathrm{o}}$.

Family of planes whose interplanar spacing is second largest is shown below,

The planes are diagonal to horizontal family of planes. As the total length between the lines is $2d_2$( where${\mathrm{d}}_2$ is second largest interplanar spacing) and it forms a right-angled triangle from which we can determine using$d_2$ Pythagoras theorem,

$\begin{array}{c}{\left(2{\mathrm{d}}_2\right)}^2={\left({\mathrm{a}}_{\mathrm{o}}\right)}^2+{\left({\mathrm{a}}_{\mathrm{o}}\right)}^2\\ {\mathrm{d}}_2^2=\frac{{\mathrm{a}}_{\mathrm{o}}^2}{2}\\ {\mathrm{d}}_2=\frac{{\mathrm{a}}_{\mathrm{o}}}{\sqrt{2}}\end{array}$

The Family of planes whose interplanar spacing is third largest is shown below,

As the total length between the lines is$5d_3$( where $d_3$is third largest interplanar spacing) and it forms a right-angled triangle from which we can determine role="math" localid="1663405458726" $d_3$using Pythagoras theorem,

$\begin{array}{c}{\left(5{\mathrm{d}}_3\right)}^2={\left({\mathrm{a}}_{\mathrm{o}}\right)}^2+{\left(2{\mathrm{a}}_{\mathrm{o}}\right)}^2\\ 25{\mathrm{d}}_3^2=5{\mathrm{a}}_{\mathrm{o}}^2\\ {\mathrm{d}}_3=\frac{{\mathrm{a}}_{\mathrm{o}}}{\sqrt{5}}\end{array}$

The Family of planes whose interplanar spacing is fourth largest is shown below,

As the total length between the lines is $10d_4$( where $d_4$is fourth largest interplanar spacing) and it forms a right-angled triangle from which we can determine ${\mathrm{d}}_4$using Pythagoras theorem,

$\begin{array}{c}{\left(10{\mathrm{d}}_4\right)}^2={\left({\mathrm{a}}_{\mathrm{o}}\right)}^2+{\left(3{\mathrm{a}}_{\mathrm{o}}\right)}^2\\ 100{\mathrm{d}}_4^2=10{\mathrm{a}}_{\mathrm{o}}^2\\ {\mathrm{d}}_4=\frac{{\mathrm{a}}_{\mathrm{o}}}{\sqrt{10}}\end{array}$

The family of planes which have fifth largest spacing is shown below

As the total length between the lines is $17{\mathrm{d}}_5$( where $d_5$is fifth largest interplanar spacing) and it forms a right-angled triangle from which we can determine $d_5$using Pythagoras theorem,

$\begin{array}{c}{\left(17{\mathrm{d}}_5\right)}^2={\left({\mathrm{a}}_{\mathrm{o}}\right)}^2+{\left(4{\mathrm{a}}_{\mathrm{o}}\right)}^2\\ 289{\mathrm{d}}_5^2=17{\mathrm{a}}_{\mathrm{o}}^2\\ {\mathrm{d}}_5=\frac{{\mathrm{a}}_{\mathrm{o}}}{\sqrt{17}}\end{array}$

The family of planes which have sixth largest spacing is shown below

As the total length between the lines is $26d_6$( where $d_6$is sixth largest interplanar spacing) and it forms a right-angled triangle from which we can determine ${\mathrm{d}}_6$using Pythagoras theorem,

$\begin{array}{c}{\left(26{\mathrm{d}}_6\right)}^2={\left({\mathrm{a}}_{\mathrm{o}}\right)}^2+{\left(5{\mathrm{a}}_{\mathrm{o}}\right)}^2\\ 676{\mathrm{d}}_6^2=26{\mathrm{a}}_{\mathrm{o}}^2\\ {\mathrm{d}}_6=\frac{{\mathrm{a}}_{\mathrm{o}}}{\sqrt{26}}\end{array}$

In the figures, the distance between the family of lines can represented as position vector with its lower end of the segment as origin $(0,0)$. Just like position vector, this segment can be represented as a linear combination of $\widehat{x}$and $\widehat{\mathrm{y}}$as

Where, $\widehat{\mathrm{x}}$and $\widehat{\mathrm{y}}$are smallest unit of distance in x and y direction whose magnitude in this case is ${\mathrm{a}}_{\mathrm{o}}$and $\mathrm{h}\&\mathrm{k}$are integers.

For the second largest, the distance can be written as

${\mathrm{d}}_2=\widehat{\mathrm{x}}+\widehat{\mathrm{y}}$

With values of $\mathrm{h}=1\&\mathrm{k}=1$. Putting these values in the given formula

$\begin{array}{c}{\mathrm{d}}_2=\frac{{\mathrm{a}}_{\mathrm{o}}}{\sqrt{{\left(1\right)}^2+{\left(1\right)}^2}}\\ =\frac{{\mathrm{a}}_{\mathrm{o}}}{\sqrt{2}}\end{array}$

For third largest spacing, the distance can be written as

${\mathrm{d}}_3=\widehat{\mathrm{x}}+2\widehat{\mathrm{y}}$

With values of $\mathrm{h}=1\&\mathrm{k}=2$. Putting these values in the given formula

$\begin{array}{c}{\mathrm{d}}_3=\frac{{\mathrm{a}}_{\mathrm{o}}}{\sqrt{{\left(1\right)}^2+{\left(2\right)}^2}}\\ =\frac{{\mathrm{a}}_{\mathrm{o}}}{\sqrt{5}}\end{array}$

For fourth largest spacing, the distance can be written as

${\mathrm{d}}_4=\widehat{\mathrm{x}}+3\widehat{\mathrm{y}}$

With values of $\mathrm{h}=1\&\mathrm{k}=3$. Putting these values in the given formula

$\begin{array}{c}{\mathrm{d}}_4=\frac{{\mathrm{a}}_{\mathrm{o}}}{\sqrt{{\left(1\right)}^2+{\left(3\right)}^2}}\\ =\frac{{\mathrm{a}}_{\mathrm{o}}}{\sqrt{10}}\end{array}$

For fifth largest spacing, the distance can be written as

${\mathrm{d}}_5=\widehat{\mathrm{x}}+4\widehat{\mathrm{y}}$

With values of $\mathrm{h}=1\&\mathrm{k}=4$. Putting these values in the given formula

$\begin{array}{c}{\mathrm{d}}_5=\frac{{\mathrm{a}}_{\mathrm{o}}}{\sqrt{{\left(1\right)}^2+{\left(4\right)}^2}}\\ =\frac{{\mathrm{a}}_{\mathrm{o}}}{\sqrt{17}}\end{array}$

For sixth largest spacing, the distance can be written as

${\mathrm{d}}_6=\widehat{\mathrm{x}}+5\widehat{\mathrm{y}}$

With values of $\mathrm{h}=1\&\mathrm{k}=5$. Putting these values in the given formula

$\begin{array}{c}{\mathrm{d}}_6=\frac{{\mathrm{a}}_{\mathrm{o}}}{\sqrt{{\left(1\right)}^2+{\left(5\right)}^2}}\\ =\frac{{\mathrm{a}}_{\mathrm{o}}}{\sqrt{26}}\end{array}$

Thus, all values of interplanar distance are verified using the given formula.