91Ó°ÊÓ

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

$2\mathrm{»å²õ¾\pm ²Ôθ}=\mathrm{³¾Î»};\text{ }\mathrm{m}=1,2,\mathrm{3...}$(Bragg’s equation)

Where $d$is the interplanar distance,$\mathrm{θ}$ is the angle of the incident beam,$\mathrm{λ}$ 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 $\mathrm{h}=1\text{ }\&\text{ }\mathrm{k}=1$. 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 $\mathrm{h}=1\text{ }\&\text{ }\mathrm{k}=2$. 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 $\mathrm{h}=1\text{ }\&\text{ }\mathrm{k}=3$. 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 $\mathrm{h}=1\text{ }\&\text{ }\mathrm{k}=4$. 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 $\mathrm{h}=1\text{ }\&\text{ }\mathrm{k}=5$. 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.