Chapter 15, Problem 15A.7E

Interpretation Introduction

Interpretation:

Value of unit cell parameter a and c have to be calculated.

Concept Introduction:

Bragg’s law:

The spacing between the layers of atoms is given by the formula,

  $\begin{array}{l}\text{nλ=2dsinθ}\\ \\ \text{d=}\frac{\text{nλ}}{\text{2sinθ}}\end{array}$

The separation of planes can be calculated using miller indices as,

    $\frac{\text{1}}{{\text{d}}^{\text{2}}}\text{=}\frac{{\text{h}}^{\text{2}}}{{\text{a}}^{\text{2}}}\text{+}\frac{{\text{k}}^{\text{2}}}{{\text{b}}^{\text{2}}}\text{+}\frac{{\text{l}}^{\text{2}}}{{\text{c}}^{\text{2}}}$

Where,

h, l, and k are miller indices which indicates each plane.

d is distance of separation of plane.

For a tetragonal primitive system, the relationship is given as below:

$\frac{\text{1}}{{\text{d}}^{\text{2}}}\text{=}\frac{{\text{h}}^{\text{2}}+{\text{k}}^{\text{2}}}{{\text{a}}^{\text{2}}}\text{+}\frac{{\text{l}}^{\text{2}}}{{\text{c}}^{\text{2}}}$

Explanation of Solution

Given,

  $\begin{array}{l}{\text{h}}_{\text{1,}}{\text{k}}_{\text{1,}}{\text{l}}_{\text{1}}\text{=}\left(\text{1,1,0}\right)\\ \\ {\text{d}}_{\text{1}}\text{=}\text{257}\text{pm}\\ \\ {\text{h}}_{\text{2,}}{\text{k}}_{\text{2,}}{\text{l}}_{\text{2}}\text{=}\left(\text{2,1,1}\right)\\ \\ {\text{d}}_{\text{2}}\text{=}\text{156}\text{pm}\end{array}$

The relationship between the distance between the layers, the miller indices and the unit cell parameters in tetragonal system is given below:

  $\frac{\text{1}}{{\text{d}}^{\text{2}}}\text{=}\frac{{\text{h}}^{\text{2}}+{\text{k}}^{\text{2}}}{{\text{a}}^{\text{2}}}\text{+}\frac{{\text{l}}^{\text{2}}}{{\text{c}}^{\text{2}}}$

The equation can be rearranged as,

  $\begin{array}{l}\frac{{\text{d}}^{\text{2}}}{1}\text{=}\frac{{\text{a}}^{\text{2}}}{{\text{h}}^{\text{2}}{\text{+k}}^{\text{2}}}\text{+}\frac{{\text{c}}^{\text{2}}}{{\text{l}}^{\text{2}}}\\ \\ \frac{\text{d}}{1}\text{=}\frac{\text{a}}{{\left(\text{h+k}\right)}^{1/2}}\text{+}\frac{\text{c}}{\text{l}}\end{array}$

The first plane given is (1,1,0). Hence, $\text{h,k,l}\text{=}\text{1,1,0}$

  $\begin{array}{c}\frac{\text{d}}{1}\text{=}\frac{\text{a}}{{\left(\text{h+k}\right)}^{1/2}}\text{+}\frac{\text{c}}{\text{l}}\\ \\ \text{257}\text{pm}\text{=}\frac{\text{a}}{{\left(\text{1+1}\right)}^{1/2}}\text{+}\frac{\text{c}}{\text{0}}\\ \\ \text{257}\text{pm}\text{=}\frac{\text{a}}{1.414}\\ \\ \text{a}\text{=}1.414\text{×}\text{257}\text{pm}\text{=}363.4\text{pm}\end{array}$

The second plane given is (2,1,1). Hence, $\text{h,k,l}\text{=}2\text{,1,1}$

  $\begin{array}{c}\frac{\text{d}}{\text{1}}\text{=}\frac{\text{a}}{{\left(\text{h+k}\right)}^{1/2}}\text{+}\frac{\text{c}}{\text{l}}\\ \\ \text{156}\text{pm}\text{=}\frac{\text{a}}{{\left(\text{2+1}\right)}^{1/2}}\text{+}\frac{\text{c}}{\text{1}}\\ \\ \text{156}\text{pm}\text{=}\frac{363.4\text{pm}}{\sqrt{\text{3}}}\text{+}\frac{\text{c}}{\text{1}}\\ \\ \text{156}\text{pm}\text{=}\frac{363.4\text{pm+}\sqrt{\text{3}}\text{c}}{\sqrt{\text{3}}}\\ \\ 270\text{pm}\text{=}363.4\text{pm+}\sqrt{\text{3}}\text{c}\\ \\ \text{c}\text{=}\text{-54}\text{pm}\end{array}$

The value of unit cell parameter, a is calculated to be $\text{514}\text{pm}$ and c is calculated to be $\text{-54}\text{pm}$