Chapter 21, Problem 21.38E

Interpretation Introduction

Interpretation:

The angles of diffraction of X rays having $\text{λ}=1.54056\stackrel{\circ }{\text{A}}$ by $\text{KBr}$ are to be predicted.

Concept introduction:

A unit cell of the crystal is the three-dimensional arrangement of the atoms present in the crystal. The unit cell is the smallest and simplest unit of the crystal which on repetition forms an entire crystal. Unit cell can be a cubic unit cell or hexagonal unit cell. The classification of a unit cell depends on the lattice site occupied by the atoms.

Answer to Problem 21.38E

The angles of diffraction of X rays having $\text{λ}=1.54056\stackrel{\circ }{\text{A}}$ by $\text{KBr}$ are,

$\begin{array}{r}h,k,l\\ \hfill 111\\ \hfill 200\\ \hfill 220\\ \hfill 311\\ 222\\ 400\end{array}\begin{array}{r}\theta \\ \hfill 11.7°\\ \hfill 13.5°\\ \hfill 19.3°\\ \hfill 22.8°\\ 23.9°\\ 27.9°\end{array}$

Explanation of Solution

The $d$ spacing is calculated by formula as shown below.

$d=\frac{a}{\sqrt{h^2+k^2+l^2}}$ …(1)

Where,

• $h,k,l$are the Miller indices.

• $a$ is the side of the unit cell.

The Bragg’s equation for diffraction of X rays is given by an expression as shown below.

$n\lambda =2d\sin \theta$ …(2)

Where,

• $\lambda$ is the wavelength of incident ray.

• $d$ is the distance between two parallel planes.

• $\theta$ is the Bragg’s angle.

• $n$ is the order of diffraction.

Substitute equation (1) in equation (2).

$\begin{array}{rll}n\lambda & =& 2\left(\frac{a}{\sqrt{h^2+k^2+l^2}}\right)\sin \theta \\ \sin \theta & =& \frac{n\lambda \sqrt{h^2+k^2+l^2}}{2a}\end{array}$

The above formula can be written as follows:

$\theta ={\sin }^{-1}\left(\frac{n\lambda \sqrt{h^2+k^2+l^2}}{2a}\right)$ …(3)

The structure of $\text{KBr}$ is similar to that of sodium chloride. Thus, it is face-centered cubic cell. From the table 21.3, the planes diffracted by face-centered cubic cell are $\left(111\right)$, $\left(200\right)$, $\left(220\right)$, $\left(311\right)$, $\left(222\right)$ and $\left(400\right)$. The side of the unit cell is $6.59\stackrel{\circ }{\text{A}}$. The wavelength of X rays is $1.54056\stackrel{\circ }{\text{A}}$.

Substitute $\left(111\right)$ side and wavelength in equation (3).

$\begin{array}{rll}\theta & =& {\sin }^{-1}\left(\frac{1\times 1.54056\stackrel{\circ }{\text{A}}\times \sqrt{1^2+1^2+1^2}}{2\times 6.59\stackrel{\circ }{\text{A}}}\right)\\ \theta & =& {\sin }^{-1}0.20245\\ \theta & =& 11.7°\end{array}$

Substitute $\left(200\right)$ side and wavelength in equation (3).

$\begin{array}{rll}\theta & =& {\sin }^{-1}\left(\frac{1\times 1.54056\stackrel{\circ }{\text{A}}\times \sqrt{2^2+0^2+0^2}}{2\times 6.59\stackrel{\circ }{\text{A}}}\right)\\ \theta & =& {\sin }^{-1}0.23377\\ \theta & =& 13.5°\end{array}$

Substitute $\left(220\right)$ side and wavelength in equation (3).

$\begin{array}{rll}\theta & =& {\sin }^{-1}\left(\frac{1\times 1.54056\stackrel{\circ }{\text{A}}\times \sqrt{2^2+2^2+0^2}}{2\times 6.59\stackrel{\circ }{\text{A}}}\right)\\ \theta & =& {\sin }^{-1}0.33060\\ \theta & =& 19.3°\end{array}$

Substitute $\left(311\right)$ side and wavelength in equation (3).

$\begin{array}{rll}\theta & =& {\sin }^{-1}\left(\frac{1\times 1.54056\stackrel{\circ }{\text{A}}\times \sqrt{3^2+1^2+1^2}}{2\times 6.59\stackrel{\circ }{\text{A}}}\right)\\ \theta & =& {\sin }^{-1}0.38767\\ \theta & =& 22.8°\end{array}$

Substitute $\left(222\right)$ side and wavelength in equation (3).

$\begin{array}{rll}\theta & =& {\sin }^{-1}\left(\frac{1\times 1.54056\stackrel{\circ }{\text{A}}\times \sqrt{2^2+2^2+2^2}}{2\times 6.59\stackrel{\circ }{\text{A}}}\right)\\ \theta & =& {\sin }^{-1}0.40490\\ \theta & =& 23.9°\end{array}$

Substitute $\left(400\right)$ side and wavelength in equation (3).

$\begin{array}{rll}\theta & =& {\sin }^{-1}\left(\frac{1\times 1.54056\stackrel{\circ }{\text{A}}\times \sqrt{4^2+0^2+0^2}}{2\times 6.59\stackrel{\circ }{\text{A}}}\right)\\ \theta & =& {\sin }^{-1}0.46754\\ \theta & =& 27.9°\end{array}$

Thus, the angles of diffraction of X rays at different planes are shown below.

$\begin{array}{r}h,k,l\\ \hfill 111\\ \hfill 200\\ \hfill 220\\ \hfill 311\\ 222\\ 400\end{array}\begin{array}{r}\theta \\ \hfill 11.7°\\ \hfill 13.5°\\ \hfill 19.3°\\ \hfill 22.8°\\ 23.9°\\ 27.9°\end{array}$

Conclusion

The angles of diffraction of X rays having $\text{λ}=1.54056\stackrel{\circ }{\text{A}}$ by $\text{KBr}$ are,

$\begin{array}{r}h,k,l\\ \hfill 111\\ \hfill 200\\ \hfill 220\\ \hfill 311\\ 222\\ 400\end{array}\begin{array}{r}\theta \\ \hfill 11.7°\\ \hfill 13.5°\\ \hfill 19.3°\\ \hfill 22.8°\\ 23.9°\\ 27.9°\end{array}$