Chapter 21, Problem 21.39E

Interpretation Introduction

Interpretation:

The reason as to why the $\left(111\right)$ plane of a body-centered cubic lattice does not cause detectable diffraction of X rays is to be stated using geometric arguments.

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 21.39E

The $\left(111\right)$ plane of a body-centered cubic lattice does not cause detectable diffraction of X rays because light would experience only destructive interference and it would be cancelled out.

Explanation of Solution

Consider a cube having a side of $1$ unit. It contains three axis $x$, $y$ and $z$ that corresponds to length $\left(a\right)$, breadth $\left(b\right)$ and height $\left(c\right)$ from the origin. Consider the $\left(111\right)$ plane in a cube that is away from the origin at $1$ unit from each side as shown below.

Figure 1

Similarly, consider another $\left(111\right)$ plane in a cube that is away from first plane as shown below.

Figure 2

Both the planes are redrawn by using the atoms as shown below.

Figure 3

The repeating pattern of Figure 3 is drawn and the path of the diffracted light is shown below.

Figure 4

In the body-centered cubic lattice, the atoms are present at the center as well as at the corners. The light is diffracted by the center atoms. The path of diffracted light is shown below.

Figure 5

The above diagram is drawn according to the Bragg’s law. The representation of above diagram is shown below.

Figure 6

From the Figure 6, the extra distance covered by the light when it is diffracted by the yellow atom (D) is calculated as follows.

$n_{\text{yellow}}\lambda =\overrightarrow{AD}+\overrightarrow{DE}-\overrightarrow{A{E}^{\text{'}}}$…(1)

Where,

• $n_{\text{yellow}}$ represents the number of repeated waves that are diffracted by the yellow atom.

• .

• $\lambda$ is the wavelength of the light.

The distance between two layers is $d$. Thus, the $\overrightarrow{AD}$ is equal to $\overrightarrow{DE}$.

Substitute $\overrightarrow{AD}=\overrightarrow{DE}$ in equation (1).

$n_{\text{yellow}}\lambda =\overrightarrow{AD}+\overrightarrow{AD}-\overrightarrow{A{E}^{\text{'}}}$

$n_{\text{yellow}}\lambda =2\overrightarrow{AD}-\overrightarrow{A{E}^{\text{'}}}$…(2)

Similarly, the extra distance covered by the light when it is diffracted by the blue atom (B) is calculated as follows.

$n_{\text{blue}}\lambda =\overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{A{C}^{\text{'}}}$…(3)

Where,

• $n_{\text{blue}}$ represents the number of repeated waves that are diffracted by the blue atom.

• $\lambda$ is the wavelength of the light.

The distance between two layers is $d/2$. Thus, the $\overrightarrow{AB}$ is equal to $\overrightarrow{BC}$.

Substitute $\overrightarrow{AB}=\overrightarrow{BC}$ in equation (3).

$n_{\text{blue}}\lambda =\overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{A{C}^{\text{'}}}$

$n_{\text{blue}}\lambda =2\overrightarrow{AB}-\overrightarrow{A{C}^{\text{'}}}$…(4)

Trigonometry is used to calculate the length of $\overrightarrow{AD}$ and $\overrightarrow{AB}$. Consider a $\Delta ADC$ as shown below.

Figure 7

The expression for $\overrightarrow{AD}$ is given as follows.

$\sin \theta =\frac{d}{\overrightarrow{AD}}$

$\overrightarrow{AD}=\frac{d}{\sin \theta }$…(5)

The expression for $\overrightarrow{AB}$ is given as follows.

$\begin{array}{rll}\sin \theta & =& \frac{d/2}{\overrightarrow{AB}}\\ \overrightarrow{AB}& =& 1/2\left(\frac{d}{\sin \theta }\right)\end{array}$

Substitute equation (5) in above expression.

$\overrightarrow{AB}=1/2\overrightarrow{AD}$…(6)

The trigonometry is used to calculate the length of $\overrightarrow{A{C}^{\text{'}}}$. Consider a quadrilateral $ABC{C}^{\text{'}}$ as shown below.

Figure 8

The expression for $\overrightarrow{A{C}^{\text{'}}}$ is given as follows.

$\cos \theta =\frac{\overrightarrow{A{C}^{\text{'}}}}{\overrightarrow{AC}}$

$\overrightarrow{A{C}^{\text{'}}}=\overrightarrow{AC}\cos \theta$…(7)

The expression for $\overrightarrow{A{C}^{\text{'}}}$ can also be given as follows.

$\tan \theta =\frac{1/2d}{1/2\overrightarrow{AC}}$

$\overrightarrow{AC}=\frac{d}{\tan \theta }$…(8)

Substitute equation (8) in equation (7).

$\begin{array}{rll}\overrightarrow{A{C}^{\text{'}}}& =& \frac{d}{\tan \theta }\cos \theta \\ & =& \frac{d}{\frac{\sin \theta }{\cos \theta }}\left(\cos \theta \right)\end{array}$

$\overrightarrow{A{C}^{\text{'}}}=\frac{d}{\sin \theta }\left({\cos }^2\theta \right)$…(9)

Similarly, trigonometry is used to calculate the length of $\overrightarrow{A{E}^{\text{'}}}$. Consider a quadrilateral $ADE{E}^{\text{'}}$ as shown below.

Figure 9

The expression for $\overrightarrow{A{E}^{\text{'}}}$ is given as follows.

$\cos \theta =\frac{\overrightarrow{A{E}^{\text{'}}}}{\overrightarrow{AE}}$

$\overrightarrow{A{E}^{\text{'}}}=\overrightarrow{AE}\cos \theta$…(10)

The expression for $\overrightarrow{A{C}^{\text{'}}}$ can also be given as follows.

$\tan \theta =\frac{d}{1/2\overrightarrow{AE}}$

$\overrightarrow{AE}=\frac{2d}{\tan \theta }$…(11)

Substitute equation (11) in equation (10).

$\begin{array}{rll}\overrightarrow{A{E}^{\text{'}}}& =& \frac{2d}{\tan \theta }\cos \theta \\ & =& \frac{2d}{\frac{\sin \theta }{\cos \theta }}\left(\cos \theta \right)\\ & =& 2\left(\frac{d}{\sin \theta }\left({\cos }^2\theta \right)\right)\end{array}$

Substitute equation (8) in above expression.

$\overrightarrow{A{E}^{\text{'}}}=2\overrightarrow{A{C}^{\text{'}}}$

$\overrightarrow{A{C}^{\text{'}}}=1/2\overrightarrow{A{E}^{\text{'}}}$…(12)

Substitute equation (12) and equation (6) in equation (4).

$\begin{array}{rll}{n}_{\text{blue}}\lambda & =& 2\overrightarrow{AB}-\overrightarrow{A{C}^{\text{'}}}\\ & =& 2\left(1/2\overrightarrow{AD}\right)-1/2\overrightarrow{A{E}^{\text{'}}}\\ & =& 1/2\left(2\overrightarrow{AD}-\overrightarrow{A{E}^{\text{'}}}\right)\end{array}$

Substitute equation (2) in the above expression.

$n_{\text{blue}}\lambda =1/2n_{\text{yellow}}\lambda$

The above expression shows that the length of thelight passed through the improperly diffracted path is half of the length of the light passed through the properly diffracted path. This indicates that only destructive interference is experienced by the light and it would be cancelled out. Therefore, the $\left(111\right)$ plane of a body-centered cubic lattice does not cause detectable diffraction of X rays.

Conclusion

The $\left(111\right)$ plane of a body-centered cubic lattice does not cause detectable diffraction of X rays because light would experience only destructive interference and it would be cancelled out.