Chapter 21, Problem 21.34E

The aluminum-nickel alloy $\text{AlNi}$ has a simple cubic lattice with a unit cell parameter of $2.88\stackrel{\circ }{\text{A}}$. If X rays having a wavelength of $1.544\stackrel{\circ }{\text{A}}$ were used, at what angles would the X rays be diffracted by (a) the $\left(100\right)$ plane of atoms; (b) the $\left(110\right)$ plane of atoms; (c) the $\left(210\right)$ plane of atoms?

Interpretation Introduction

(a)

Interpretation:

The angle by which the X rays would be diffracted by the $\left(100\right)$ plane of atoms is to be determined.

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 21.34E

The angle by which the X rays would be diffracted by the $\left(100\right)$ plane of atoms $15.5°$.

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 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 given plane is $\left(100\right)$. The side of the unit cell is $2.88\stackrel{\circ }{\text{A}}$. The wavelength of X rays is $1.544\stackrel{\circ }{\text{A}}$.

Substitute the value of $\left(h,k,l\right)$, side and wavelength in equation (3).

$\begin{array}{rll}\theta & =& {\sin }^{-1}\left(\frac{1\times 1.544\stackrel{\circ }{\text{A}}\times \sqrt{1^2+0^2+0^2}}{2\times 2.88\stackrel{\circ }{\text{A}}}\right)\\ \theta & =& {\sin }^{-1}0.26805556\\ \theta & =& 15.5°\end{array}$

Thus, the angle by which the X rays would be diffracted by the $\left(100\right)$ plane of atoms $15.5°$.

Conclusion

The angle by which the X rays would be diffracted by the $\left(100\right)$ plane of atoms $15.5°$.

Interpretation Introduction

(b)

Interpretation:

The angle by which the X rays would be diffracted by the $\left(110\right)$ plane of atoms is to be determined.

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 21.34E

The angle by which the X rays would be e diffracted by the $\left(110\right)$ plane of atoms $22.3°$.

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 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 given plane is $\left(110\right)$. The side of the unit cell is $2.88\stackrel{\circ }{\text{A}}$. The wavelength of X rays is $1.544\stackrel{\circ }{\text{A}}$.

Substitute the value of $\left(h,k,l\right)$, side and wavelength in equation (3).

$\begin{array}{rll}\theta & =& {\sin }^{-1}\left(\frac{1\times 1.544\stackrel{\circ }{\text{A}}\times \sqrt{1^2+1^2+0^2}}{2\times 2.88\stackrel{\circ }{\text{A}}}\right)\\ \theta & =& {\sin }^{-1}0.3790878\\ \theta & =& 22.3°\end{array}$

Thus, the angle by which the X rays would be diffracted by the $\left(110\right)$ plane of atoms $22.3°$.

Conclusion

The angle by which the X rays would be diffracted by the $\left(110\right)$ plane of atoms $22.3°$.

Interpretation Introduction

(c)

Interpretation:

The angle would the X rays be diffracted by the $\left(100\right)$ plane of atoms is to be determined.

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 21.34E

The angle by which the X rays would be diffracted by the $\left(210\right)$ plane of atoms $36.8°$.

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 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 given plane is $\left(210\right)$. The side of the unit cell is $2.88\stackrel{\circ }{\text{A}}$. The wavelength of X rays is $1.544\stackrel{\circ }{\text{A}}$.

Substitute the value of $\left(h,k,l\right)$, side and wavelength in equation (3).

$\begin{array}{rll}\theta & =& {\sin }^{-1}\left(\frac{1\times 1.544\stackrel{\circ }{\text{A}}\times \sqrt{2^2+1^2+0^2}}{2\times 2.88\stackrel{\circ }{\text{A}}}\right)\\ \theta & =& {\sin }^{-1}0.59939044\\ \theta & =& 36.8°\end{array}$

Thus, the angle by which the X rays would be diffracted by the $\left(210\right)$ plane of atoms $36.8°$.

Conclusion

The angle by which the X rays would be diffracted by the $\left(210\right)$ plane of atoms $36.8°$.