Chapter 16, Problem 1ASA

Interpretation Introduction

(a)

Interpretation:

The length of the face diagonal in terms of $d_{\text{O}}$, in the given cubic cell having an edge with a length equal to $d_{\text{O}}$ is to be calculated.

Concept introduction:

The crystal can exist in various shapes but the most common one is cubic. The FCC unit cell has lattice points at eight corners of the unit cell along with the face centers of unit cell. The total number of atoms in FCC unit cell is four.

Answer to Problem 1ASA

The length of the face diagonal in terms of $d_{\text{O}}$, in the given cubic cell having an edge with a length equal to $d_{\text{O}}$ is $\sqrt{2}{d}_{\text{O}}$.

Explanation of Solution

The face diagonal BC, represented by a, is shown below.

Figure 1

In the Figure 1, the face diagonal, represented by a, is the hypotenuse of right angle triangle ABC. The given unit cell is cube. Therefore, the length of AB will be equal to length of AC.

Therefore,

$\text{AC}=\text{AB}=d_{\text{O}}$

The length of face diagonal is calculated by using Pythagoras theorem as shown below.

$\begin{array}{rll}{\left(\text{BC}\right)}^2& =& {\left(\text{AB}\right)}^2+{\left(\text{AC}\right)}^2\\ \text{BC}& =& \sqrt{{\left(\text{AB}\right)}^2+{\left(\text{AC}\right)}^2}\\ \text{a}& =& \sqrt{{\left(\text{AB}\right)}^2+{\left(\text{AC}\right)}^2}\end{array}$ …(1)

Substitute the value of AB and AC in equation (1)

$\begin{array}{rll}\text{a}& =& \sqrt{{\left(\text{AB}\right)}^2+{\left(\text{AC}\right)}^2}\\ & =& \sqrt{{\left(d_{\text{O}}\right)}^2+{\left(d_{\text{O}}\right)}^2}\\ & =& \sqrt{2{\left(d_{\text{O}}\right)}^2}\\ & =& \sqrt{2}{d}_{\text{O}}\end{array}$

Therefore, the length of the face diagonal in terms of $d_{\text{O}}$ is $\sqrt{2}{d}_{\text{O}}$.

Conclusion

The length of the face diagonal in terms of $d_{\text{O}}$ is $\sqrt{2}{d}_{\text{O}}$.

Interpretation Introduction

(b)

Interpretation:

The length of the cube diagonal in terms of $d_{\text{O}}$, in the given cubic cell having an edge with a length equal to $d_{\text{O}}$ is to be calculated.

Concept introduction:

The crystal can exist in various shapes but the most common one is cubic. The FCC unit cell has lattice points at eight corners of the unit cell along with the face centers of unit cell. The total number of atoms in FCC unit cell is four.

Answer to Problem 1ASA

The length of the cube diagonal in terms of $d_{\text{O}}$, in the given cubic cell having an edge with a length equal to $d_{\text{O}}$ is $\sqrt{3}{d}_{\text{O}}$.

Explanation of Solution

A line that runs from one corner to other corner of the cube via centre of the cube is termed as cube diagonal. The cube diagonal is represented by CE as shown below.

Figure 2

In the Figure 2, the cube diagonal, represented by b, is the hypotenuse of right angle triangle CDE. The edge length DE is given as $d_{\text{O}}$. The given unit cell is cube. Therefore, the length of face diagonal BC will be equal to length of face diagonal CD.

Therefore,

$\text{BC}=\text{CD}=\sqrt{2}{d}_{\text{O}}$

The length of face diagonal is calculated by using Pythagoras theorem as shown below.

$\begin{array}{rll}{\left(\text{CE}\right)}^2& =& {\left(\text{CD}\right)}^2+{\left(\text{DE}\right)}^2\\ \text{CE}& =& \sqrt{{\left(\text{CD}\right)}^2+{\left(\text{DE}\right)}^2}\\ \text{b}& =& \sqrt{{\left(\text{CD}\right)}^2+{\left(\text{DE}\right)}^2}\end{array}$ …(2)

Substitute the value of edge length DE and face diagonal CD in equation (2).

$\begin{array}{rll}\text{b}& =& \sqrt{{\left(\text{CD}\right)}^2+{\left(\text{DE}\right)}^2}\\ & =& \sqrt{{\left(\sqrt{2}{d}_{\text{O}}\right)}^2+{\left(d_{\text{O}}\right)}^2}\\ & =& \sqrt{2{\left(d_{\text{O}}\right)}^2+{\left(d_{\text{O}}\right)}^2}\\ & =& \sqrt{3}{d}_{\text{O}}\end{array}$

Therefore, the length of the cube diagonal in terms of $d_{\text{O}}$ is $\sqrt{3}{d}_{\text{O}}$.

Conclusion

The length of the cube diagonal in terms of $d_{\text{O}}$ is $\sqrt{3}{d}_{\text{O}}$.