Chapter 22, Problem 22.35E

Interpretation Introduction

(a)

Interpretation:

The closest ${\text{Na}}^{+}-{\text{Na}}^{+}$ distance for a surface made by the $\left(100\right)$ plane is to be calculated.

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 22.35E

The closest ${\text{Na}}^{+}-{\text{Na}}^{+}$ distance for a surface made by the $\left(100\right)$ plane is $3.988\stackrel{\circ }{\text{A}}$.

Explanation of Solution

The structure of a face-centered cubic lattice is shown below.

Figure 1

The plane of face-centered cubic lattice that has $\left(100\right)$ Miller indices is the outer plane. The outer plane of the face-centered cubic lattice is shown below.

Figure 2

The closest ${\text{Na}}^{+}-{\text{Na}}^{+}$ distance is denoted by the symbol $b$ in the Figure (2).

The lattice parameter $\left(a\right)$ of face-centered cubic lattice is $5.640\stackrel{\circ }{\text{A}}$.

The Pythagoras theorem is shown below.

$H^2=B^2+P^2$…(1)

Where,

• $H$ is the hypotenuse of the triangle.

• $B$ is the base of the triangle.

• $P$ is the perpendicular of the triangle.

The hypotenuse of the triangle shown in Figure (2) is $2b$.

The base and perpendicular of the triangle shown in Figure (2) are $a$.

Substitute the value of hypotenuse, base, and perpendicular in the equation (1).

$\begin{array}{rll}{\left(2b\right)}^2& =& a^2+a^2\\ 4b^2& =& 2a^2\\ 2b^2& =& a^2\\ b& =& \frac{a}{\sqrt{2}}\end{array}$

Substitute the value of $a$ in the above expression.

$\begin{array}{rll}b& =& \frac{5.640\stackrel{\circ }{\text{A}}}{\sqrt{2}}\\ & =& 3.988\stackrel{\circ }{\text{A}}\end{array}$

Therefore, the closest ${\text{Na}}^{+}-{\text{Na}}^{+}$ distance for a surface made by the $\left(100\right)$ plane is $3.988\stackrel{\circ }{\text{A}}$.

Conclusion

The closest ${\text{Na}}^{+}-{\text{Na}}^{+}$ distance for a surface made by the $\left(100\right)$ plane is $3.988\stackrel{\circ }{\text{A}}$.

Interpretation Introduction

(b)

Interpretation:

The closest ${\text{Na}}^{+}-{\text{Na}}^{+}$ distance for a surface made by the $\left(110\right)$ plane is to be calculated.

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 22.35E

The closest ${\text{Na}}^{+}-{\text{Na}}^{+}$ distance for a surface made by the $\left(110\right)$ plane is $3.988\stackrel{\circ }{\text{A}}$.

Explanation of Solution

The structure of a face-centered cubic lattice is shown below.

Figure 1

The plane of face-centered cubic lattice that has $\left(110\right)$ Miller indices is the diagonal plane. The diagonal plane of the face-centered cubic lattice is shown below.

Figure 3

The lattice parameter $\left(a\right)$ of face-centered cubic lattice is $5.640\stackrel{\circ }{\text{A}}$.

The relation between the length of the edge of a cube $\left(a\right)$ and diagonal $\left(d\right)$ of the cube is shown below.

$d=\sqrt{2}a$

The diagonal of the triangle shown in Figure (3) is $2b$.

Substitute the value of $d$, and $a$ in the above equation.

$\begin{array}{rll}2b& =& \left(\sqrt{2}\right)\left(5.640\stackrel{\circ }{\text{A}}\right)\\ b& =& \frac{5.640\stackrel{\circ }{\text{A}}}{\sqrt{2}}\\ & =& 3.988\stackrel{\circ }{\text{A}}\end{array}$

Therefore, the closest ${\text{Na}}^{+}-{\text{Na}}^{+}$ distance for a surface made by the $\left(110\right)$ plane is $3.988\stackrel{\circ }{\text{A}}$.

Conclusion

The closest ${\text{Na}}^{+}-{\text{Na}}^{+}$ distance for a surface made by the $\left(110\right)$ plane is $3.988\stackrel{\circ }{\text{A}}$.

Interpretation Introduction

(c)

Interpretation:

The closest ${\text{Na}}^{+}-{\text{Na}}^{+}$ distance for a surface made by the $\left(111\right)$ plane is to be calculated.

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 22.35E

The closest ${\text{Na}}^{+}-{\text{Na}}^{+}$ distance for a surface made by the $\left(111\right)$ plane is $3.988\stackrel{\circ }{\text{A}}$.

Explanation of Solution

The structure of a face-centered cubic lattice is shown below.

Figure 1

The plane of face-centered cubic lattice that has $\left(111\right)$ Miller indices is the trigonal plane. The outer plane of the face-centered cubic lattice is shown below.

Figure 4

The side of the trigonal plane is same as the diagonal of the of the cube.

The lattice parameter $\left(a\right)$ of face-centered cubic lattice is $5.640\stackrel{\circ }{\text{A}}$.

The relation between the length of the edge of a cube $\left(a\right)$ and diagonal $\left(d\right)$ of the cube is shown below.

$d=\sqrt{2}a$

The diagonal of the triangle shown in Figure (4) is $2b$.

Substitute the value of $d$, and $a$ in the above equation.

$\begin{array}{rll}2b& =& \left(\sqrt{2}\right)\left(5.640\stackrel{\circ }{\text{A}}\right)\\ b& =& \frac{5.640\stackrel{\circ }{\text{A}}}{\sqrt{2}}\\ & =& 3.988\stackrel{\circ }{\text{A}}\end{array}$

Therefore, the closest ${\text{Na}}^{+}-{\text{Na}}^{+}$ distance for a surface made by the $\left(111\right)$ plane is $3.988\stackrel{\circ }{\text{A}}$.

Conclusion

The closest ${\text{Na}}^{+}-{\text{Na}}^{+}$ distance for a surface made by the $\left(111\right)$ plane is $3.988\stackrel{\circ }{\text{A}}$.