Chapter 3, Problem 3.62P

Interpretation Introduction

(a)

Interpretation:

The direction within the hexagonal unit cell for $\left[\text{0 }1\text{ <mover accent="true"><mo>1</mo><mo>¯</mo></mover><mo> 0</mo>}\right]$ needs to be sketched.

Concept Introduction:

Miller indices at plane pass through the three points at edges of the hexagonal lattice. It has a four-digit system $\left[h,k,i,l\right].$ The additional indices are used to show the direction in three-dimensional vector.

Answer to Problem 3.62P

Figure shows the direction within the hexagonal unit cell for $\left[\text{0 }1\text{ <mover accent="true"><mo>1</mo><mo>¯</mo></mover><mo> 0</mo>}\right]$

  

Explanation of Solution

Procedure for drawing direction for $\left[\text{0 }1\text{ <mover accent="true"><mo>1</mo><mo>¯</mo></mover><mo> 0</mo>}\right]$ :

1. First draw empty hexagonal unit cell.
2. There is no displacement along a₁ direction as coordinate is zero.
3. Draw the line in a₂ direction by 1 unit.
4. Draw the line in opposite direction of a₃ by $\overline{1}$ units.
5. There is no displacement along z-axis direction as coordinate is zero.
6. Then join the origin to the point to obtain desired direction

  

Interpretation Introduction

(b)

Interpretation:

The direction within the hexagonal unit cell for $\left[11\overline{2}\text{ 0}\right]$ needs to be sketched.

Concept Introduction:

Miller indices at plane pass through the three points at edges of the hexagonal lattice. It has a four-digit system $\left[h,k,i,l\right].$ The additional indices are used to show the direction in three-dimensional vector.

Answer to Problem 3.62P

Figure shows the direction within the hexagonal unit cell for $\left[11\overline{2}\text{ 0}\right]$

  

Explanation of Solution

  $\left[11\overline{2}0\right]$

Procedure for drawing direction for $\left[11\overline{2}\text{ 0}\right]$ :

1. First draw empty hexagonal unit cell.
2. Draw the line in direction of a₁ by 1 units which indicates 1.
3. Draw the line in a₂ direction by 1 unit.
4. Draw the line in opposite direction of a₃ by $\overline{2}$ units.
5. There is no displacement along z-axis direction as coordinate is zero.
6. Then join the origin to the point to obtain desired direction

  

Interpretation Introduction

(c)

Interpretation:

The direction within the hexagonal unit cell for $\left[\overline{1}01\text{ 1}\right]$ needs to be sketched.

Concept Introducton:

Miller indices at plane pass through the three points at edges of the hexagonal lattice. It has a four-digit system $\left[h,k,i,l\right].$ The additional indices are used to show the direction in three-dimensional vector.

Answer to Problem 3.62P

Figure shows the direction within the hexagonal unit cell for $\left[\overline{1}01\text{ 1}\right]$

  

Explanation of Solution

Procedure for drawing direction for $\left[\overline{1}01\text{ 1}\right]$ :

1. First draw empty hexagonal unit cell.
2. Draw the line in opposite direction of a₁ by 1 unit which indicates $\overline{1}$ .
3. There is no displacement along a₂ direction as coordinate is zero.
4. Draw the line in direction of a₃ by 1 unit which indicates 1.
5. Draw the line in direction along z-axis by 1 unit which indicates 1.
6. Then join the origin to the point to obtain desired direction_.

  

Interpretation Introduction

(d)

Interpretation:

The direction within the hexagonal unit cell for $\left[0003\right]$ needs to be sketched.

Concept Introduction:

Miller indices at plane pass through the three points at edges of the hexagonal lattice. It has a four-digit system $\left[h,k,i,l\right].$ The additional indices are used to show the direction in three-dimensional vector.

Answer to Problem 3.62P

Figure shows the plane within the hexagonal unit cell for $\left[\text{ 0}003\right]$

  

Explanation of Solution

Procedure for drawing planes for $\left[\text{ 0}003\right]$ :

1. Converting four index miller indices of $\left[\text{ 0}003\right]$_. into three indices.
$\begin{array}{l}\left(hkil\right)= \left(0003\right)\\ h=0\\ k=0\\ i=0\\ l=3\end{array}$
2. Now taking reciprocals of plane and putting lattice parameters
$\begin{array}{l}{a}_1=\infty \\ a_2=\infty \\ a_3=\infty \\ z=3a\end{array}$

  

Interpretation Introduction

(e)

Interpretation:

The direction within the hexagonal unit cell for $\left[\overline{1}010\right]$ needs to be sketched.

Concept Introduction:

Miller indices at plane pass through the three points at edges of the hexagonal lattice. It has a four-digit system $\left[hkil\right]$ The additional indices are used to show the direction in three-dimensional vector.

Answer to Problem 3.62P

Figure shows the plane within the hexagonal unit cell for $\left[\text{ <mover accent="true"><mo>1</mo><mo>¯</mo></mover>}\text{0 }10\right]$

  

Explanation of Solution

Procedure for drawing planes for $\left[\text{ <mover accent="true"><mo>1</mo><mo>¯</mo></mover>}\text{0 }10\right]$ :

1. Converting four index miller indices of $\left[\text{ <mover accent="true"><mo>1</mo><mo>¯</mo></mover>}\text{0 }10\right]$_. into three indices.
$\begin{array}{l}\left(hkil\right)= \left(\overline{1}010\right)\\ h=\overline{1}\\ k=0\\ i=1\\ l=0\end{array}$
2. Now taking reciprocals of plane and putting lattice parameters
$\begin{array}{l}{a}_1=-a\\ a_2=\infty \\ a_3=a\\ z=\infty \end{array}$

  

Interpretation Introduction

(f)

Interpretation:

The direction within the hexagonal unit cell for $\left[01\overline{1}1\right]$ needs to be sketched.

Concept Introduction:

Miller indices at plane pass through the three points at edges of the hexagonal lattice. It has a four-digit system $\left[h,k,i,l\right].$ The additional indices are used to show the direction in three-dimensional vector.

Answer to Problem 3.62P

Figure shows the plane within the hexagonal unit cell for $\left[01\overline{1}1\right]$

  

Explanation of Solution

Procedure for drawing planes for $\left[01\overline{1}1\right]$ :

1. Converting four index miller indices of $\left[01\overline{1}1\right]$_. into three indices.
$\begin{array}{l}\left(hkil\right)= \left(01\overline{1}1\right)\\ h=0\\ k=1\\ i=\overline{1}\\ l=1\end{array}$
2. Now taking reciprocals of plane and putting lattice parameters
$\begin{array}{l}{a}_1=\infty \\ a_2=a\\ a_3=-a\\ z=a\end{array}$