Chapter 3.15, Problem 30AAP

(a)

To determine

Show the direction vector in a FCC unit cell for the cubic direction $\left[\overline{1}\overline{1}1\right]$. Write the position coordinates of the atoms with their centers intersected by each plane. Find the repeat distance in plane direction $\left[\overline{1}\overline{1}1\right]$ in terms of lattice constant.

Explanation of Solution

Show the direction vector in the FCC unit cell for the cubic direction $\left[\overline{1}\overline{1}1\right]$ as in Figure (1).

Position coordinates:

For the cubic direction $\left[\overline{1}\overline{1}1\right]$, the position coordinates of the atoms are $\left(0,0,0\right)$, and $\left(-1,-1,1\right)$.

Repeat distance:

For the cubic direction $\left[\overline{1}\overline{1}1\right]$, the repeat distance is the distance between the corner atoms which is $\sqrt{3}a$. Therefore, the repeat distance is $\sqrt{3}a$.

(b)

To determine

Show the direction vector in a FCC unit cell for the cubic direction $\left[10\overline{1}\right]$. Write the position coordinates of the atoms with their centers intersected by each plane. Find the repeat distance in plane direction $\left[10\overline{1}\right]$ in terms of lattice constant.

Explanation of Solution

Show the direction vector in the FCC unit cell for the cubic direction $\left[10\overline{1}\right]$ as in Figure (2).

Position coordinates:

For the cubic direction $\left[10\overline{1}\right]$, the position coordinates of the atoms are $\left(0,0,0\right)$, $\left(1,0,-1\right)$, and $\left(\frac{1}{2},0,-\frac{1}{2}\right)$.

Repeat distance:

For the cubic direction $\left[10\overline{1}\right]$, the repeat distance is the distance between the corner atom and the face centered atom which is $\frac{\sqrt{2}}{2}a$. Therefore, the repeat distance is $\frac{\sqrt{2}}{2}a$.

(c)

To determine

Show the direction vector in a FCC unit cell for the cubic direction $\left[2\overline{1}\overline{1}\right]$. Write the position coordinates of the atoms with their centers intersected by each plane. Find the repeat distance in plane direction $\left[2\overline{1}\overline{1}\right]$ in terms of lattice constant.

Explanation of Solution

Show the direction vector in the FCC unit cell for the cubic direction $\left[2\overline{1}\overline{1}\right]$ as in Figure (3).

Position coordinates:

For the cubic direction $\left[2\overline{1}\overline{1}\right]$, the position coordinates of the atoms are $\left(0,0,0\right)$, $\left(1,-\frac{1}{2},-\frac{1}{2}\right)$, and $\left(2,-1,-1\right)$.

Repeat distance:

For the cubic direction $\left[2\overline{1}\overline{1}\right]$, the repeat distance is the distance between the face-centered atom and the corner atom which is $\frac{\sqrt{3}}{2}a$. Therefore, the repeat distance is $\frac{\sqrt{3}}{2}a$.

(d)

To determine

Show the direction vector in a FCC unit cell for the cubic direction $\left[\overline{1}3\overline{1}\right]$. Write the position coordinates of the atoms with their centers intersected by each plane. Find the repeat distance in plane direction $\left[\overline{1}3\overline{1}\right]$ in terms of lattice constant.

Explanation of Solution

Show the direction vector in the FCC unit cell for the cubic direction $\left[\overline{1}3\overline{1}\right]$ as in Figure (4).

Position coordinates:

For the cubic direction $\left[\overline{1}3\overline{1}\right]$, the position coordinates of the atoms are $\left(0,0,0\right)$, and $\left(-1,3,-1\right)$.

Repeat distance:

For the cubic direction $\left[\overline{1}3\overline{1}\right]$, the repeat distance is the distance between the corner atom and the corner atom two lattice cells over which is $\sqrt{11}a$. Therefore, the repeat distance is $\sqrt{11}a$.

(e)

To determine

The angle between the direction vectors $\left[10\overline{1}\right]$ and $\left[\overline{1}3\overline{1}\right]$.

Answer to Problem 30AAP

The angle between the direction vectors $\left[10\overline{1}\right]$ and $\left[\overline{1}3\overline{1}\right]$ is $\text{90}°$.

Explanation of Solution

Write the expression to calculate angle between the direction vectors $\left(\theta \right)$.

  $\theta ={\cos }^{-1}\left[\frac{h_1{h}_2+k_1{k}_2+l_1{l}_2}{\sqrt{h_1^2+k_1^2+l_1^2}\sqrt{h_2^2+k_2^2+l_2^2}}\right]$                                                                    (I)

Here, Miller indices of the cubic plane 1 are $h_1$, $k_1$ and $l_1$ respectively and Miller indices of the cubic plane 2 are $h_2$, $k_2$ and $l_2$ respectively.

Conclusion:

Substitute 1 for $h_1$, 0 for $k_1$, $-1$ for $l_1$, $-1$ for $h_2$, 3 for $k_2$ and $-1$ for $l_2$ in Equation (I).

 $\begin{array}{c}\theta ={\cos }^{-1}\left[\frac{\left(1\right)\left(-1\right)+\left(0\right)\left(3\right)+\left(-1\right)\left(-1\right)}{\sqrt{1^2+0^2+{\left(-1\right)}^2}\sqrt{{\left(-1\right)}^2+3^2+{\left(-1\right)}^2}}\right]\\ ={\cos }^{-1}\left[\frac{0}{\left(\sqrt{2}\right)\left(\sqrt{11}\right)}\right]\\ ={\cos }^{-1}\left(0\right)\\ =\text{90}°\end{array}$

Thus, the angle between the direction vectors $\left[10\overline{1}\right]$ and $\left[\overline{1}3\overline{1}\right]$ is $\text{90}°$.