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    from matplotlib import pyplot as plt 
import numpy as np
from math import pi

Solutions for lecture 9 exercises

Warm-up exercises

  1. A primitive unit cell is a unit cell that contains a single lattice point. Because it contains only one lattice point, it has the smallest volume of all unit cells.

  2. Since the primitive unit cell should contain one lattice point, its volume is \(a^3/4\).

  3. Let us first consider the FCC lattice. Its primitive lattice vectors are

    \[\begin{align*} \mathbf{a}_1 &= \frac{a}{2}(\mathbf{\hat{x}} + \mathbf{\hat{y}})\\ \mathbf{a}_2 &= \frac{a}{2}(\mathbf{\hat{x}} + \mathbf{\hat{z}})\\ \mathbf{a}_3 &= \frac{a}{2}(\mathbf{\hat{y}} + \mathbf{\hat{z}}). \end{align*}\]

    With respect to the conventional unit cell, the basis in fractional coordinates is \(\bigcirc(1/2,1/2,0)\), \(\bigcirc(1/2,0,1/2)\), \(\bigcirc(0,1/2,1/2)\) and \(\bigcirc(0,0,0)\). With respect to the primitive unit cell, the basis is \(\bigcirc(0,0,0)\). Let us now consider the BCC lattice. A possible set of primitive lattice vectors is

    \[\begin{align*} \mathbf{a}_1 &= \frac{a}{2}(\mathbf{\hat{x}} + \mathbf{\hat{y}} - \mathbf{\hat{z}})\\ \mathbf{a}_2 &= \frac{a}{2}(\mathbf{\hat{x}} - \mathbf{\hat{y}} + \mathbf{\hat{z}})\\ \mathbf{a}_3 &= \frac{a}{2}(-\mathbf{\hat{x}} + \mathbf{\hat{y}} + \mathbf{\hat{z}}). \end{align*}\]

    The basis of the conventional unit cell is \(\bigcirc(0,0,0)\) and \(\bigcirc(1/2,1/2,1/2)\). For the primitive unit cell the basis is \(\bigcirc(0,0,0)\).

  4. You would need at least two. A diatomic crystal could require more than two basis vectors if the atoms are distinguished by their spatial positions. One needs to be able to construct all atoms in the crystal using a basis plus a lattice.

  5. The filling factor is

    \[ F = \frac{\pi}{6} \]

  6. See lecture notes

Exercise 1: Diatomic crystal

y = np.repeat(np.arange(0,8,2),4)
x = np.tile(np.arange(0,8,2),4)
plt.figure(figsize=(5,5))
plt.axis('off')

# WZ
plt.plot([5,5,7,7,5],[5,7,7,5,5], color='k',ls=':')
plt.annotate('WZ',(6,6.5),fontsize=14,ha='center')

# PUC1 
plt.plot([0,2,4,2,0],[4,6,6,4,4], color='k',ls=':')

# UPC2
plt.plot([6,4,2,4,6],[0,0,2,2,0], color='k',ls=':')

plt.plot(x,y,'ko', markersize=15)
plt.plot(x+1,y+1, 'o', markerfacecolor='none', markeredgecolor='k', markersize=15);

png

  1. See plot above

  2. The area of the primitive unit cell is \(A = a^2\). If the filled and empty circles are identical particles, the nearest-neighbour distance becomes \(a^* = \frac{a}{\sqrt{2}}\) and thus the area \(A^* = {a^*}^2 = \frac{a^2}{2} = \frac{A}{2}\).

  3. One set of primitive lattice vectors is

    \[ \mathbf{a_1} = a \hat{\mathbf{x}}, \quad \mathbf{a_2} = a \hat{\mathbf{y}}. \]

    With respect to the primitive lattice vectors, the basis is

    \[ \huge \bullet \normalsize(0,0), \quad \bigcirc(\frac{1}{2},\frac{1}{2}). \]

  4. The lattice is a cubic lattice. The basis of the crystal is

    \[ \huge \bullet \normalsize(0,0,0), \quad \bigcirc(\frac{1}{2},\frac{1}{2},\frac{1}{2}). \]

    An example of such a material is cesium chloride (CsCl).

  5. We obtain the BCC lattice. Example: Sodium (Na)

  6. The filling factor is

    \[ F = \frac{\sqrt{3}\pi}{8} \]

Exercise 2: Diamond lattice

  1. The conventional unit cell of diamond consists of two intertwined fcc lattices. A possible set of primitive lattice vectors is

    \[\begin{align*} \mathbf{a_1} &= \frac{a}{2} \left(\hat{\mathbf{x}}+\hat{\mathbf{y}} \right) \\ \mathbf{a_2} &= \frac{a}{2} \left(\hat{\mathbf{x}}+\hat{\mathbf{z}} \right) \\ \mathbf{a_3} &= \frac{a}{2} \left(\hat{\mathbf{y}}+\hat{\mathbf{z}} \right). \end{align*}\]

    The volume of the primitive unit cell is

    \[ V = \left| \mathbf{a_1} \cdot \left(\mathbf{a_2} \times \mathbf{a_3} \right) \right| = \frac{a^3}{4} \]

    as expected because the conventional unit cell contains 4 lattice points.

  2. The primitive unit cell contains 2 atoms. With respect to the set of primitive lattice vectors, the basis is $ \mathrm{C}(0,0,0)$ and \(\mathrm{C}(\frac{1}{4},\frac{1}{4},\frac{1}{4})\).

  3. The conventional unit cell of the FCC lattice contains 4 atoms. Because the diamond conventional unit cell contains two shifted FCC lattices, it contains 8 atoms. The volume of the conventional unit cell is \(V = a^3\).

  4. We identify a nearest-neighbour pair consisting of the atom at the origin and the atom at \((0.25a, 0.25a, 0.25a)\). Therefore, the distance between nearest-neighbouring atoms is \(d = \frac{\sqrt{3}a}{4}\).

  5. The atom density is \(\rho=N_\text{atom}/a^3 = 175/\) nm\(^3\). The filling factor is

    \[ F = \frac{\sqrt{3}\pi}{16} \]

Exercise 3: Directions and spacings of Miller planes

  1. A Miller plane is one member of a family of parallel lattice planes, specified by a set of three integers called Miller indices.

  2. The \((hkl)\) plane intersects the lattice vectors at position vectors of \(\frac{\mathbf{a_1}}{h}\), \(\frac{\mathbf{a_2}}{k}\), and \(\frac{\mathbf{a_3}}{l}\). To span the plane, choose any two non-parallel vectors lying in it; their cross product gives a normal vector.

  3. The procedure described in the previous subquestion also applies here.

  4. We can calculate the distance between two planes by projecting any vector connecting them onto a unit vector that is normal to the planes.