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 which contains a single lattice point. Because it only contains a lattice point, it is the unit cell with the smallest volume out of all unit cells.
2. Let us first consider the FCC lattice. Its primitive lattice vectors 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. The primitive lattice vectors are \begin{align} \mathbf{a}_1 &= \frac{a}{2}(\mathbf{\hat{x}} + \mathbf{\hat{y}} + \mathbf{\hat{z}})\ \mathbf{a}_2 &= a\mathbf{\hat{x}}\ \mathbf{a}_3 &= a\mathbf{\hat{y}}. \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 \(\bigcirc(0,0,0)\).
3. You would need at least two. A diatomic crystal could require more than two basis vectors if the proportion between the two elements is not one to one.
4. $$ F = \frac{\pi}{6} $$
5. See lecture notes
Exercise 1: Diatomic crystal¶¶
1.
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);
2. The area of the primitive unit cell is \(A = a^2\). If the filled and empty circles are identical particle, 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. $$ F = \frac{\sqrt(3)\pi}{8} $$
Exercise 2: Diamond lattice¶
1. The conevtional unit cell of diamond consists out of two shifted fcc lattices. One 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 $$ V = \left| \mathbf{a_1} \cdot \left(\mathbf{a_2} \times \mathbf{a_3} \right) \right| = \frac{a^3}{4} $$
2. The primitive unit cell contains 2 atoms. With respect to the set of primitive lattcie vectors, the basis is $ \mathrm{C}(0,0,0)$ and \(\mathrm{C}(\frac{1}{4},\frac{1}{4},\frac{1}{4})\).
3. One FCC lattice contains 4 atom. Because the diamond conventional unit cell contains two shifted FCC lattices, it will contain 8 atoms. The volume of the conventional unit cell is \(V = a^3\).
4.
Visual hint
Consider the atom at (0.25,0.25,0.25) coordinates in the interactive diamond lattice image and look at its neighbours.
The distance between neirest neighbouring atoms is \(d = \frac{\sqrt{3}a}{4}\)
5. $$ F = \frac{\sqrt{3}\pi}{16} $$
Exercise 3: Directions and Spacings of Miller planes¶
1. Miller plane is a plane that intersects an infinite number of lattice points. Miller indeces are a set of 3 integers which specify a set of parallel planes.
2.
Small hint
The \((hkl)\) plane intersects lattice at position vectors of \(\frac{\mathbf{a_1}}{h}, \frac{\mathbf{a_2}}{k}, \frac{\mathbf{a_1}}{l}\). Can you define two general vector inside the \((hkl)\) plane that span the plane?
3. The same hint as mentioned in the previous subquestion also applies here.
Subquestion 4¶
Big hint
There is always a neighbouring lattice plane which intersects the (0,0,0) lattice point.
Small hint
Do not forget to normalize your unit vector.