UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
PHY3707: Solid State Physics
Assignment 1 — Semester 1, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
PHY3707
Module Code:
Solid State Physics
Module Name:
Assignment 1
Assignment Number:
127027
Unique Number:
17 April 2026
Due Date:
30
Total Marks:
Submitted in partial fulfilment of the requirements for PHY3707 — UNISA 2026
,UNISA | PHY3707 Assignment 1 – Solid State Physics
Contents
1 Problem 1.6: Tetrahedral Bond Angle in Diamond via Vector Analysis 3
1.1 1.1 Setting Up the Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 1.2 Applying the Dot-Product Formula . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 1.3 Geometric Interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2 Problem 1.7(a): Indices of (100) and (001) Planes in FCC Referred to Primitive Axes 6
2.1 2.1 Primitive Basis Vectors of the FCC Lattice . . . . . . . . . . . . . . . . . . . . 6
2.2 2.2 Expressing Conventional Axes in Terms of Primitive Axes . . . . . . . . . . . 6
2.3 2.3 Re-indexing the (100) Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2.4 2.4 Re-indexing the (001) Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
3 Problem 1.7(b): Proof that [hkl] is Perpendicular to (hkl) in a Cubic Crystal 8
3.1 3.1 Setup and Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
3.2 3.2 Defining the Direction [hkl] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
3.3 3.3 Characterising the Plane (hkl) . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
3.4 3.4 Proving Perpendicularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3.4.1 Dot product with v1 : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3.4.2 Dot product with v2 : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3.5 3.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
Reference List 11
Page 2 of 11
, UNISA | PHY3707 Assignment 1 – Solid State Physics
Problem 1.6: Tetrahedral Bond Angle in Diamond via Vector Analysis
The diamond crystal structure places each carbon atom at the centre of a regular tetrahedron
formed by its four nearest neighbours. Because the tetrahedral bonds of diamond are geo-
metrically identical to the body diagonals of a cube, finding the angle between any two body
diagonals of a cube gives the tetrahedral bond angle directly (Srivastava, 2014:15).
1.1 Setting Up the Vectors
Place a cube with one corner at the origin and side length a = 1. The eight corners lie at all
combinations of (0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 0), (1, 0, 1), (0, 1, 1), and (1, 1, 1).
A cube has four body diagonals. Taking the diagonal that runs from the origin (0, 0, 0) to the
opposite corner (1, 1, 1), and a second diagonal from (1, 0, 0) to (0, 1, 1), the corresponding
direction vectors are:
d1 = (1, 1, 1) − (0, 0, 0) = (1, 1, 1) (1)
d2 = (0, 1, 1) − (1, 0, 0) = (−1, 1, 1) (2)
1 1 1
These two body diagonals share the interior point 2, 2, 2 , which is exactly where a central
atom of the diamond basis sits.
1.2 Applying the Dot-Product Formula
The angle θ between two vectors follows from the standard dot-product definition:
d1 · d 2
cos θ = (3)
|d1 | |d2 |
Computing each piece:
Page 3 of 11
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
PHY3707: Solid State Physics
Assignment 1 — Semester 1, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
PHY3707
Module Code:
Solid State Physics
Module Name:
Assignment 1
Assignment Number:
127027
Unique Number:
17 April 2026
Due Date:
30
Total Marks:
Submitted in partial fulfilment of the requirements for PHY3707 — UNISA 2026
,UNISA | PHY3707 Assignment 1 – Solid State Physics
Contents
1 Problem 1.6: Tetrahedral Bond Angle in Diamond via Vector Analysis 3
1.1 1.1 Setting Up the Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 1.2 Applying the Dot-Product Formula . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 1.3 Geometric Interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2 Problem 1.7(a): Indices of (100) and (001) Planes in FCC Referred to Primitive Axes 6
2.1 2.1 Primitive Basis Vectors of the FCC Lattice . . . . . . . . . . . . . . . . . . . . 6
2.2 2.2 Expressing Conventional Axes in Terms of Primitive Axes . . . . . . . . . . . 6
2.3 2.3 Re-indexing the (100) Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2.4 2.4 Re-indexing the (001) Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
3 Problem 1.7(b): Proof that [hkl] is Perpendicular to (hkl) in a Cubic Crystal 8
3.1 3.1 Setup and Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
3.2 3.2 Defining the Direction [hkl] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
3.3 3.3 Characterising the Plane (hkl) . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
3.4 3.4 Proving Perpendicularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3.4.1 Dot product with v1 : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3.4.2 Dot product with v2 : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3.5 3.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
Reference List 11
Page 2 of 11
, UNISA | PHY3707 Assignment 1 – Solid State Physics
Problem 1.6: Tetrahedral Bond Angle in Diamond via Vector Analysis
The diamond crystal structure places each carbon atom at the centre of a regular tetrahedron
formed by its four nearest neighbours. Because the tetrahedral bonds of diamond are geo-
metrically identical to the body diagonals of a cube, finding the angle between any two body
diagonals of a cube gives the tetrahedral bond angle directly (Srivastava, 2014:15).
1.1 Setting Up the Vectors
Place a cube with one corner at the origin and side length a = 1. The eight corners lie at all
combinations of (0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 0), (1, 0, 1), (0, 1, 1), and (1, 1, 1).
A cube has four body diagonals. Taking the diagonal that runs from the origin (0, 0, 0) to the
opposite corner (1, 1, 1), and a second diagonal from (1, 0, 0) to (0, 1, 1), the corresponding
direction vectors are:
d1 = (1, 1, 1) − (0, 0, 0) = (1, 1, 1) (1)
d2 = (0, 1, 1) − (1, 0, 0) = (−1, 1, 1) (2)
1 1 1
These two body diagonals share the interior point 2, 2, 2 , which is exactly where a central
atom of the diamond basis sits.
1.2 Applying the Dot-Product Formula
The angle θ between two vectors follows from the standard dot-product definition:
d1 · d 2
cos θ = (3)
|d1 | |d2 |
Computing each piece:
Page 3 of 11
