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PHY3707 Assignment 1 2026 (127027) Due 17 April 2026

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UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology



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PHY3707 ASSIGNMENT 1
Elements of Solid State Physics — Semester 1, 2026


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Module Code: PHY3707

Module Name: Elements of Solid State Physics

Student Name: [Student Name]

Student Number: [Student Number]

Assignment No.: 1

Due Date: 17 April 2026

Semester: Semester 1, 2026

Unique Number: 127027




Submitted in partial fulfilment of the requirements for PHY3707
at the University of South Africa.

,UNISA | PHY3707 Assignment 1 — Solid State Physics



Contents


1 Problem 1.6: Tetrahedral Bond Angle in Diamond 2

1.1 1.1 Setting Up the Body Diagonal Vectors . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 1.2 Applying the Dot Product Formula . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.3 1.3 Verification with a Different Pair of Diagonals . . . . . . . . . . . . . . . . . . 3


2 Problem 1.7(a): Miller Indices Referred to Primitive FCC Axes 5

2.1 1.1 Primitive Basis Vectors of the FCC Lattice . . . . . . . . . . . . . . . . . . . . 5

2.2 1.2 Relating Conventional and Primitive Axes . . . . . . . . . . . . . . . . . . . . 5

2.3 1.3 Miller Index Transformation Rule . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.4 1.4 Index of the (100) Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.5 1.5 Index of the (001) Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6


3 Problem 1.7(b): Direction [hkl] is Perpendicular to Plane (hkl) in Cubic Symmetry 8

3.1 1.1 Setting Up the Primitive Basis Vectors . . . . . . . . . . . . . . . . . . . . . . 8

3.2 1.2 The Direction Vector [hkl] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3.3 1.3 Vectors Lying in the Plane (hkl) . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3.4 1.4 Proving t ⊥ u . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.5 1.5 Proving t ⊥ v . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.6 1.6 Conclusion of the Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10


Reference List 11




Page 1 of 11

, UNISA | PHY3707 Assignment 1 — Solid State Physics



Problem 1.6: Tetrahedral Bond Angle in Diamond

The diamond crystal structure is built on an FCC lattice with a two-atom basis, and its tetra-
hedral bonding geometry is one of the most studied configurations in solid state physics.
Because the bond directions in diamond are identical to the body diagonal directions of a
cube, the angle between any two such bonds is directly calculable using elementary vector
analysis (Srivastava, 2014:12).


1.1 Setting Up the Body Diagonal Vectors


Place a cube with one corner at the origin and side length a. The cube has four body diag-
onals. Each one connects a corner of the cube to the diagonally opposite corner. The four
distinct body diagonals, expressed as vectors from the origin, are:




d1 = a(+x̂ + ŷ + ẑ) (1)

d2 = a(−x̂ − ŷ + ẑ) (2)

d3 = a(−x̂ + ŷ − ẑ) (3)

d4 = a(+x̂ − ŷ − ẑ) (4)



Each of these four vectors represents one of the four tetrahedral bond directions in diamond.
To be precise about the geometry: in the actual diamond structure, the central atom sits at
the cube corner and bonds to four neighbours, each displaced by one of these diagonal direc-
tions. The angle between any two body diagonals is therefore the tetrahedral bond angle.


1.2 Applying the Dot Product Formula


Take two of these diagonal vectors, say d1 and d2 . The angle θ between them satisfies:


d1 · d 2
cos θ = (5)
|d1 | |d2 |

Step 1: Compute the dot product.

The component form of d1 is (+1, +1, +1) and that of d2 is (−1, −1, +1), where the common



Page 2 of 11

Connected book
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Philip Hofmann Solid State Physics
Publisher: 2015 ISBN: 9783527412822 Edition: Unknown

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