Question

Which of the three-dimensional primitive lattices has a unit cell where none of the internal angles is \(90^{\circ}\) ? (a) Orthorhombic, (b) hexagonal, (c) rhombohedral, (d) triclinic, (e) both rhombohedral and triclinic.

Short answer

The correct answer is (e) both rhombohedral and triclinic lattices, as they both have unit cells where none of the internal angles is equal to \(90^{\circ}\).

Step-by-step solution

Understanding 3D primitive lattices

Here's a brief overview of the given lattice types and their unit cell properties: (a) Orthorhombic: Lot has three cell edges (a, b, and c) with different lengths and all internal angles equal to \(90^{\circ}\). (b) Hexagonal: Lattice has three cell edges of equal length (a=a=b), and only two internal angles equal to \(90^{\circ}\) while the other is equal to \(120^{\circ}\). (c) Rhombohedral: Lattice has three cell edges of equal length (a=a=b) and all internal angles are equal but not equal to \(90^{\circ}\). (d) Triclinic: Lattice has three cell edges (a, b, and c) with different lengths, and no internal angles are equal to \(90^{\circ}\). Now, let's analyze which of these options meet the requirement of having no internal angle equal to \(90^{\circ}\).

Comparing lattice properties with given requirements

Looking at the properties mentioned earlier, we can rule out option (a) - Orthorhombic and option (b) - Hexagonal because they have at least one internal angle equal to \(90^{\circ}\). Option (c) - Rhombohedral lattice has all internal angles equal but not equal to \(90^{\circ}\), which satisfies the requirements of this problem. Option (d) - Triclinic lattice has no internal angles equal to \(90^{\circ}\), which also meets the requirements of this problem. Therefore, both rhombohedral and triclinic lattices fulfill the given condition.

Choosing the correct answer

Since both option (c) - Rhombohedral and option (d) - Triclinic lattices fulfill the given condition, the correct answer is (e) both rhombohedral and triclinic.