Imagine the primitive cubic lattice. Now imagine grabbing opposite corners and stretching it along the body diagonal while keeping the edge lengths equal. The three angles between the lattice vectors remain equal but are no longer \(90^{\circ}\) . What kind of primitive lattice have you made?

Short Answer

Expert verified
After stretching the primitive cubic lattice along the body diagonal while maintaining equal edge lengths and equal angles, the resulting lattice is a rhombohedral (or trigonal) lattice which belongs to the hexagonal crystal system. This lattice has equal edge lengths (a = b = c) and equal angles between the lattice vectors, but the angles are no longer \(90^{\circ}\).

Step by step solution

01

Identify key features of the initial primitive cubic lattice

A primitive cubic lattice has three lattice vectors (a, b, and c) that are mutually perpendicular, meaning all the angles between them are \(90^{\circ}\). The edge lengths of the unit cell are all equal: a = b = c.
02

Visualize the transformation

Now, we need to stretch the lattice along the body diagonal while maintaining equal edge lengths. This means that we will still have a = b = c, but the angles between the lattice vectors will no longer remain \(90^{\circ}\).
03

Identify the type of lattice after the transformation

After the transformation, we still have equal edge lengths a = b = c, and the angles between the lattice vectors are no longer \(90^{\circ}\) but remain equal. This type of lattice structure is known as a "rhombohedral" (or "trigonal") lattice which belongs to the hexagonal crystal system. So, the resulting primitive lattice after the transformation is a rhombohedral lattice.

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