Question

Of the seven three-dimensional primitive lattices, which ones have a unit cell where all three lattice vectors are of the same length?

 

 

Answer 1

Given

Three dimensional lattices, in three dimensions a lattice is defined by three vectors a, b, and c. These lattice vectors a unit cell that is parallelepiped (a six-sided figure whose faces are all parallelograms) and is described by the lengths a, b, c of the cell edges and the angles α, β, γ between these edges. There are seven possible shapes for a three dimensional unit cell.

 

If we place a lattice point at each corner of a unit cell, we get primitive lattice. All seven lattices are primitive lattices.

 

 

From the given data, the unit cell has two lattice vectors a, and b which are form a rectangular base and lattice vector c is perpendicular to lattice vectors a, and b with different length. According to this data, the given unit cell is Orthorhombic. In orthorhombic, the three lengths are unequal and the angles between these edges are equal. So, the orthorhombic unit cell is

a ≠ b ≠ c and α = β = γ = 90°

 

 

 

From the given data, the unit cell has two lattice vectors a, and b which are form a square base and lattice vector c is perpendicular to lattice vectors a, and b with different length. According to this data, the given unit cell is tetragonal. In tetragonal, for three lengths two are equal and third one is different and the angles between these edges are equal. So, the tetragonal unit cell is

a = b ≠ c and α = β = γ = 90°

Given

Three dimensional lattices, in three dimensions a lattice is defined by three vectors a, b, and c. These lattice vectors a unit cell that is parallelepiped (a six-sided figure whose faces are all parallelograms) and is described by the lengths a, b, c of the cell edges and the angles α, β, γ between these edges. There are seven possible shapes for a three dimensional unit cell.