Consider a hypothetical metal that has the following lattice parameters: α = β = γ = 90° and a = b = 0.298 nm and c = 0.480 nm. Given that atoms are located at all corners of the unit cell, and that one atom is situated at the unit cell's center, determine the following:a). The crystal system to which the unit cell belongs:

b). The density of this material given that its atomic weight is 123 g/mol.

When α = β = γ = 90 and a = b = c, we have a cubic lattice, but if we make c larger than a and b (a = b) we have a tetragonal lattice.

You can see that this lattice is similar to the cubic lattice, the simplest one, which can be simple or primitive cubic or body centered. Similarly the tetragonal lattice can be primitive and body centered lattice.

When they tell us that there are atoms at all the corners and also one at the center of the unit cell we deduce that the unit cell is body centered lattice.

b). For the density calculation, we need to know the mass and the volume of the unit cell.

Number of atoms / unit cell:

8 atoms x 1/8 = 1

1 center

Therefore we have 2 atoms per unit cell.

The mass is:

= 123 g / mol x 1 mol / 6.022 x 10²³ x 2 atoms/unit cell

= 4.1 g x 10⁻²² g/unit cell

The volume is given by V = a x b x c

Lets convert the dimensions to cm since the density is commonly expressed in g/cm³:

0.298 nm x 1 cm / 10⁷ cm = 2.98 x 10⁻⁸ cm

Then,

V = 2.98 x 10⁻⁸ cm x 2.98 x 10⁻⁸ cm x 4.80 x 10⁻⁸ cm

V = 4.3 x 10⁻²³ cm³

Finally the density is

d = m/V = 4.1 x 10⁻²² g / 4.3 x 10⁻²³ cm³ = 9.5 g/cm³