Calculation of the Volume of the Unit Cell of Magnesium

Given:
- Structure: Hexagonal Close-Packed (hcp)
- Radius of Magnesium atom: 0.1605 nanometer (nm)

Step 1: Determining the Lattice Parameters
In an hcp structure, the lattice is composed of two stacked triangular lattices. To calculate the volume of the unit cell, we need to determine the lattice parameters.

1.1 Lattice Constant (a)
The lattice constant (a) can be calculated using the radius of the atom. In an hcp structure, the relation between the lattice constant and the atomic radius is given by the following equation:

a = 2r√(2/3)

Substituting the given radius of the Magnesium atom (0.1605 nm) into the equation, we can calculate the lattice constant (a).

a = 2 * 0.1605 nm * √(2/3)
a ≈ 0.3245 nm

1.2 Height of the Unit Cell (c)
In an hcp structure, the height of the unit cell (c) can be calculated using the lattice constant (a) and the relation:

c = 2a√(2/3)

Substituting the calculated lattice constant (0.3245 nm) into the equation, we can determine the height of the unit cell (c).

c = 2 * 0.3245 nm * √(2/3)
c ≈ 0.5292 nm

Step 2: Calculating the Volume of the Unit Cell
The unit cell in an hcp structure can be visualized as a hexagonal prism. To calculate its volume, we need to determine the area of the base (A) and multiply it by the height (c).

2.1 Area of the Base (A)
The area of the base of the hexagonal prism can be calculated using the formula for the area of a regular hexagon:

A = (3√3/2) * (a^2)

Substituting the calculated lattice constant (0.3245 nm) into the equation, we can determine the area of the base (A).

A = (3√3/2) * (0.3245 nm)^2
A ≈ 0.2807 nm^2

2.2 Volume of the Unit Cell
The volume of the unit cell can be calculated by multiplying the area of the base (A) by the height (c).

Volume = A * c
Volume ≈ 0.2807 nm^2 * 0.5292 nm
Volume ≈ 0.1485 nm^3

Therefore, the volume of the unit cell of Magnesium in the hcp structure is approximately 0.1485 nm^3.