Question

Calculate the radius of a vanadium atom, given that V has a BCCcrystal structure, a density of 5.96 g/cm³, and anatomic weight of 50.9 g/mol.

nm

Answer 1

Concepts and reason

Crystal structure:

It describes the arrangement of atoms or ions in a crystalline material. This structure is important to determine the properties of the material.

Metallic crystal structures:

Metals consist of metallic bonds. Atoms or ions are arranged in crystal structure with metallic bonds in metallic crystal structures.

Body centered cubic (BCC):

In this one atom is present in the center a unit cell. Another eight atoms are present on each corner of the unit cell.

Atom on the center of the unit cell is not shared by any other atom.

Each atom in the corner is shared by seven other unit cells which are in contact with this unit cell. So, atoms on corners are shared by eight unit cells.

Atomic packing factor (APF):

It is the ratio of sum of the volume of each atom present in the unit cell to the total volume of the unit cell.

It gives the number of atoms present in one mole of matter.

Fundamentals

The formula for calculation of Atomic packing factor (APF) is as follows:

Mass of the unit cell is given by ratio of product of number of atoms and atomic weight with Avogadro number.

Avogadro number,

Density of the unit cell is as follows:

Here, Avogadro’s number is N, number of atoms in unit cell is n, and atomic mass of the vanadium is , and volume of the unit cell is V.

Conversions used are as follows:

Calculate the total number of unit cells in BCC structure.

Equate the length of the cube diagonal in terms of unit cell length, S and radius of atom, R.

Consider the volume of unit cell.

Consider the density of vanadium atom.

Here, Avogadro’s number is N, number of atoms in unit cell is n, and atomic mass of the vanadium is , and volume of the unit cell is V.

Substitute for V, for N, 2 for n, for , and for .

Ans:

The radius of the vanadium atom is .