Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm–3, calculate atomic radius of niobium using its atomic mass 93 u.

To find the atomic radius of niobium, we can use its density and atomic mass along with the properties of the body-centered cubic (BCC) structure. The BCC structure has a specific relationship between the atomic radius, the volume of the unit cell, and the density of the material. Let's break this down step by step.

Understanding the Body-Centered Cubic Structure

In a BCC structure, there are two atoms per unit cell. The unit cell is a cube where one atom is located at each corner and one atom is at the center. The relationship between the atomic radius (r) and the edge length (a) of the cube in a BCC structure can be expressed as:

• Diagonal of the cube = 4r
• Using Pythagoras' theorem, the diagonal can also be expressed as: √3 * a

From these relationships, we can derive:

Calculating the Edge Length

Setting the two expressions for the diagonal equal gives us:

4r = √3 * a

From this, we can solve for the edge length:

a = (4r) / √3

Relating Density to Atomic Mass and Volume

The density (ρ) of a material is defined as its mass (m) per unit volume (V). For a BCC crystal structure, the volume of the unit cell can be expressed as:

V = a³

Since there are two atoms in a BCC unit cell, the mass of the unit cell can be calculated using the atomic mass (M) of niobium:

m = 2 * (M / N_A)

where N_A is Avogadro's number (approximately 6.022 x 10²³ mol^-1).

Putting It All Together

Now, we can express density in terms of atomic mass and volume:

ρ = m / V = (2 * (M / N_A)) / a³

Rearranging this gives us:

a³ = (2 * (M / N_A)) / ρ

Substituting the values we have:

• M = 93 g/mol (atomic mass of niobium)
• ρ = 8.55 g/cm³ (density of niobium)
• N_A = 6.022 x 10²³ mol^-1

Now, we can calculate a³:

a³ = (2 * (93 g/mol) / (6.022 x 10²³ mol^-1)) / (8.55 g/cm³)

Calculating this gives:

a³ ≈ 1.29 x 10^-22 cm³

Taking the cube root to find a:

a ≈ 2.34 x 10^-8 cm

Finding the Atomic Radius

Now that we have the edge length, we can find the atomic radius using the earlier relationship:

r = (√3 * a) / 4

Substituting the value of a:

r ≈ (√3 * (2.34 x 10^-8 cm)) / 4

Calculating this gives:

r ≈ 1.01 x 10^-8 cm or 1.01 Å (angstroms)

Final Thoughts

Thus, the atomic radius of niobium is approximately 1.01 Å. This calculation illustrates how we can derive atomic dimensions from macroscopic properties like density and atomic mass, showcasing the fascinating connection between the microscopic and macroscopic worlds in materials science.