Question

Niobium crystallises in body-centred cubic structure. If the atomic radius is 143.1 pm, calculate the density of Niobium. (Atomic mass = 93u).

Answer 1

A/c to question,

       Niobium crystallise in body - cantered cubic structure.

so, radius of Niobium  atom = √3/4 × edge length of unit cell

given, atomic radius of Niobium = 143.1pm

so, edge length of  unit cell  = 4 × 143.1/√3 = 330.4 pm

Now use formula, ρ = zM/a³N

where ρ is density of Niobium,

    z is number of atoms per unit cell

    a is edge length of unit cell

N is Avogadro number

Here, z = 2 , M = 93u, a = 330.4 × 10⁻¹² m , and N = 6.023 × 10²³

So, ρ = (2 × 93 )/{330.4 × 10⁻¹²)³× 6.023 × 10²³} = 8.56g/cm³

Answer 2

According to question,

Niobium crystallise in body - cantered cubic structure.

For BCC crystalline structure,

edge length of unit cell=4/√3 *radius of Niobium  atom

According to question,

atomic radius of Niobium = 143.1pm

Hence, edge length of  unit cell  = 4 * 143.1/√3 = 330.4 pm

Now using the density formula, ρ = zM/a³N

where ρ =density of Niobium,

          z = number of atoms per unit cell

          a =edge length of unit cell  

          N =Avogadro number

According to question, z = 2 , M = 93u, a = 330.4 × 10⁻¹² m , and N = 6.023 × 10²³

So, ρ = (2 * 93 )/{330.4 * 10⁻¹²)³* (6.023 *10²³} = 8.56g/cm³

The density of Niobium is 8.56g/cm³.