Question

Niobium has a BCC crystal structure, an atomic radius of 0.143 nm and an atomic weight of 92.91 g/mol. Calculate the theoretical density for Nb.

Answer 1

The theoretical density for Niobium is
`1.87 g/cm^3`.

Step-by-step explanation:

Formula used :

`\rho=(Z* M)/(N_(A)* a^(3))`

where,

`\rho` = density of the unit cell

Z = number of atom in unit cell

M = atomic mass

`(N_(A))` = Avogadro's number

a = edge length of unit cell

We have :

Z = 2 (BCC)

M = 92.91 g/mol ( Niobium)

Atomic radius for niobium = r = 0.143 nm

Edge length of the unit cell = a

r = 0.866 a (BCC unit cell)

`a=(0.143 nm)/(0.866)=0.165 nm=0.165 * 10^(-7) cm`

`1 nm = 10^(-7) cm`

On substituting all the given values , we will get the value of 'a'.

`\rho=(2* 92.91)/(6.022* 10^(23) mol^(-1)* (0.165 * 10^(-7) cm)^(3))`

`\rho =1.87 g/cm^3`

The theoretical density for Niobium is
`1.87 g/cm^3`.