#### Question

Chromium metal crystallises with a body centred cubic lattice. The edge length of the unit cell is found to be 287 pm. Calculate the atomic radius: What would be the density of chromium in g/cm^{3}? (atomic mass of Cr = 52 ·99)

#### Solution

Given :

Crystal structure of Cr = BCC

Edge length (a) = 287 pm

To find

Atomic radius = r

Density = ρ

Solution:

For BCC crystal system

Edge length (a) = `(4r)/sqrt3`

`:. r = (sqrt3a)/4`

= 0.433 x 287 pm

r = 124.27 pm

Now density of a crystal :

`rho = (zM)/(a^3N_A)`

Where

z = number of atom in BCC unit cell 2

M = Molar mass ofCr = 52.99 g

`a^3 = (287 "pm")^3`

`= (287 xx 10^(-12) m)^3`

`=(287 xx 10^(-10)cm)^3`

`= 2.37 xx 10^(-23) cm^3`

`N_A = 6.022 xx 10^23 atoms `

Density = `(zM)/(a^3N_A)`

`= (2 xx 52.99)/(2.37 xx 10^(-23) xx 6.022 xx 10^23)`

` = 105.98/14.27`

Density of crystal = 7.42 `"g/cm"^3`