Question

Chromium metal crystallises with a body-centred cubic lattice. The length of the unit cell edge is found to be 287 pm. Calculate the atomic radius. What would be the density of chromium in g cm⁻³?

Answer 1

Given: a = 287 pm

= 287 × 10⁻¹² m

= 287 × 10⁻¹⁰ cm

For a body-centred cubic unit cell

a = `4/sqrt 3 r`

Therefore, the atomic radius

r = `(a sqrt 3)/4`

= `(287 xx 10^-10 xx sqrt 3)/4`

= `(287 xx 10^-10 xx 1.73)/4`

= `(496.51 xx 10^-10)/4`

= 124.12 × 10⁻¹⁰

= 1.242 × 10⁻⁸ cm

For a body-centred cubic lattice, Z = 2. For chromium, M = 51.99. Therefore, the density of the metal,

`rho = (Z xx M)/(a^3 xx N_A)`

= `(2 xx 51.99)/((287 xx 10^-10)^3 xx 6.02 xx 10^23`

= `(103.98)/(23.6 xx 10^-24 xx 6.022 xx 10^23)`

= `(103.98)/(142.1 xx 10^-1)`

= `(103.98)/(14.21)`

= 7.31 g cm⁻³