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/cm³? (atomic mass of Cr = 52.99)

Answer 1

Given: a = 287 pm

= 287 × 10⁻¹² m

= 287 × 10⁻¹⁰ cm

= 2.87 × 10⁻⁸ cm

M = 25.99

Formula: For a body-centred cubic unit cell

a = `4/sqrt 3 r`

Therefore, the atomic radius

r = `(a sqrt 3)/4`

= `(2.87 xx 10^-8 xx sqrt 3)/4`

= `(2.87 xx 10^-8 xx 1.732)/4`

= `4.97/4 xx 10^-8`

= 1.24 × 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 52.99)/((2.87 xx 10^-8)^3 xx 6.022 xx 10^23`

= `105.98/(23.63 xx 10^-24 xx 6.022 xx 10^23`

= `105.98/(23.63 xx 6.022 xx 10^-1`

= `105.98/(2.363 xx 6.022)`

= `105.98/14.22`

= 7.45 g cm⁻³