Question

Crystalline CsBr has a bcc structure. Calculate the unit cell edge length if the density of CsBr crystal is 4.24 g cm⁻³. (Atomic masses: Cs = 133; Br =  80)

Answer 1

Given: Density ρ = 4.24 g/cm³

Crystal Structure: BCC → Z = 2 atoms per unit cell

Molar mass of CsBr: M = 133 (Cs) + 80 (Br) = 213 g/mol

Avogadro’s number: N_A ​= 6.022 × 10²³ mol⁻¹

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

∴ `a^3 = (Z xx M)/(rho xx N_A)`

= `(2 xx 213)/(4.24 xx 6.022 xx 10^23)`

= `426/(25.534 xx 10^23)`

= 16.68 × 10⁻²³ cm³

= 1.668 × 10⁻²² cm³

Now take the cube root:

a = `root 3(1.67 xx 10^-22) ≈ 4.37 xx 10^-8` cm

a = 4.37 × 10⁻⁸ × 10¹⁰ pm

a = 437 pm