Question

3. The electrical conductivity of a compound AX is controlled by the mobility of A+ ions (4.8 x 10-14m2/volt.sec). Suppose that AX has the NaCl structure with radii of A+ = 0.88 Å and X- = 1.61Å. Calculate (i) the volume of AX unit cell (ii) the electrical conductivity of AX.

Answer 1

(i) radius of A+ = 0.88

Radius of X- = 1.61

NaCl is fcc lattice, where Cl- lies at edges and Na+ lies in the centre. So, total edge length is sum of diameter of Na+ and Cl-.

edge length of compound AX = 2(0.88)+2(1.61) = 4.98

Volume of fcc lattice = (edge length)³ = (4.98)³ = 123.51 ³

Volume of AX = 123.51 ³

(ii) Electrical conductivity of AX = = nqu

n = carrier density

q = charge = 1.6*10^-19 C

u = mobility = 4.8*10^-14 m²/V.s

If volume of 1 unit cell = 123.51 ³ = 123.51*10^-30 m³, then number of unit cells in 1 m³ = 1/(123.51*10^-30) = 8.1*10²⁷ carriers

carrier density = 8.1*10²⁷ carriers/m³

So, conductivity = nqu = 8.1*10²⁷*1.6*10^-19*4.8*10^-14 = 62.21*10^-6 (ohm.m)^-1