Question

The interplanar spacing of (220) planes of a FCC structure is 1.7458 Å. Calculate the lattice constant.

(a) 4.983 Å

(b) 2.458 Å

(c) 0

(d) 5.125 Å

Answer 1

Correct option is (a) 4.983 Å

Explanation: D = 1.7458 Å = 1.7458 × 10^-10 m

h = 2; k = 2; l = 0

D=a/√(h^2+k^2+l^2)

a = 4.983 Å.