Question

An element of atomic mass $98.5gmo{l}^{-1}$ occurs in FCC structure. If its density is $5.22gc{m}^{-3}$, determine the edge length of unit cell.

Answer 1

For FCC, density is given by

$d=\frac{Z\times M}{a^3\times N}$ where $z$ is $4$, i.e. Number of atoms per unit cell.

$a^3=\frac{4\times 98.5}{5.22}$ given: $d=5.22gc{m}^{-3}$and atomic mass is $98.5gmo{l}^{-1}$

$a^3=75.48\therefore a=25.16cm$