Question

a. Calculate packing efficiency simple cubic lattice.

b. An element having atomic mass $107.9$u has FCC lattice. The edge length of its unit cell is $408.6$ pm. Calculate the density of the unit cell.

[Given, $N_A=6.022\times {10}^{23}$ $mo{l}^{-1}$].

Answer 1

(a) In a simple cubic unit cell, atoms are located only at corners of the cube. The particles touch one another along the edge.

Then, the edge length 'a' is related to the radius 'r' by the expression a$=2$r.

Volume of cubic unit cell $=a^3=(2r{)}^3=8r^3$

A simple cubic unit cell has 1 atom per unit cell.

Volume occupied by each sphere is $\frac{4}{3}\pi r^3$

Packing efficiency $=\frac{\text{Volume occupied by each atom}\times \text{total number of atoms per unit cell}}{\text{Volume of cubic unit cell}}\times 100$

$=\frac{\frac{4}{3}\pi r^3\times 1}{8r^3}\times 100=52.36\%$

b. A fcc unit cell has 4 atoms per unit cell. $\therefore Z=4$

The density of the unit cell,

$d=\frac{ZM}{a^3{N}_A}=\frac{4\times 107.9\times {10}^{-3}}{(408.6\times {10}^{-12}{)}^3\times 6.022\times {10}^{23}}$

$d=10.5\times {10}^3$ kg$/m^3$