Question

A metal packs with 68% efficiency. If the edge length of the unit cell is 0.291nm, find the colume of all the atoms in this unit cell in cm^3.

Answer 1

Solution.

$APF = \frac{n \times V(atom) \times 100 \%}{V(unit \ cell)}$

APF = 68 %, so it is the body-centred unit cell.

$V(atoms \ in \ unit \ cell) = \frac{APF \times V(unit \ cell)}{100 \%}$

$V(unit \ cell) = a^3$

$V(unit \ cell) = (0.291 \times 100 \times 10^{-9})^3 = 2.46 \times 10^{-23} \ cm^3$

$V(atoms \ in \ unit \ cell) = \frac{68 \times 2.46 \times 10^{-23}}{100 \%} = 1.68 \times 10^{-23} \ cm^3$

Answer:

$V(atoms \ in \ unit \ cell) = 1.68 \times 10^{-23} \ cm^3$