Question

Stannum has a body centered tetragonal with lattice constant, a = b = 5.83A and c = 3.18A. If the atomic radius is 0.145 nm, determine the atomic packing factor of Sn.​

Answer 1

Answer:

the atomic packing factor of Sn is 0.24

Explanation:

a = b = 5.83A and c = 3.18A.

Volume of unit cell = a²c

= (5.83)² *  3.18 * 10⁻²⁴ cm³

= 1.08 * 10⁻²²cm³

Volume of atoms =

$2 \times \frac{4}{3} \pi r^3$

(∴ BCC, effective number of atom is 2)

Volume of atoms =

$2 * \frac{4}{3} *3.14*(0.145*10^-^7cm)^3$

= 2.55*10⁻²³cm³

$\text {Atomic packing factor}=\frac{\text {volume occupied by atom}}{\text {volume of unit cell }}$

$=\frac{2.55*10^-^2^3}{1.08*10^-^2^2} \\\\=0.24$

therefore, the atomic packing factor of Sn is 0.24