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

the atomic packing factor of Sn is 0.24

Step-by-step 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 * (4)/(3) \pi r^3`

(∴ BCC, effective number of atom is 2)

Volume of atoms =

`2 * (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 }}`

`=(2.55*10^-^2^3)/(1.08*10^-^2^2) \\\\=0.24`

therefore, the atomic packing factor of Sn is 0.24