Answer
$APF = 0.536$
Work Step by Step
Given:
Uranium has orthorhombic symmetry with a= 0.286, b= 0.587 and c= 0.495 nm. Its density= 19.05 $\frac{g}{cm^{3}}$, atomic weight = 238.03 g/mol, and atomic radius= 0.1385 nm.
Required:
Atomic Packing Factor
Solution:
$APF = \frac{V_{s}}{V_{c}}$
In order to solve the APF, calculate first the ${V_{s}}$ and ${V_{c}}$. To calculate ${V_{s}}$, determine the value of $n$ or the number of spheres in the unit cell.
$n = \frac{pV_{c}N_{A}}{A_{U}} = \frac{(19.05 g/cm^{3})(2.86 \times 5.87 \times 4.95 \times 10^{-24} cm^{3})(6.022 \times 10^{23} atoms/mol)}{238.03 g/mol} = 4.0 atoms/unit cell$
Computing for APF,
$APF = \frac{V_{s}}{V_{c}} = \frac{(4) (\frac{4}{3}πR^{3})}{abc} = \frac{(4) (\frac{4}{3}π(1.385 \times 10^{-8} cm)^{3})}{(2.86 \times 5.87 \times 4.95 \times 10^{-24} cm^{3})} = 0.536$
