Answer
n= 4.0 atoms/unit cell
Work Step by Step
Given:
Indium has tetragonal unit cell with a= 0.459 and c= 0.495 nm. The APF is 0.693 and atomic radius is 0.1625 nm
Required:
number of atoms in each unit cell
Solution:
$APF = \frac{V_{s}}{V_{c}} = \frac{(n) (\frac{4}{3}πR^{3})}{a^{2}c}$
Solving for n and incorporating the given values,
$n = \frac{(APF)(a^{2}c)}{(\frac{4}{3}πR^{3})} = \frac{(0.693)((4.59 \times 10^{-8} cm)^{2}(4.95 \times 10^{-8} cm))}{(\frac{4}{3}π(1.625 \times 10^{-8} cm)^{3})} = 4.0 atoms/unit cell $
