## Materials Science and Engineering: An Introduction

$p= 7.31 g/cm^{3}$
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. From 3.18a, n= 4.0 atoms/unit cell Required: theoretical density if atomic weight = 114.82 g/mol Solution: $p = \frac{nA_{In}}{a^{2}cN_{A}} = \frac{(4.0 atoms/unit cell)(114.82 g/mol)}{[(4.59 \times 10^{-8} cm)^{2}(4.95 \times 10^{-8} cm)/unit cell] (6.022 \times 10^{23} atoms/mol)} = 7.31 g/cm^{3}$