## Materials Science and Engineering: An Introduction

$1.10\times{10}^{-8}N$
The equilibrium separation distance is the sum of the atomic radii, so we have: $r_{0}=r_{Mg^{2+}}+r_{F^{-}}$ Using equation 2.14, and letting $Z_{1}=+2$ for $Mg^{2+}$ and $Z_{1}=-1$ for $F^{-}$, $F_{A}=\frac{(2.31\times{10}^{-28}N.m^{2})(|+2|)(|-1|)}{(0.205\times{10}^{-9}m)^{2}}= 1.10\times{10}^{-8}N$