Materials Science and Engineering: An Introduction

Published by Wiley
ISBN 10: 1118324579
ISBN 13: 978-1-11832-457-8

Chapter 2 - Atomic Structure and Interatomic Bonding - Questions and Problems - Page 48: 2.16a

Answer

$1.10\times{10}^{-8}N$

Work Step by Step

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 $
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