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: 2.15

Answer

$5.91\times10^{-10}N$

Work Step by Step

Using equation 2.14, we obtain: $Z_{1}=+2$ for $Ca^{2+}$ and $Z_{2}=-2$ for $Zn^{2-}$ and $r=1.25 nm = 1.25 \times 10^{-9} m $ So the force of attraction between the two ions is: $F_{A}=\frac{(2.31\times10^{-28})(|+2|)(|-2|)}{{(1.25 \times 10^{-9} m)}^{2}}= 5.91\times10^{-10}N $
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