College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 15 - Electric Charge, Forces, and Fields - Learning Path Questions and Exercises - Exercises - Page 556: 14

Answer

Attractive electric force between the ions is $2.8972\times10^{-9}N$

Work Step by Step

Singly charge sodium means one electron is removed from the sodium atom so charge on sodium ion is $q_{Na^+}= +e=+1.6\times10^{-19}C$ similarly singly charged chlorine ions means one electron is given to chlorine atom so charge on chlorine ion $q_{Cl^-}= -e=-1.6\times10^{-19}C$ Distance between sodium and chlorine ions in crystal is given as $r= 2.82\times10^{-10}m$ so from coulombs law magnitude of attractive force between the ions will be $F=k\frac{q_{Na^+}\times q_{Cl^-}}{r^2}$ putting magnitudes of $q_{Cl^-}$, $q_{Na^+}$ as as $1.6\times10^{-19}C$ , $k=9.0\times10^{9}N.m^2/C^2$, $r= 2.82\times10^{-10}m$ $F=9.0\times10^{9}N.m^2/C^2\times \frac{1.6\times10^{-19}C\times 1.6\times10^{-19}C}{(2.82\times10^{-10}m)^2}$ =$2.8972\times10^{-9}N$
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