Question

How much energy will be released when a sodium ion and a chloride ion, originally at infinite distance are brought together to a distance of $2.76\mathring{A}$ (the shortest distance of approach in a sodium chloride crystal)? Assume that ions act as point charges, each with a magnitude of $1.6\times {10}^{-19}C$. Permitivity constant of the medium is $9\times {10}^{9}N{m}^2{C}^{-2}$.

Answer 1

Given, $e=1.6\times {10}^{-19}CK=9\times {10}^9{Nm}^2{C}^{-2}$

Shortest Distance=r=$2.76\times {10}^{-10}$

$Energy=\frac{-K{q}_1{q}_2}{r}=-\frac{9\times {10}^9\times {(1.6\times {10}^{-19})}^2}{2.76\times {10}^{-10}}=-8.35\times {10}^{-19}J$