Answer:
binding energy, E = 0.513 eV
Explanation:
It is given that,
Distance between ions, [tex]r=0.28\ nm=0.28\times 10^{-9}\ m[/tex]
Charge, [tex]q=1e=1.6\times 10^{-19}\ C[/tex]
Binding energy can be calculated using potential energy as :
[tex]E=\dfrac{kq^2}{r}[/tex]
[tex]E=\dfrac{9\times 10^8\times (1.6\times 10^{-19})^2}{0.28\times 10^{-9}}[/tex]
[tex]E=8.22\times 10^{-20}\ J[/tex]
or
E = 0.513 eV
So, the binding energy of a KCl molecule is 0.513 eV. Hence, this is the required solution.
