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Calculate the first 3 contributions to the electrostatic potential energy of an ion in the CsCl lattice.

I believe the formula I'm supposed to use is

[tex]

U_{c}\,=\,-\alpha\frac{e^{2}}{4\pi\epsilon_{0}R}

[/tex]

Just from looking in the chapter I can see this is a bcc type lattice with an [itex]\alpha[/itex] of 1.7627, but I'm not sure how they're getting the answer in the back of the book which is;

[tex]

U_{c}\,=\,-\frac{e^{2}}{4\pi\epsilon_{0}R}\left(8-\frac{6}{2/\sqrt{3}} + \frac{24}{\sqrt{11/3}}\right)

[/tex]

There's an example in the book showing the same procedure for an fcc lattice (NaCl) and that converging term is,

[tex]

6-\frac{12}{\sqrt{2}}+\frac{8}{\sqrt{3}}-\ldots

[/tex]

but it doesn't derive it, so I'm not really sure how they get it.

Just looking for a few pointers, thanks.