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## Homework Statement

The energy per ion in for CsCl is nearly – (αe 2 /(4πε0)) + 8Ae -(R/ρ) , where α is the Madelung constant and A = 5.64 x 103 eV and ρ = 0.34 Å. Calculate the nearest neighbour equilibrium distance.

## Homework Equations

alpha = 2 ln 2

## The Attempt at a Solution

I think that CsCl is a simple cubic structure

I found online

For a simple cubic lattice, it is clear that the nearest neighbor distance is just the lattice parameter, a. Therefore, for a simple cubic lattice there are six (6) nearest neighbors for any given lattice point.

so then my answer would be 0.34 A ? Is this correct ?