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.
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 ?