Question

The equilibrium separation of the Rb+ and Cl ions in RbCl is about 0.267 nm a) Calculate the potential energy of attraction of the ions, assuming them to be point particles. Find the dissociation energy, neglecting the energy of repulsion. from the exclusion p b) The ionization energy of rubidium is 4.18 eV, and the electron affinity of Cl is 3.62 eV. c) The measured dissociation energy is 4.37 eV. What is the repulsion energy that arises

Answer 1

2 137 = 5.4--(4.18-3.62) eV 4.84 ev 0 47 ev est