Consider two oppositely charged atoms with charges of +1 and -1 units, respectively. The two atoms interact with each other through electrostatic (Coulombic) and van der Waals forces, and reach an equilibrium at a distance of 2.2 . At this distance, the attractive electrostatic force is balanced by the repulsion part of the van der Waals force. The van der Waals energy is given by: R U=E min r 12 R 2 min r Here we assume that the well depth (E) is 1.0 kJ/mol, and the distance of minimum energy, Rmin, is 2.9 . The electrostatic energy is given by: 199 U .1391 kJ/mol here, q; and q; are the elementary charges, and & is the dielectric constant of the environment. (i) Derive an expression for the van der Waals force on one atom, starting from the energy equation given above. (ii) Derive an expression for the electrostatic force on one atom, starting from the energy equation given above. (iii) Calculate the value of , the dielectric constant, given that the van der Waals repulsion and the electrostatic attraction are balanced at 2.2 . Based on your answer, do you think that the environment is (a) water, (b) interior of a protein, or (c) vacuum?