How to calculate the energy to dissociate a bond into neutral atoms? I am self studying chemistry through MiT ocw 5.111 . On practice exam 2 problem 2 there is a question which states the following

    element           ionization energy          electron   affinity
    Potassium(K)      418 kJ/mol                 48 kJ/mol
    Fluorine(F)       1680 kJ/mol                328 kJ/mol
    Chlorine(Cl)      1255 kJ/mol                349 kJ/mol



  (a) (12 points)   The ionic bond  length for KF is 0.217 nm. Calculate
  the energy (in units  of kJ/mol required  to  dissociate  a single
  molecule of KF into the neutral atoms K and F, using information
  provided above. For this calculation, assume that the potassium and
  fluorine ions are point   charges.


I proceeded to calculate the bond dissociation energy using the following formula $U(r)=\frac{z_1z_2e^2}{4\pi\varepsilon_0r}$  from which I obtained a dissociation energy of $-640$ kJ/mol. Then left with two ions F$^{-1}$ and K$^{+1}$.
 I went on to determine their ionization energies and electron affinities respectively. 

To make F$^{-1}\to$ F + $e^{-}$ an electron must be lost thus warranting an ionization energy of $-1680$ kJ/mol and. To make K$^{+1}$+ $e^{-} \to$ K and electron must be gained thus referring to the election affinity of K of $18$ kJ/mol. When equated as (Energy Required - Energy Released) I arrive at     

 (Dissociation Energy + Ionization Energy) - Electron Affinity = (-640 kJ/ mol - 1680 kJ/mol) + 48 kJ/mol = -2272 kJ/mol required


However I see in the answer key this is incorrect could someone guide me as to where I am going wrong here thank you.
Your $\pu{640 kJ mol^-1}$ value is correct, so the next step is to neutralize the ions. First add an electron to a potassium ion to get a potassium atom. This releases $\pu{418 kJ mol^-1}$ because it is the reverse of the potassium atom ionization. Then take away an electron from fluoride ion to get a fluorine atom. This requires an input of $\pu{328 kJ mol^-1}$ because it is the reverse of the fluorine electron affinity. So the result is $\pu{550 kJ mol^-1}$, i.e., $\pu{640 kJ mol^-1}$ - $\pu{418 kJ mol^-1}$ + $\pu{328 kJ mol^-1}$.

This figure makes it clear:



Figure copyright information: D.W. Oxtoby, H.P. Gillis, L.J. Butler, Principles of Modern Chemistry, 8th Ed., Cengage Learning, Section 3.8, p. 87, © 2016 Cengage Learning.