In simple terms, the collision of two atoms $\ce{A}$ and $\ce{B}$ will result in ions $\ce{A^+}$ and $\ce{B^-}$ if $$I_a(\ce{A})+E_a(\ce{A})<I_a(\ce{B})+E_a(\ce{B})$$ where $I_a$ and $E_a$ are the ionisation energies and electron affinities, respectively. Let us view $\ce{A^+B^-}$ as a molecule of sorts, having ions of rigid, non-polarisable spheres. Say $r$ is the sum of the radii of such spheres. The potential energy $E_p$ of such a system would then be $$E_p=I_a(\ce{A})-E_a(\ce{B})-\frac{e^2}{r}+\frac{b}{r^n}.$$ It is intuitive that - the term $I_a(\ce{A})-E_a(\ce{B})$ characterises the energy required to form isolated ions $\ce{A^+}$ and $\ce{B^-}$ in the gas phase; - $-e^2/r$ takes into account culonic attraction between two ions. It is *not* intuitive for me why ${b}/{r^n}$ is how it is. > The final term, introduced by Max Born, encompasses the repulsion generated by shells of electrons at either ion. Of course, such effects cannot be ignored. But I am interested in why is it $b/r^n$. What is the justification? Feel free to provide the rigorous approach, even though it will probably$^{[1]}$ be above my abilities of comprehension. --- Browsing on [Wikipedia](https://en.wikipedia.org/wiki/Born%E2%80%93Land%C3%A9_equation), the information might be given in the book [*Advanced Inorganic Chemistry*](http://bib.convdocs.org/v20061/?download=1) by F. Albert Cotton, Geoffrey Wilkinson, Carlos A. Murillo, and Manfred Bochmann. The pertinent section “1-6. Energetics of Ionic Crystals” starts at page 18$^{[2]}$, however I cannot find the derivation. There is only that the value of $n$ relies on a compressibility measurement, given as fractional change in volume per unit change in pressure, or $$\frac{\Delta V}{V\Delta P}.$$ Also, this is seems somewhat contradictory to the earlier assumption of rigid spheres as written in another book$^{[3]}$. So, if anything, the linked book confused me further. $^{[1]}$ Most certainly. $^{[2]}$ Page 28 in the PDF. $^{[3]}$ U. Palm, V. Past. *Physical Chemistry*. (1974) [To my knowledge, not available in English.]