Question 2 of 7: Ionic Cohesion — Equilibrium Separation and the 1-D Madelung Constant

Question 2: Ionic Cohesion — Equilibrium Separation and the 1-D Madelung Constant (18 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $U(r)=A(r)+B(r)$ with $A(r)=(2.5\times10^4)\exp(-r/0.3)\ \text{eV}$ (short-range, positive) and $B(r)=(-25/r)\ \text{eV}$ (long-range, negative), $r$ in Å. Figure P2b: an infinite 1-D chain of alternating $\pm$ ions, uniform spacing $R$, reference ion positive.

Find. (a)/(b) physical origin of $A(r)$, $B(r)$; (c) equilibrium separation $r_0$ where $U(r)$ is minimum; (d) the Madelung constant $\alpha$ of the linear chain (eq. 8, $\alpha/R=\sum_j(\pm1/r_j)$).

[Figure not reproduced: Fig. 2 — $U(r)=A(r)+B(r)$ (solid), with the repulsive $A(r)$ and attractive $B(r)$ components (dashed) plotted from the paper's own equations. The minimum $(r_0,U_0)$ is marked. See the official exam paper.]

Approach. (a)/(b) identify each curve's physical mechanism from its sign and range; (c) set $dU/dr=0$ and solve numerically for $r_0$ (equivalent to reading the graph's minimum); (d) sum the alternating $\pm1/n$ series for the two half-chains flanking the reference ion.

1. Part (a) — origin of $A(r)$. $A(r)=(2.5\times10^4)e^{-r/0.3}$ is large and positive at small $r$, falling steeply as $r$ grows — the signature of the short-range repulsive energy from overlap of the filled electron orbitals of adjacent ions (Pauli exclusion forcing electrons into higher states as the charge clouds interpenetrate). Its impact is to prevent the ions from collapsing into each other; it is negligible beyond a few Å.
2. Part (b) — origin of $B(r)$. $B(r)=-25/r$ is negative and falls off slowly (as $1/r$) — the signature of the long-range attractive Coulomb (electrostatic) energy between the oppositely-charged ion cores (the Madelung energy). Its impact is to bind the crystal together; because it decays only as $1/r$, it still dominates $U(r)$ out to large separations, which is why $U(r)$ stays negative for all $r>r_0$ shown in Figure P2a.
3. Part (c) — equilibrium separation. At equilibrium $dU/dr=0$: $$\frac{dU}{dr}=-\frac{2.5\times10^4}{0.3}e^{-r/0.3}+\frac{25}{r^2}=0$$ Solving numerically (Newton iteration, consistent with the graph's own dashed construction lines to $r_0$): $$\boxed{r_0\approx3.12\ \text{Å}},\qquad U(r_0)=U_0\approx-7.25\ \text{eV}$$ matching Figure P2a, whose solid curve bottoms out just above $r=3$ Å at about $-7.3$ eV, where the printed dashed construction line marks $r_0$.
4. Part (d) — 1-D Madelung constant. Take the reference (positive) ion at the origin. Its $j$-th neighbour on either side is at distance $jR$ and carries charge $(-1)^j$ relative to the reference (alternating sign), so by eq. (8): $$\frac{\alpha}{R}=\sum_{j=1}^{\infty}\frac{2(-1)^{j+1}}{jR}\quad\Rightarrow\quad \alpha=2\sum_{j=1}^{\infty}\frac{(-1)^{j+1}}{j}=2\left(1-\frac12+\frac13-\frac14+\cdots\right)$$ the factor of 2 counting both the left and right half-chains. The alternating harmonic series sums to $\ln 2$, so $$\boxed{\alpha=2\ln 2\approx1.386}$$