# Homework Help: Dissociation energy of NaCl, finding parameters and more questions

1. Apr 5, 2012

### fluidistic

1. The problem statement, all variables and given/known data
The distance between the nuclei Na and Cl is $R_0=0.236 nm$. The curvature of the potential energy is $109J/m^2$. The potential of ionization of the Na is 5.1 eV while the electron affinity of the Cl is 3.65 eV. From these data, find the parameters $a$ and $\alpha$ in the potential $V(r)=\frac{\alpha exp (-ar)}{r}-\frac{e^2}{4\pi \varepsilon _0r}$ and find the energy of dissociation D. Compare this result with the experimental one: 4.22 eV.

2. Relevant equations
Not sure.
$109J/m² =680.16 \frac{eV}{nm^2}$.

3. The attempt at a solution
First, I want to get $R_0$ in terms of alpha and a. So I derivated V(r) and equated to 0.
This gave me $\alpha e^{-aR_0}(aR_0+1)=\frac{e^2}{4\pi \varepsilon _0}$. I must isolate $R_0$, though this doesn't look like easy at all. Am I missing something?
Second, when they talk about the "curvature", I assume they mean the curvature of $V(R_0)$, right? Should I approximate V(r) by a Taylor expansion of order 2 centered at R_0 to find the curvature in order to get K in terms of alpha and a? The curvature of a parabola is defined to be $K=\frac{1}{2a(1+r)^{3/2}}$.
I've plotted V(r), more or less in http://www.wolframalpha.com/input/?i=plot+9*exp+(-1.7*x)/x+-+7/x+from+0+to+1. Since I don't know the constant alpha and a and that the graph changes drastically with a minor change in these parameters, the graph could be very wrong but at least it seems I can distinguish the minimum ($R_0$).
Also I have no idea how to find the dissociation energy D from the given data. I know it's the energy required to separate the 2 atoms infinitely far away. Wouldn't that just be $V(R_0)$? That looks like too simple.
Thanks for any help.