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

  1. Apr 5, 2012 #1


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

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

    3. The attempt at a solution
    First, I want to get [itex]R_0[/itex] in terms of alpha and a. So I derivated V(r) and equated to 0.
    This gave me [itex]\alpha e^{-aR_0}(aR_0+1)=\frac{e^2}{4\pi \varepsilon _0}[/itex]. I must isolate [itex]R_0[/itex], 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 [itex]V(R_0)[/itex], 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 [itex]K=\frac{1}{2a(1+r)^{3/2}}[/itex].
    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 ([itex]R_0[/itex]).
    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 [itex]V(R_0)[/itex]? That looks like too simple.
    Thanks for any help.
  2. jcsd
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