Modern Physics Question: Corrected potential for NaCl

[Deleted member 333884]

Homework Statement

The potential energy "V" of NaCl can be described empirically by:

V = -(e²/(4*Pi*epsilon)) + A*exp[-R/rho]

where R is the internuclear separation. The equilibrium separation of the nuclei Ro is 2.4 Angstroms and the dissociation energy is 3.6 eV.
a) Calculate A and rho/Ro, neglecting zero-point vibrations.
b) Sketch V and each of the terms in V on one graph.
c) Give the physical significance of A and rho.

Homework Equations

Electrostatic Potential: V = V = -(e²/(4*Pi*epsilon))

The Attempt at a Solution

I found the uncorrected electrostatic potential to be -6 eV, so I plugged that in and found the following:

2.4 = A*exp[-2.4*10^-10/Rho]

The chapter talked about A and rho for the photon and phonon gas, but I'm not sure how to apply it to NaCl. Do I have to find the degeneracy of states for Sodium, Chlorine and then somehow combine the two? Thanks.

By the way: This is problem 4 in Chapter 12 of Eisberg-Resnick "Quantum Physics of Atoms-Molecules-Nuclei-Particles"

 

[ideasrule]

Homework Statement

The potential energy "V" of NaCl can be described empirically by:

V = -(e²/(4*Pi*epsilon)) + A*exp[-R/rho]

Are you sure that's the right function? The first term doesn't have the right units. Once you get the correct equation, the equilibrium position should be where the potential energy is at a minimum. That will give you one equation. Disassociation energy should be the magnitude of the minimum potential energy. That should give you a second equation, so you can solve for A and rho.

 

[Deleted member 333884]

Are you sure that's the right function? The first term doesn't have the right units. Once you get the correct equation, the equilibrium position should be where the potential energy is at a minimum. That will give you one equation. Disassociation energy should be the magnitude of the minimum potential energy. That should give you a second equation, so you can solve for A and rho.

Thank you for the help! There was an R in the denominator of the first quotient, so I'll try that out and see where it leads.

Thanks again.

Doug