Determining Spring Constant in Ion Pair

  • Thread starter PEZenfuego
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  • #1
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Homework Statement



The force between an ion pair is given by [itex]F=-k\alpha\frac{e^{2}}{r^{2}}[1-\left(\frac{r_{\circ}}{r}\right)^{m-1}] [/itex]
Find the value of [tex]r[/tex] where the equilibrium position is.

Determine the effective spring constant for small oscillations from the equilibrium.

Using [itex]m=8~\text{and}~\alpha=1.7476[/itex] estimate the frequency of vibration of a Na+ ion in NaCl

Homework Equations



Binomial expansion theorem


The Attempt at a Solution



The first question is easy as you set the force equal to 0 and it is no surprise that the answer is [itex]r=r_{\circ}.[/itex] When I try using the binomial expansion theorem, I always end up with a dependence on [itex]r_{\circ}[/itex]. But in the next portion, I have to find the frequency of vibration for Na+ ion and am only given alpha and m. Thanks for any help.
 
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Answers and Replies

  • #2
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Where do you get a binomial expansions? Please show your work, otherwise it is hard to find out what went wrong.
 
  • #3
Maybe use that [itex]\left(1-x\right)^n\approx 1-nx[/itex] where x is much less than 1.
 
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  • #4
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Maybe use that [itex]\left(1-x\right)^n\approx 1-nx[/itex] where x is much less than 1.
I did this! So I converted [itex]r[/itex] to [itex]r_{\circ}+\Delta r[/itex] and get [itex]F=-k\alpha \frac{e^{2}}{r_{\circ}+\Delta r}[1-\left(1-\frac{\Delta r}{r_{\circ}}\right)^{1-m}][/itex] and say that [itex]\frac{\Delta r}{r_{\circ}}[/itex] is much less than 1 (which is reasonable for small angles). I was hoping that this would get rid of the r and [itex]r_{\circ}[/itex] dependence this way, but even if I do, what do I do about F?
 
  • #5
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A right, with ##r \approx r_0## you get that binomial expansion, okay.

but even if I do, what do I do about F?
How does F vary for very small Δr? In particular, what about its derivatives?
 
  • #6
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Finding the derivative of force will show the points at which it is minimized and maximized, but I all ready know that. What good is it?
 
  • #7
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It will also show you how the force varies for small deviations from the equilibrium point. This gives the effective spring constant.
 
  • #8
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It will also show you how the force varies for small deviations from the equilibrium point. This gives the effective spring constant.

Well, I end up with the spring constant being [itex]\frac{k\alpha e^{2} \left(m-1\right)}{r_{0}^{3}}[/itex] and plugging this into mathematica shows the tangent, which is in good agreement for small deviations. Great...so now what?
 
  • #9
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So I took r0 to be the ion separation between na and cl which was 0.28 nm. Using this and all of the other information, I ended up with a frequency of 1.19*10^13 hertz. Is this reasonable?
 
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  • #10
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12,527
Well, I end up with the spring constant being [itex]\frac{k\alpha e^{2} \left(m-1\right)}{r_{0}^{3}}[/itex]
I agree with that result.

Looks like you have to use the external value for r0.
I don't know about the frequency, but it does not look completely wrong.
 
  • #11
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For the units to work out I HAVE to know r0. Maybe my value is crap, but at least my solution shows the understanding is there. It turned out to not be too hard...maybe a little convoluted. Thanks for the help!
 

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