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Solid State Physics - Point Defects in Gold

  1. Nov 21, 2006 #1
    Sooo, as this is my first time posting I thought I'd start off by saying hello to you all=)

    Anyway, the problem is as follows:

    Point defects in metals can cause additional electrical resistivity at low temperatures due to extra electron scattering which is proportional to the number of defects. The table gives the relative change in the resistivity at 78 K of a gold wire when it is quenched from various temperatures:

    Temp K 920 970 1020 1060 1220
    Resistivity Change 0.41% 0.7% 1.4% 2.3% 9.0%

    Calculate the energy of formation of a vacancy in gold.


    So, I've spent some time on this problem, using the equation n/N = exp(-E/(k*T)), where n is the number of defects, N the number of atoms and E is the energy of formation.

    Seeing as resistivity change is directly proportional to n/N, I divided the percentages by a hundred and put them into the formula. Putting in the different temperatures and resistivity changes gives me five different values for the energy of formation in the ordner of magnitue of 10^-20.

    Now, I think the question asks me to calculate the energy of a single vacancy, but I'm not too sure about that. And if that is indeed the case, how could I calculate it? In that case n would be 1, but seeing as I don't have N, how do I solve the equation?

    Any help would be greatly appreciated!
     
  2. jcsd
  3. Nov 21, 2006 #2

    OlderDan

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    If there is an energy per defect, the total energy should be proportional to the number of defects. Your equation says the energy only depends on the proportion of defects to the number of atoms. So either the E in the equation is the energy per defect, or the idea of an energy per defect makes no sense.
     
  4. Nov 21, 2006 #3
    So I should theoretically get the same result for all five different cases? I don't, though...
     
  5. Nov 21, 2006 #4

    OlderDan

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    No they should not be the same. The energy per defect would depend on how dense the defects are. Think of an analogy. Suppose you have an ensemble of like charges arranged in a grid. There would be an electrical potential energy for the system and each charge would have a potential energy because of the electrical potential from all the other charges. Electric potential (and potential energy) is inversely proportional to distance from a point charge, so if you were to shrink the grid the potential energy of each charge would increase and the total energy would increase. Alternatively, if you added additional charges to the grid the potential energy of every charge in the grid would increase. Either way, the potential energy per charge increases with the charge density (charge per unit volume). All that your equation is saying is that the energy of a defect increases with the density of defects.
     
  6. Nov 21, 2006 #5
    So effectively it's either my equation that's not quite right (so I'm not using the right one) or the point of that problem was just to solve for E and stick a few numbers in?
     
  7. Nov 21, 2006 #6

    OlderDan

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    I am not familiar with the equation you have, but I would guess the latter statement is correct. If the resistivity change is indeed proportional to n/N you should be able to fit the RC vs T data to a curve

    RC = βn/N = βexp(-E/(k*T))

    You might consider taking a logarithm of this equation.
     
  8. Nov 22, 2006 #7
    So if I fit the data to a curve, would E be the gradient then?
     
  9. Nov 22, 2006 #8

    OlderDan

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    If you fit it to the exponential, the curve will not be linear. If you take the log, you can get a linear equation with a slope (gradient) proportional to E.
     
  10. Nov 22, 2006 #9
    That makes sense... don't know if I've got a plotting programme on my computer though, but that problem can be solved. So is there any way for me to get to E without knowing the proportionality constant that links n/N to the resistivity change?
     
  11. Nov 22, 2006 #10

    OlderDan

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    Fitting the data will give you both E and the proportionality constant. TI graphing calculators have a built in exponential regression in the statistics functions. If you use the log you can graph it by hand if you have to.
     
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