Say we got an alloy made of $90$% Au and $10$%Cu. The atomic mass of gold is $$A_{Au}=197g/mole$$ and of copper $$A_{Cu}=63.5g/mole$$ The specific resistivity of gold is $$\rho=22.8n\Omega m$$ and the Nordheim coefficient is $$C=450n\Omega m$$ Then by the Nordheim rule we have that the resistivity of this alloy is given as $$ \rho=\rho _{Au}+CX(1-X)$$ where $X=0.1$, and represents the percentage of Copper in the alloy. Using the above I got that $$ \rho=63.3 n\Omega m$$, but data shows that the actual resistivity is $ \rho=108 n\Omega m$. What about X? Is it the percentage of atoms in the alloy, or is it the percentage of mass in the alloy?
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$\begingroup$ I have updated my answer. $\endgroup$– FarcherCommented Sep 15, 2016 at 18:50
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2 Answers
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The resistivity is dependent on the number of atoms and so you must find the ratio of copper atoms to the total number of atoms to find $X$ and hence the resistivity of the alloy.
If you do this correctly you should find that the value you have calculated is in agreement with the book value.
Update
The molar fraction of copper (fraction of copper atoms to the copper and gold atoms) is given by
$X = \dfrac{\left( \dfrac{10}{63.5} \right)}{\left( \dfrac{10}{63.5}+\dfrac{90}{197} \right)} = \dfrac{394}{1537}$
This comes from the idea that 10 g of copper is $\dfrac{10}{63.2}$ moles of copper which is $\dfrac{10}{63.2} \times N_{\rm A}$ atoms of copper where $N_{\rm A}$ is Avagadro's constant.
$\rho = 22.8 + 450 \times \dfrac{394}{1537} \left ( 1 - \dfrac{394}{1537}\right ) = 108.6$
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$\begingroup$ I think to find C must be a typo because C is a given constant. It should be to find X, which is the mole fraction of Cu atoms. $\endgroup$ Commented Sep 7, 2016 at 18:14
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$\begingroup$ What does mole fraction mean? $\endgroup$ Commented Sep 8, 2016 at 20:17
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An accurate physical theory of the electrical resistivity of metal alloys does not exist as of year 2017. I am planning to develop one in 2018. So, there is no answer to your question, yet.
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$\begingroup$ Whatever it is that you're smoking, I want some of it. $\endgroup$ Commented Aug 27, 2018 at 18:33