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?
[Physics] Determining the resistivity of gold and copper alloy
electrical-resistance
Related Solutions
Assuming that the expressions given for a 2D electron gas here are valid, there really is no contradiction. Remember that there is an implicit sum in a repeated index (Einstein summation convention). Equivalently, an inverse matrix need not be the component-by-component inverse of the original matrix. For example, we get: $$[\sigma\cdot\rho]_{xx} = \sigma_{xx}\rho_{xx} + \sigma_{xy}\rho_{yx}$$ $$ = \frac{\rho_{xx}^2}{\rho_{xx}^2+\rho_{xy}^2} - \frac{\rho_{xy}\rho_{yx}}{\rho_{xx}^2+\rho_{xy}^2}$$ Together with $\rho_{xy} = -\rho_{yx}$, we get: $$[\sigma\cdot\rho]_{xx} = 1$$ Similarly, $$[\sigma\cdot\rho]_{xy} = \frac{\rho_{xx}\rho_{xy}}{\rho_{xx}^2+\rho_{xy}^2} - \frac{\rho_{xy}\rho_{yy}}{\rho_{xx}^2+\rho_{xy}^2}$$ And using $\rho_{xx} = \rho_{yy}$, we get: $$[\sigma\cdot\rho]_{xy} = 0$$
And so on for the remaining two components. Overall, we then get: $$[\sigma\cdot\rho]_{ij} = \delta_{ij}$$ which is precisely what we require.
We can also derive the required expression directly from the matrix: $$\rho = \begin{pmatrix} \rho_{xx} & \rho_{xy} \\ -\rho_{xy} & \rho_{xx}\end{pmatrix}$$ The determinant is: $$\det\rho = \rho_{xx}^2+\rho_{xy}^2$$ and its adjugate is: $$\operatorname{adj}\rho = \begin{pmatrix} \rho_{xx} & -\rho_{xy} \\ \rho_{xy} & \rho_{xx}\end{pmatrix}$$ It's inverse is then: $$\sigma = \rho^{-1} = \frac{1}{\det\rho}\operatorname{adj}\rho$$ leading to the expression for $\sigma$.
Update:
The reason why $\rho_{xy} = -\rho_{yx}$ is because the magnetic field, $\mathbf{B} = B_z\hat{\mathbf{e}}_z$, always has a certain sense of orientation, being related to the force through a cross product. Let's say that it deflects the current anticlockwise (right handed coordinates). Then, a current in the $+x$ direction would be deflected in the $+y$ direction; that in the $+y$ direction would be deflected in the $-x$ direction. Thus, the $x$ component in the latter case is negative.
In the general case, this is contained in the relation $\rho_{xy} = -\rho_{yx}$, which effectively says that an electric field in one direction always results in a deflected (perpendicular) current which is in a specific orientation (clockwise or anti-clockwise) from it, with the same magnitude for a given electric field magnitude.
In copper there are mobile (free) electrons which are not attached to any particular nucleus and these free electrons are responsible for the conduction process in copper and other metals.
So you can think of a lump of copper as having copper ions held in position in a structure called a lattice and the ions vibrating about fixed positions.
The free electrons move around these ions within the metal just like gas molecules which are in a box.
When a voltage is applied across copper the free electrons start moving from the negative terminal to the positive terminal and in doing so gain kinetic energy.
However the free electrons do not have a free passage through the copper and collide with the vibrating copper ions losing some of their energy and making the copper ions vibrate more. Thus the temperature of the copper increases.
After a collision with a copper ion a free electron again gains kinetic energy and the process repeats itself.
The more the ions vibrate the greater the impediment to the passage of the free electrons.
So as the temperature of the copper gets less the copper ions vibrate less and so there are fewer impediments to the passage of the free electrons – the resistance of the copper is lower.
This is a simple model which does illustrate what happens but perhaps now it is better to make the model slightly more sophisticated and say that if all the copper ions in the lattice were arranged in perfect order then free electrons would not interact with the copper ions (lattice) and the resistance of the copper would be zero.
So you can think of the thermal vibrations of the copper ions as introducing irregularity in the copper lattice and these irregularities are responsible for the resistance of copper.
As your graph shows, reducing the temperature reduces the resistance and that is because the irregularities due to the ions vibrating are reduced.
The variation of resistance of copper with temperature can be predicted fairly accurately and it is found that theory so far and experiment agree until the temperatures get closer to 0 kelvin.
The predicted resistance due to the lattice vibrations becomes much less than the actual resistance which is measured.
So there must be other processes which are responsible for the copper having resistance at very low temperatures.
Again this resistance is due to irregularities in the copper lattice but not because the copper ions are vibrating.
The irregularities are there because there are impurity atoms in the copper and so where there should be a copper ion there is an atom of another element.
It is these impurities which interact with the free electrons and cause resistance.
But it is not just impurity atoms which are responsible for the resistance. It could be the lack of a copper ion at a particular location, it could be that the copper ions are not lined up perfectly as in a perfect crystal (these are called dislocations) and the copper is made up not just of one but many crystals, it is polycrystalline. The boundaries between the crystals are also irregularities which the free electrons interact with and this causes resistance.
So at very low temperature the purity and the structure of the copper are mainly responsible for the resistance of copper rather than the thermal vibrations of the ions.
It does not stop there because even if the sample of copper is very, very pure and one single almost perfect crystal the outer surfaces of the crystal would have an effect on the resistance of copper at very low temperature. the surface of the crystal being an irregularity.
You will find in more advanced texts that the lattice vibrations are thought of as bundles of energy and momentum and are called phonons. This is similar to calling a bundle of energy and momentum related to an electromagnetic wave a photon. The interactions between the lattice and the free electrons are thought of as being interactions/collisions between the phonons and the free electrons.
One of the reasons that copper is a better conductor than lead is that the phonon – electron interaction in copper is not as strong (weaker) than the phonon – interaction in lead.
This means that the lattice vibrations impede free electrons in copper less than in lead.
So it is rather strange that metals which are relatively poor conductors are more likely to become superconductors at low temperature?
Superconductivity is due to pairs of electron (Cooper pairs) coupling together with the aid of phonons. If the phonon – electron interaction is weak as it is in copper this means that Copper pairs are less likely to occur and so copper will never become a superconductor no matter how low the temperature. There are other reasons as to why superconductivity does or does not happen eg how many free electrons are produced by each atom. Copper only produces one free electron per atom whereas for lead there are more free electrons per atom available than in copper and the photon – electron interaction also being stronger gives lead more of a chance of becoming a superconductor.
Best Answer
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.
$\rho = 22.8 + 450 \times \dfrac{394}{1537} \left ( 1 - \dfrac{394}{1537}\right ) = 108.6$