# How to calculate Hall coefficient in gold and silver?

BACKGROUND - Hall Effect

Using the free-electron model, the Hall coefficient is calculated as $$R_H = \frac{1}{ne},$$ where $e$ is the elementary charge and $n$ is the carrier density. For a metal $X$, we can also write it in the following way: $$R_H (X)= \frac{m_X N_X}{\rho_X e},$$ where $m_X$ stands for the mass of a single atom, $N_X$ is the number of valent electrons in that metal, $\rho_X$ is its mass density. Since gold and silver are both monovalent, we set $N_X = 1$.

Using this model, I obtain: $$R_H(Au)=105.8 \times 10^{-12} \, \text{m}^3 \, \text{A} \, \text{s}$$ $$R_H(Ag)=106.6 \times 10^{-12} \, \text{m}^3 \, \text{A} \, \text{s}$$ The values I found in literature (The Hall Effect in Metals and Alloys, Colin Hurd, 1972) for this model are $$R_H(Au)=106.0 \times 10^{-12} \, \text{m}^3 \, \text{A} \, \text{s}$$ $$R_H(Ag)=106.5 \times 10^{-12} \, \text{m}^3 \, \text{A} \, \text{s}$$ so I guess it's not wrong.

THE PROBLEM

Those coefficients don't agree with observed values (same source): $$R_H(Au)=71.6 \times 10^{-12} \, \text{m}^3 \, \text{A} \, \text{s}$$ $$R_H(Ag)=88.1 \times 10^{-12} \, \text{m}^3 \, \text{A} \, \text{s}$$ Those are both monovalent 1B metals with FCC structure so I understand that the two-band model would be a better approximation to reality. In that model, two types of carriers are used, electrons and holes. The book gives a simple formula for the Hall coefficient in the low-field limit, but it includes cyclotron frequency $\omega$, relaxation time $\tau$ and effective mass $m^{*}$ for both the holes and electrons.

I spent hours searching all over the web, but I can't seem to find a simple expression for those quantities that would be applicable to $Au$ and $Ag$. Anyone?

• Are you sure about $N_X = 1$? I think the free electrons per atom of silver, for example, is specified at about 1.3.. – John David Gordon Sep 15 '18 at 15:02