Question

How to calculate Hall coefficient in gold and silver?

Answer 1

hey your ans is
Using the free-electron model, the Hall coefficient is calculated as
RH=1ne,
RH=1ne,
where ee is the elementary charge and nn is the carrier density. For a metal XX, we can also write it in the following way:
RH(X)=mXNXρXe,
RH(X)=mXNXρXe,
where mXmX stands for the mass of a single atom, NXNX is the number of valent electrons in that metal, ρXρX is its mass density. Since gold and silver are both monovalent, we set NX=1NX=1.

Using this model, I obtain:
RH(Au)=105.8×10−12m3As
RH(Au)=105.8×10−12m3As
RH(Ag)=106.6×10−12m3As
RH(Ag)=106.6×10−12m3As
The values I found in literature (The Hall Effect in Metals and Alloys, Colin Hurd, 1972) for this model are
RH(Au)=106.0×10−12m3As
RH(Au)=106.0×10−12m3As
RH(Ag)=106.5×10−12m3As
RH(Ag)=106.5×10−12m3As
so I guess it's not wrong.

THE PROBLEM

Those coefficients don't agree with observed values (same source):
RH(Au)=71.6×10−12m3As
RH(Au)=71.6×10−12m3As
RH(Ag)=88.1×10−12m3As
RH(Ag)=88.1×10−12m3As
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 ωω, relaxation time ττ and effective mass m∗m∗ for both the holes and electrons.

I hope its help you plzz mark me brainliest ans and follow me

Answer 2

To keep them moving straight, charges are necessary to establish an electric field in the y-direction to counterbalance the magnetic force, and these charges are shown by the small + and - symbols. ... For most metals, the Hall coefficient is negative, as expected if the charge carriers are electrons.