Metals — Question 8

Question 8

The Hall voltage identifies the sign and density of metallic carriers

A single-electron-band slab has thickness $t=.100\,\mathrm{mm}$ along z. Current I flows along +x and $B_z=+.500\,\mathrm T$. Define $V_H=V(y=0)-V(y=w)=E_yw$. At $I=2.00\,\mathrm A$, measure $V_H=-1.25\,\mu\mathrm V$. Use $e=1.60\times10^{-19}\,\mathrm C$.

1. Derive the Hall coefficient and electron density.

2. If longitudinal resistivity is $2.00\times10^{-8}\,\Omega\,\mathrm m$, find the mobility.

Question 8 - Solution

The Hall voltage identifies the sign and density of metallic carriers

(a) Method and calculation

$$E_y=v_xB_z,\quad J_x=-nev_x\Rightarrow R_H=E_y/(J_xB_z)=-1/(ne).$$$$\boxed{R_H=V_Ht/(IB_z)=-1.25\times10^{-10}\,\mathrm{m^3/C},\quad n=5.00\times10^{28}\,\mathrm{m^{-3}}}.$$

(b) Interpretation and further calculation

$$\sigma=ne\mu=1/\rho\Rightarrow\mu=|R_H|/\rho.$$$$\boxed{\mu=6.25\times10^{-3}\,\mathrm{m^2/(V\,s)}}.$$

Physical check

The voltage definition fixes the sign. Multiband metals need not obey the single-band density interpretation used here.