= Solution
A completely filled <valence band> carries no net current: its occupied crystal-momentum states have <velocities> that cancel. Near a valence-band maximum, an almost full band is conveniently described by the few missing <electrons>, or <electron holes>. Removing an <electron> of charge $-e_0$, with $e_0>0$, changes the charge relative to the filled band by $+e_0$. This is a <quasiparticle> description of electronic motion, not a mobile positive ion or a <Positron>.
For an isotropic parabolic maximum, $\epsilon_v(\mathbf k)=E_v-\hbar^2k^2/(2m_h^*)$. Assign the missing-electron state momentum $\mathbf k_h=-\mathbf k_e$ and hole <energy>
$$
\epsilon_h(\mathbf k_h)=E_v-\epsilon_v(-\mathbf k_h)
=\frac{\hbar^2k_h^2}{2m_h^*}.
$$
It has positive <effective mass> $m_h^*$. Its <velocity> equals the <velocity> of the missing <electron> state, and its positive-charge current replaces the negative of that state's electronic current. Semiclassical motion is thus $\hbar\dot{\mathbf k}_h=+e_0(\mathbf E+\mathbf v_h\times\mathbf B)$. With <momentum relaxation time> $\tau_h$, the low-field mobility is $\mu_h=e_0\tau_h/m_h^*$ and
$$
\boxed{\sigma_h=e_0p\mu_h=\frac{p e_0^2\tau_h}{m_h^*}.}
$$
If <electrons> also contribute, $\sigma=e_0(p\mu_h+n\mu_e)$: the two species' longitudinal conductivities add.
A <Hall bar> separates carrier concentration from mobility. Apply <current density> $j_x$, <magnetic field> $B_z$, and measure transverse <electric field> $E_y$. For one isotropic hole population, transverse <force> balance gives $E_y=v_xB_z$, so
$$
\boxed{R_H=\frac{E_y}{j_xB_z}=\frac1{e_0p}>0,\qquad
p=\frac1{e_0R_H},\qquad \mu_h=\sigma R_H.}
$$
For thickness $t$, the Hall-voltage magnitude is $|V_H|=IB/(e_0pt)$. A longitudinal <four-terminal resistance measurement> supplies $\sigma$ independently. Use <ohmic contacts>, known conducting thickness, controlled <temperature> and low enough current to avoid self-heating. Reversing <magnetic field> removes even-in-field longitudinal offsets from contact misalignment; reversing current also helps reject thermoelectric offsets. These are central controls, rather than merely taking one transverse voltage reading.<Resistivity and Hall Measurements by National Institute of Standards and Technology>[Sample Geometry; Resistivity Measurements; Definitions for Hall Measurements; Hall Measurements and Hall Calculations]{ref}
The simple inference assumes one carrier type and a known <Hall scattering factor>. More generally $R_H=r_H/(e_0p)$, so $\sigma R_H$ measures <Hall mobility> rather than automatically the drift mobility. With two carrier populations at weak field,
$$
R_H=\frac{p\mu_h^2-n\mu_e^2}{e_0(p\mu_h+n\mu_e)^2}.
$$
Its sign alone need not identify the more numerous carrier. A Hall experiment also cannot determine $m_h^*$ and $\tau_h$ separately: their ratio sets mobility. <cyclotron resonance> supplies an independent <mass> measurement by sweeping <frequency> or <magnetic field> through $\omega=e_0B/m_c$ and detecting resonant absorption. The measured <cyclotron effective mass> can depend on band anisotropy and need not equal every curvature <mass>; the isotropic parabolic model identifies them. Its linewidth provides a scattering-time estimate only with a specified <damping> model.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2015/paper-2-holes.png]
{title=Positive-curvature hole energy, a Hall-bar measurement, and an acceptor level in a semiconductor}
<semiconductor doping> generates holes by adding <acceptor impurities>. In <silicon>, a substitutional group-III atom such as <boron> provides one fewer valence <electron> than <silicon>. A shallow acceptor level lies slightly above the valence-band edge: accepting a valence <electron> makes the dopant negatively ionized and leaves a mobile hole in the <valence band>. The impurity stays fixed in the lattice. <charge neutrality> and the nondegenerate <equilibrium> mass-action law give
$$
p+N_D^+=n+N_A^-,\qquad np=n_i^2.
$$
For fully ionized, uncompensated acceptors in the extrinsic regime, $p\simeq N_A$ and $n\simeq n_i^2/N_A$; with fully ionized donor compensation, $p\simeq N_A-N_D$ while holes remain dominant. The <Fermi level> moves towards the <valence band>. At low <temperature>, <semiconductor carrier freeze-out> suppresses mobile <electron holes>; at high <temperature> intrinsic electron-hole generation can dominate. Increasing dopant density also changes scattering, so a conductivity increase alone cannot distinguish increased <carrier density> from changed mobility. Temperature-dependent Hall and conductivity measurements separate these effects, subject to the carrier and scattering qualifications above.
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