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An intrinsic semiconductor has electron and hole densities $n=p=n_i$, so ${\sigma }_i=e{n}_i({\mu }_e+{\mu }_h)$. In a fully ionised, uncompensated n-type sample with $N_D\gg n_i$, $n\simeq N_D$, $p\simeq n_i^2/N_D$, andThus a carrier-density ratio alone is not generally the electrical conductivity ratio. If electron mobility is unchanged and electron transport dominates the intrinsic conductivity, it reduces to $N_D/n_i$; equal electron and hole mobilities instead give $N_D/(2n_i)$. Ionised-impurity scattering can change the doped mobility.

The positive electron mobility relates the drift velocity of conduction Electrons to an electric field: ${\mathbf{v}}_d=-{\mu }_e\mathbf{E}$. In the isotropic Drude model, ${\mu }_e=e\tau /m^{\ast }$, where $\tau$ is the momentum relaxation time and $m^{\ast }$ the band effective mass. The minus sign describes negative charge; the electron mobility itself is positive.

The atomic number density isOne fully ionised single-electron donor per million atoms givesThe usual simplified answer treats conductivity as proportional to Electron concentration and assumes unchanged electron mobility, giving about $1.46\times 10^3$ times the undoped conductivity. There is a qualification: the conductivity ratio of doped and intrinsic semiconductors is not fixed by Electron counts alone. Intrinsic material also has $p=n_i$, whereas the heavily n-type sample has $n\simeq N_D$ and $p\simeq n_i^2/N_D$. ThereforeWith unchanged Electron mobility and negligible intrinsic hole conductivity this reduces to $1.46\times 10^3$; with equal, unchanged Electron and hole mobilities it is about $7.29\times 10^2$. The problem supplies no mobilities, so the density-ratio estimate requires its stated approximation. Ionised impurities may additionally reduce mobility.

Let $e>0$ denote the elementary charge, so an Electron's charge is $-e$. In the relaxation-time Drude model with positive transport effective mass,The steady drift velocity is ${\mathbf{v}}_d=-e\tau \mathbf{E}/m^{\ast }=-\mu \mathbf{E}$, defining positive electron mobility $\mu =e\tau /m^{\ast }$. The current density is opposite to the Electron drift:By the definition of electrical conductivity, $\mathbf{J}=\sigma \mathbf{E}$, soIts units are $({\mathrm{m}}^{-3})(\mathrm{C})({\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1})=\mathrm{S}{\mathrm{m}}^{-1}$. The mobility definition alone establishes the result in the scalar linear-response regime; the relaxation-time equation supplies a microscopic approximation for $\mu$.

Treat sodium as monovalent, with one mobile Electron per atom. Its mass density is ${\rho }_m=0.97\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}=970\mathrm{kg}{\mathrm{m}}^{-3}$, and its molar mass is $M=0.023\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. Using the Avogadro constant,The electron mobility $\mu =e\tau /m_e$ relates the magnitude of drift velocity to electric field. The Drude model then givesUsing $m_e=9.109\times 10^{-31}\mathrm{kg}$,The inferred time is the momentum relaxation time. Identifying it with the mean time between collisions is the intended simple isotropic Drude model assumption; small-angle scattering need not relax directed momentum in a single event.

For a single carrier species with positive charge carrier mobility $\mu$, the Hall coefficient and longitudinal electrical conductivity areTake the electric current density along $x$, magnetic field along $z$, and no transverse current. Then $E_y=R_H{j}_{x}B$ and $j_x=\sigma E_x$, giving the Hall angle relationThe sign specifies the electron Hall polarity; the ratio of field magnitudes is $|E_H|/|E|=\mu |B|$. Converting the stated electron mobility to SI gives $\mu =900\times 10^{-4}=0.0900{\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$. ThereforeThis uses the homogeneous single-carrier Drude model; the carrier density cancels.

The drift velocity is the small ensemble-average velocity acquired by charge carriers in an applied electric field. It is distinct from their much larger random thermal or Fermi velocity. For conduction Electrons of charge $-e$, the isotropic Drude model giveswhere $\tau$ is the momentum relaxation time.

The positive electron mobility ${\mu }_e=|{\mathbf{v}}_d|/|\mathbf{E}|$ measures the ease of this drift in linear response. In the same approximation,Its units are ${\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$; in an anisotropic crystal the mobility is a tensor.

The band effective mass describes the acceleration implied by the curvature of the dispersion relation. The group velocity is $\mathbf{v}={\hslash }^{-1}{\nabla }_{\mathbf{k}}E$, while $\hslash \dot{\mathbf{k}}=-e\mathbf{E}$. ThusFor an isotropic parabolic minimum this reduces to $E=E_c+{\hslash }^2{k}^2/(2m^{\ast })$ and $m^{\ast }\dot{\mathbf{v}}=-e\mathbf{E}$. It incorporates the periodic crystal potential and need not equal the bare electron mass. Negative electronic curvature near a band maximum is conveniently described by positive-mass electron holes.