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Let the source's hydrogen-ionizing photon production rate bewhere $h{\nu }_0$ is the hydrogen ionization energy. The idealized Strömgren sphere has an almost fully ionized interior and a thin ionization front. In photoionization equilibrium, every ionization is balanced by a Case B recombination, so spherical symmetry givesFor pure hydrogen of constant number density $n$, the interior has $n_e=n_p=n$, and the Strömgren radius is therefore

Now let the effective number of dust grains per hydrogen nucleus be $f_d$, so the dust absorption coefficient is $a(r)=f_d{\sigma }_{d}n(r)$. If $Q(r)$ is the ionizing-photon rate crossing the sphere of radius $r$, recombinations and dust absorption give the linear ordinary differential equationIf $f_d$ denotes a dust mass fraction instead, the grain mass and gas mean particle mass are simply absorbed into the effective product $f_d{\sigma }_d$. Define the dust optical depthMultiplication by the integrating factor $e^{{\tau }_d(r)}$ and integration to the dusty front gives its governing equationThe exponential function weights recombinations at large optical depth by the extra source photons that dust must remove before those photons reach that radius.

For constant $n$, put $a=f_d{\sigma }_{d}n$ and $\tau =a{R}_d$. The elementary integralreduces the equation toEquivalently, with $y=R_d/R_S$ and ${\tau }_S=a{R}_S$,Its dust-free limit is $y\to 1$, while absorption makes $R_d<R_S$ for nonzero dust abundance.

For an ultrarelativistic particle, $E\simeq |\mathbf{p}|$. Setting $\mu =0$ and writing the three-dimensional momentum integral radially givesThe standard Bose integrals yieldFor fermions, the corresponding integrals differ by the familiar factorsThus the number density scales as $T^3$, while the energy density and pressure scale as $T^4$.

Summing over the $g$ internal states, the number density, energy density, and isotropic pressure areandThe factor ${\mathbf{p}}^2/(3E)$ in the kinetic pressure of an isotropic gas is the angular average of one diagonal component of the momentum flux.

For a spherically symmetric number density, the radial Laplacian is $r^{-2}{\partial }_r(r^2{\partial }_{r}n)$. Applying separation of variables and choosing the separation constant $-k^2$ givesso the ordinary differential equations areThe radial problem is an eigenvalue problem for $-\Delta$ with Dirichlet boundary conditions, hence $k^2>0$: multiplying the radial equation by $r^2\overline{R}$ and integrating gives $k^2{\int }_0^a{r}^2|R{|}^{2}dr={\int }_0^a{r}^2|R^{\prime }{|}^{2}dr>0$ for a nonzero regular mode vanishing at $a$.

To solve it explicitly, set $R=f/r$, corresponding to the suggested power $p=-1$. The equation reduces to $f^{\prime \prime }+k^{2}f=0$, soFiniteness at the origin forces $B=0$, and the Dirichlet boundary condition at $r=a$ requires $k=m\pi /a$ for an integer $m\ge 1$. The absence of interior zeros excludes $m\ge 2$, selecting the fundamental normal mode. If $n_c>0$ denotes the initial central density, the result iswhere the sine ratio is defined as $1$ at the origin. It is strictly positive in the interior for every finite time. The arbitrary positive scale $n_c$ cannot be fixed without additional initial data.

The spatial factor is time independent, so the density and the total neutron population have the same exponential growth rate. Thus the spherical reaction-diffusion growth threshold isAt equality the fundamental mode is stationary; below it the mode decays. For fixed $\lambda >0$, the equivalent critical-radius condition is $a>\pi /\sqrt{\lambda }$. This is the competition between multiplication in the bulk and diffusion loss at the surface in the stated dimensionless model.

Let $G={\gamma }_I{\Sigma }_f\varphi$ be the constant precursor production rate per volume. Here ${\gamma }_I$ is the precursor yield per nuclear fission, and $I$ and $X$ are number densities. At a steady state,For fixed homogeneous coefficients and spatial shape, the reactivity contribution of an added absorption coefficient ${\Sigma }_{aX}={\sigma }_{aX}{X}_0$ is $-{\Sigma }_{aX}/(\nu {\Sigma }_f)$. Thus the model givesWhen capture removal dominates xenon radioactive decay, $\varphi {\sigma }_{aX}\gg {\lambda }_X$, and henceThe limit is independent of the xenon neutron absorption cross section: at high neutron flux, nearly every xenon atom supplied by the precursor ultimately consumes a neutron. The assumptions include constant pre-shutdown neutron flux, sufficient time for precursor and xenon equilibrium, the supplied single-precursor model with no direct xenon fission source, and fixed parameters for translating absorption into reactor reactivity.

The key imbalance is preferential loss of self-interstitials to biased defect sinks, leaving a vacancy supersaturation that can feed void swelling. The initial displacement cascade creates both vacancies and self-interstitials, often in equal net numbers as Frenkel pairs; it need not create more vacancies than interstitials at birth.

Many nearby Frenkel pairs recombine. Others separate, become trapped or form dislocation loops. Surviving self-interstitials are usually much more mobile than vacancies over the relevant range. Dislocations also have an interstitial bias: their elastic interaction with defects makes them more efficient sinks for self-interstitials. Preferential interstitial removal to the dislocation structure can leave the mobile vacancy population above its thermal equilibrium value.

A minimal radiation damage rate theory makes the distinction explicit:Here $c_i,c_v$ are defect number densities, $G$ is the surviving mobile-pair production rate per volume, $K$ the recombination coefficient, $D_i,D_v$ the diffusion coefficients, and $k_i^2,k_v^2$ the respective sink strengths. The vacancy loss term includes thermal emission in an effective form. Equal production terms do not require equal steady populations or equal partitioning among sinks. This simplified model omits an explicit cluster-size distribution.

A stable cavity receiving a greater vacancy influx than interstitial influx and vacancy emission grows. Vacancies therefore condense into voids, producing void swelling. Helium can stabilize small cavity nuclei, but a growing vacancy void need not be a high-pressure gas bubble. Dislocation loops, precipitates and changes in the dislocation density evolve alongside the voids and modify the competing sink strengths.

There is no violation of atom conservation: atoms represented by removed self-interstitials have joined extended defects or surfaces, while the corresponding unfilled lattice volume can accumulate in cavities. The classic electron-microscope observations of cavities in irradiated type316 steel provide direct evidence for this microscopic basis of swelling.^[ref]

Assume a homogeneous Fe–Cr–Ni alloy, normal incidence, independent scattering events and the stated effective scattering cross-sections. Neglect absorption, other constituents and changes in neutron energy within the thin sheet. The atomic fractions are not supplied; take $x_{\mathrm{Fe}}=0.70$, $x_{\mathrm{Cr}}=0.20$, $x_{\mathrm{Ni}}=0.10$ as an explicit representative assumption, not a composition measured or specified in the paper.

The conventional cell of a face-centered cubic lattice containsatoms. The number density is thereforeFor a mixture, the scattering probability per unit path length is ${\Sigma }_s={\sum }_j{n}_j{\sigma }_j=n{\bar{\sigma }}_s$, whereThus ${\Sigma }_s=27.751{\mathrm{m}}^{-1}$. For thickness $t=10^{-3}\mathrm{m}$, the unscattered fraction $U$ satisfies $dU/dz=-{\Sigma }_{s}U$ with $U(0)=1$. The fraction undergoing at least one neutron scattering event isThe thin-sheet approximation gives $P_s\simeq {\Sigma }_{s}t=2.78\%$, consistent to the accuracy of the assumed composition. A simple iron-dominated estimate ${\bar{\sigma }}_s\simeq 3.2$ barns gives essentially the same about $3\%$ scattered conclusion.

For any other assumed composition, use the displayed weighted formula; mass fractions must first be converted to atomic fractions. Since the supplied elemental cross-sections span $2.8$–$4.1$ barns, any mixture of these three elements at the stated number density gives approximately $2.39$–$3.48\%$. This is a bound within the idealized mixture model, not a claim that every such composition is stable austenite.

annihilation can produce gamma rays, charged antiparticles and neutrinos, directly or through decay and hadronization of the primary products. For self-conjugate dark matter of mass $m_{\chi }$, the annihilation rate per unit volume is $⟨\sigma v⟩{\rho }_{\chi }^2/(2m_{\chi }^2)$. For a photon yield $d{N}_{\gamma }/dE$, the optically thin flux from a chosen solid angle isUse consistent mass-energy units. The factor $1/2$ assumes identical self-conjugate particles; distinct particle-antiparticle populations require their separate number densities. A gamma-ray continuum with an endpoint near the particle mass is generic for many channels, whereas a narrow line from $\chi \chi \to \gamma \gamma$ or $\gamma Z$ would be especially distinctive but often loop suppressed. dwarf spheroidal galaxies offer high dark-matter content with limited conventional emission; the Galactic centre has a potentially larger signal and much more complicated foregrounds.

Charged Positrons and antiprotons are redistributed by magnetic fields and lose energy, so their radiation spectrum depend on propagation and cannot usually point directly to the source. annihilation electrons can also produce synchrotron and inverse-Compton radiation. dark matter captured in the Sun could produce escaping high-energy neutrinos, with the signal depending on capture, annihilation equilibrium and particle interactions. Early annihilation energy injection can alter recombination and cosmic microwave background anisotropies.

A persuasive interpretation joins particle-like spectral structure with a consistent density-dependent morphology and other channels. Pulsars, cosmic-ray interactions and unresolved sources can imitate continuum excesses. Density-profile uncertainties, substructure and velocity-dependent annihilation modify the predicted signal; a lack of a detectable line does not exclude all annihilating dark matter.

supersymmetry relates bosons and fermions through fermionic symmetry generators. A main motivation is the Lorentz scalar hierarchy problem: related boson and fermion loop contributions cancel particular ultraviolet sensitivities in an exactly supersymmetric theory. soft supersymmetry breaking leaves residual sensitivity to the superpartner scale, so a very heavy spectrum does not automatically solve naturalness. Supersymmetric matter content can also improve gauge-coupling unification, and a stable neutral lightest superpartner can be a dark-matter candidate. None of these motivations is itself experimental evidence; supersymmetry must be broken if ordinary particles and their partners have different observed masses.

If R-parity is conserved, superpartners are produced in pairs and cascades end in a stable lightest supersymmetric particle. A neutralino LSP can escape the detector, giving missing transverse momentum. Colored squarks and gluinos can generate energetic jets, sometimes b-tagged jets and leptons, plus that imbalance. An inclusive supersymmetry search classifies broad event samples by jet and lepton multiplicities, transverse momentum sums and missing momentum, without requiring a fully reconstructed decay chain.

Background controls are part of the physics argument. Lost-lepton W and top events can be constrained using lepton collider control regions; invisible-Z production can be related to visible-Z or photon samples with appropriate transfer factors. Instrumental imbalance in multijet events needs detector-quality checks and data sidebands. An excess in a collider signal region should survive uncertainty in those extrapolations, efficiencies and luminosity, with independently defined validation regions and attention to the number of searches performed.

An exclusive supersymmetry search targets a more specific production mode and decay pattern. Electroweak chargino-neutralino production can yield three leptons plus missing momentum; a cascade through a slepton can yield a dilepton invariant-mass edge that constrains mass combinations. Such searches often have cleaner backgrounds but depend more strongly on branching fractions and intermediate masses. An endpoint does not necessarily determine every absolute mass without additional observables.

Different spectra require different strategies. For a compressed supersymmetric spectrum, small parent-daughter mass differences make the visible decay products soft. initial-state radiation can recoil against the pair and provide a trigger, supplemented by soft-lepton reconstruction. The ATLAS compressed-spectrum study explicitly uses small neutralino mass differences and initial-state radiation to define a dedicated search rather than applying a hard-lepton selection indiscriminately.^[ref] If R-parity is violated, the LSP can decay visibly, so high-multiplicity jet or lepton channels may matter more than large missing momentum. CMS separately studied hadronic and leptonic final states for this reason.^[ref] Long-lived particles instead motivate displaced vertices, disappearing tracks, abnormal ionization or time-of-flight measurements. Thus there is no single model-independent missing-energy cut that covers every supersymmetric spectrum.

For the hypothetical relic-density calculation, use the supplied distribution and retain its prefactor exactly. Put $M=m{c}^2$, $\Theta =k_{B}T$ and ${\Theta }_0=k_B{T}_0$. Thennumber density, denoted $N$ in the paper, is written $n$ here to distinguish it from a total particle count. The cosmological particle freeze-out temperature in energy units isAssume a stable nonrelativistic relic, negligible further annihilation and $aT=\mathrm{constant}$ after cosmological particle freeze-out, using sharp-freeze-out relic dilution. Its comoving particle number is conserved, soThe particle mass cancels from this present number density under the stated approximation. Its present rest-energy density is $M{n}_0$, not merely $n_0$:This is only 0.371% of the critical density, not a universe-closing component. The correct further comparison is with the measured dark-matter fraction, not an assumption that all critical density must be dark matter. In standard cosmology this abundance would be a small subcomponent, despite a possible collider discovery.

The exponential freeze-out sensitivity makes the result highly sensitive to $P$. Also, thermal entropy release means $aT$ is not exactly constant through the whole cosmic history. If total comoving entropy is conserved, the above number density acquires a factor $g_{\ast s,0}/g_{\ast s,f}$, involving entropy degrees of freedom. Those degrees of freedom and a full annihilation history are not given, so no precise correction is justified from the supplied data.

Establishing the nature of dark matter would require more than measuring a missing particle's collider mass. Measure its spin, charges, interactions, decay lifetime or stability limits, and annihilation or coannihilation processes; compare the resulting relic calculation with cosmological observations. Seek compatible direct nuclear-recoil signals and indirect annihilation or decay products, while controlling astrophysical and detector backgrounds. A collider-invisible lifetime need not be cosmologically long, and nonthermal production could change the abundance. A consistent particle identity, cosmological abundance and independent non-collider evidence are needed to connect the hypothetical discovery to the dark matter.

For a thickness $x$ with number density ${\rho }_n$ of scattering centres, the mean count is $N={\rho }_{n}x\sigma$. Using the total cross section in i and the variance addition in ii,ThusThe stated assumptions make the prefactor independent of $x$ and approximately independent of $p$. The lower-cutoff factors cancel except inside the slowly varying logarithm. Treating $N$ as a Poisson random variable rather than a fixed count gives the same result after replacing $N$ by its mean.

In the Hall effect, the Lorentz force deflects drifting charge carriers until a transverse electric field balances it: ${\mathbf{E}}_H=-{\mathbf{v}}_d\times \mathbf{B}$. Choose current along $x$, field along $z$ and define $R_H=E_y/(j_x{B}_z)$. Since $j_x=nq{v}_x$,The sign is negative for electrons with this convention. If the sample's mass density is ${\rho }_m$ and its molar mass is $M$, the atomic number density is $n_{\mathrm{at}}={\rho }_m{N}_A/M$, soMeasure the transverse voltage, specimen thickness, current and magnetic field; reversing field and current helps remove longitudinal-voltage pickup and offsets. This free-electron count from a Hall coefficient assumes one approximately free carrier band. Multiple bands, unequal mobilities or a complicated Fermi surface can make a Hall coefficient differ from $-1/(ne)$.

Take the electron's signed charge to be $e=-q$, where $q>0$ is the elementary-charge magnitude. Conventional electric current flows along $+x$, while the mean electron drift velocity is negative. With $\mathbf{B}=B_z\hat{\mathbf{z}}$, the magnetic Lorentz force separates charge across the width until a transverse electric field balances it. The accumulated edge charge then sustains a Hall voltage without a transverse steady current. In a one-carrier model, the Hall coefficient identifies the sign and number density of carriers; real multiband materials require a more careful interpretation.

The acceleration term contains the effective mass describing the local band response. The relaxation term removes drift momentum on the mean momentum-relaxation time $\tau$. The remaining two terms are the electric and magnetic Lorentz force on signed charge $e$; the magnetic force is perpendicular to velocity and does no instantaneous work. The relaxation-time approximation replaces individual scattering events by an averaged drift equation.

Let $n$ be the electron number density, distinct from the negative charge $e$. In transverse open circuit $⟨\mathbf{v}⟩=(v_x,0,0)$ and $j_x=ne{v}_x=I_x/(wt)$. The steady component equations givehenceConsequentlyThere is no sign ambiguity in this field-defined $R_H$ once axes and carrier charge are fixed.

A measured voltage ratio additionally needs probe polarities, which the geometric bar diagram does not label. To reproduce the positive ratio specified for electrons, define the longitudinal voltage drop $V_x=\phi (0)-\phi (l)=E_{x}l$ and the transverse rise $V_y=\phi (w)-\phi (0)=-E_{y}w$. ThenReversing either voltage's probes reverses the measured ratio, without changing the physical carrier density. With both voltages instead defined as drops along their positive axes, the algebraic ratio would have the opposite sign. The figure records the convention used here.

For the density, geometry gives $\rho =Rwt/l=Rt/(l/w)$ and $|R_H|=t|V_y|/(I_x|B_z|)=tR|V_y/V_x|/|B_z|$. ThusandThe mobility is $\mu =|R_H|/\rho =0.0130{\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$. With $m^{\ast }=m_e$,

A monovalent copper model assigns roughly one delocalized conduction electron per atom to a broad, approximately free-electron-like band. Using $m_e$ is therefore a sensible first estimate for its transport scale. Monovalence alone does not prove that every band curvature or orbit has exactly the free mass: periodic-potential mixing, the nonspherical Fermi surface and direction-dependent scattering limit the model. The calculated $\tau$ is conditional on the stipulated effective-mass approximation.

Experimentally, reverse both current and magnetic field and extract the part of the transverse voltage odd in each. This rejects fixed contact offsets and much longitudinal pickup; use four-terminal longitudinal sensing, measure dimensions independently and check heating before assigning material parameters. These controls are implemented in a reusable Hall/conductivity teaching experiment, although that experiment uses germanium rather than the copper sample here.^[ref]

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.

For donor number density $N_D$, the mean separation is of order $\ell \sim N_D^{-1/3}$. A donor-state overlap estimate sets this scale comparable with the effective radius:Relative to the atomic density, this is $3.77\times 10^{-5}$, about $38$ parts per million. If “start to overlap” is defined geometrically by centre separation $2a_B^{\ast }$, the estimate is instead $2.06\times 10^{23}{\mathrm{m}}^{-3}$, about $4.7$ parts per million. Thus the geometric estimate is of order $10^{23}$–$10^{24}{\mathrm{m}}^{-3}$. Real donor wavefunctions have tails, not hard orbit edges, so this does not define a sharp threshold or by itself locate a metal-insulator transition.

Under the one-carrier Drude model, the magnitude of the Hall coefficient gives an effective carrier densityConvert the material mass density to $1850\mathrm{kg}{\mathrm{m}}^{-3}$ and its molar mass to $0.009\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. The atomic number density isHere $N_A=6.02214076\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$ is the Avogadro constant. The positive Hall coefficient gives a hole-like response in a one-carrier interpretation. The Hall carrier count versus valence distinction matters: beryllium has two valence electrons, but the result is neither two free electrons nor a literal fractional atom donating charge. Its energy bands and Fermi surface invalidate the simple isotropic one-carrier inference. Electron-like and hole-like contributions with different charge carrier mobilities can yield this sign and effective count.

Doping changes the mobile carrier population by introducing controlled impurity levels. In an intrinsic semiconductor, the filled valence band is separated from the nearly empty conduction band by a band gap $E_g$. Thermal excitation produces an electron-hole pair, so the electron and hole number densities obey $n=p=n_i$. A electron hole represents a missing valence-band electron and behaves as a mobile positive charge carrier. For nondegenerate bands in thermal equilibrium,Here $N_C,N_V$ are the effective density of states, $E_C,E_V$ the band edges, and $E_F$ denotes the chemical potential, conventionally called the Fermi level. The semiconductor mass-action law holds in this nondegenerate equilibrium regime.

In a tetrahedrally bonded elemental semiconductor such as silicon, a substitutional group-V atom has an extra outer electron relative to the host. It is a donor impurity: a shallow level just below $E_C$ can ionise, releasing an electron to the conduction band and leaving a fixed positively charged donor. The resulting n-type semiconductor has electrons as majority carriers and holes as minority carriers. Conversely a group-III substitution supplies one fewer bonding electron. An acceptor impurity can take an electron from the valence band, leaving a mobile electron hole and a fixed negative acceptor; the resulting p-type semiconductor has holes as majority carriers.

The impurity state is shallow because the electron or hole responds to a screened Coulomb potential and an effective mass, rather than the bare atomic potential. A simple hydrogenic donor estimate iswhere ${ϵ}_r$ is the relative permittivity. This is a useful scaling model, with real band anisotropy and central-cell effects omitted. Dopants are fixed in the lattice; their ionisation changes mobile carrier numbers without making a uniformly doped bulk sample carry a large net electric charge.

The charge neutrality condition isWhen shallow dopants are almost fully ionised and $N_D-N_A\gg n_i$, an n-type region has $n\simeq N_D-N_A$ and $p\simeq n_i^2/n$. A p-type region similarly has $p\simeq N_A-N_D$ and $n\simeq n_i^2/p$. Thus compensation by the opposite impurity type reduces the majority-carrier density; the total dopant concentration is not generally the mobile carrier density. At sufficiently high impurity density the degenerate semiconductor regime requires the full Fermi-Dirac distribution and these exponential approximations fail.

There are three useful temperature regimes. At low temperature, dopant freeze-out leaves many shallow levels un-ionised. In the extrinsic regime, the dopants are largely ionised and set the majority-carrier density. At high enough temperature, intrinsic pair creation dominates and $n,p$ approach $n_i$. The Fermi level moves towards the conduction band with donor doping and towards the valence band with acceptor doping. The electrical conductivity isDoping therefore controls conductivity and its carrier type, but ionised impurities also scatter carriers and can reduce charge carrier mobility. Carrier density and mobility must both be considered.

A p-n junction forms when p-type and n-type regions are joined within a semiconductor. Initially the concentration gradients drive diffusion current: electrons diffuse from the n region to the p region and holes in the opposite direction. Recombination removes mobile majority carriers near the join, exposing fixed positive donors on the n side and fixed negative acceptors on the p side. This forms a depletion region and an electric field directed from n to p. Its drift current opposes diffusion. At equilibrium the net current vanishes and the Fermi level is uniform, although the band edges vary spatially through band bending.

For an abrupt nondegenerate junction with fully ionised dopants, its built-in potential isUnder the depletion approximation, the space-charge densities are $-e{N}_A$ on a p-side width $w_p$ and $+e{N}_D$ on an n-side width $w_n$. Integrating Poisson equation and imposing zero field outside gives $N_A{w}_p=N_D{w}_n$ andMost of the depletion width therefore extends into the more lightly doped side. The schematic shows the impurity levels before contact and the resulting space charge and potential after contact; there is no externally extractable equilibrium battery voltage from the built-in field alone.

A forward bias, with p positive relative to n, reduces the barrier from $V_{\mathrm{bi}}$ towards $V_{\mathrm{bi}}-V$ and injects minority carriers across the junction. In the ideal low-injection, diffusion-controlled regime this produces the Shockley diode equation, $I=I_s[e^{eV/(k_{B}T)}-1]$. Reverse bias increases the barrier and the depletion region; only a small thermally generated minority-carrier current flows before breakdown. Reverse-biased depletion regions also give a voltage-dependent junction capacitance. Real devices have recombination, leakage, series resistance and breakdown mechanisms beyond the ideal law. Controlled doping creates majority and minority carriers, and their junction produces a depletion barrier that makes rectification possible.

The Hall effect is a transverse electric response to a magnetic force on moving charge carriers. Take conventional current density $\mathbf{J}=J_x\hat{\mathbf{x}}$ and magnetic field $\mathbf{B}=B_z\hat{\mathbf{z}}$ in a conducting slab. For one carrier type of number density $n$ and signed electric charge $q$, the drift velocity is ${\mathbf{v}}_d=\mathbf{J}/(nq)$. Carriers initially deflect sideways until a Hall electric field balances the magnetic Lorentz force:The Hall coefficient is thereforeFor electrons it is $-1/(ne)$, whereas a simple hole-dominated conductor gives $+1/(pe)$. The sign reveals the effective carrier sign; the magnitude estimates their density in the one-carrier approximation.

For width $w$ along $y$ and thickness $t$ along $z$, $I=J_{x}wt$. Define the signed Hall voltage by $V_H=E_{y}w$, namely the electric potential at the lower-$y$ edge minus that at the higher-$y$ edge. ThenThis explicit contact convention fixes the voltage sign. A Hall bar with longitudinal and transverse contacts measures both electrical resistivity and the Hall response. Reverse the field and take the part of the transverse voltage odd in $B_z$ to suppress longitudinal-voltage pickup from misaligned contacts; reversing current provides another useful check.

A Hall measurement is useful for carrier identification, semiconductor doping measurements and magnetic-field sensing. Combining a one-carrier $R_H$ with electrical conductivity gives a mobility magnitude $\mu =\sigma |R_H|$. However, mixed electron and hole conduction is not described by a single density: in weak fields,The response weights mobilities differently from ordinary conductivity. Complicated energy bands, anisotropic Fermi surfaces and multiple bands can likewise change the simple interpretation in metals; a positive Hall coefficient does not mean that the material contains positively charged fundamental electrons.

In a metal, a partially filled energy band or overlapping energy bands supplies electrons with nearby unoccupied states. At thermal equilibrium, oppositely directed group velocities are equally populated and the net electric current vanishes. An applied electric field distorts the occupation distribution, producing a small mean drift velocity. A filled energy band alone has no such low-field intraband response because its velocity contributions cancel over the entire Brillouin zone.

In the Drude model, let $n$ be the mobile electron number density, let $e>0$ be the elementary charge, and let $\tau$ be the momentum relaxation time. The mean motion obeysThe steady state is ${\mathbf{v}}_d=-e\tau \mathbf{E}/m$. The current density is opposite to the electron drift velocity:This gives microscopic Ohm's law, with electrical conductivity $\sigma$ in $\mathrm{S}{\mathrm{m}}^{-1}$ and electrical resistivity ${\rho }_{\mathrm{el}}$ in $\Omega \mathrm{m}$. For a uniform wire of length $\ell$ and cross-sectional area $A$, its electrical resistance is $R={\rho }_{\mathrm{el}}\ell /A$. The dissipated power per volume is $\mathbf{J}\cdot \mathbf{E}=\sigma E^2$, the Joule heating supplied by the external source.

The free electron model with Fermi-Dirac statistics improves the microscopic picture. The Pauli exclusion principle fills states up to the Fermi energy at low temperature. For a spherical, parabolic energy band with two spin states,Only states near the Fermi surface change their occupation in a weak-field response; the large equilibrium Fermi velocity is not the small drift velocity. The same $n{e}^2\tau /m$ formula results for this isotropic parabolic model when $\tau$ is evaluated for transport near the Fermi surface. In a crystal, the energy band group velocity, effective mass and possibly direction-dependent scattering replace the simplest bare-particle picture. The mean free path is approximately ${\ell }_{\mathrm{mfp}}=v_F\tau$.

A perfectly periodic static lattice is already included in Bloch states and does not by itself give a finite electrical resistivity. Momentum relaxation comes from defects, impurities, boundaries, and electron-phonon scattering. Independent scattering mechanisms approximately add their rates, giving Matthiessen's rule,The approximately temperature-independent ${\rho }_0$ is the residual resistivity. In ordinary simple metals, the phonon contribution is approximately proportional to temperature at high temperature; a three-dimensional acoustic-phonon transport model gives a $T^5$ contribution well below the relevant Debye temperature. Other mechanisms can dominate in particular materials or regimes, so neither power law is universal.

The classical Drude model incorrectly treats the equilibrium carrier statistics. Fermi-Dirac statistics explains the small electronic heat capacity, while electrons carrying both charge and heat give the approximate Wiedemann-Franz law ${\kappa }_e/(\sigma T)={\pi }^2{k}_B^2/(3e^2)$ for degenerate carriers with appropriate nearly elastic scattering. Both the simple electrical conductivity formula and this thermal relation depend on the assumptions about energy bands and scattering, rather than being exact for all metals.

For the one-mole convention, $b$ is the excluded molar volume: finite molecular size reduces the volume available for translational motion from $V$ to $V-b$. It is an effective excluded volume, not simply the sum of the bare molecular volumes. $a/V^2$ corrects for attractive intermolecular forces, which reduce the pressure exerted on the wall. The attraction per molecule scales with number density, and the number of affected molecules also scales with density, producing the density-squared correction. The Van der Waals equation consequently raises the measured pressure to the corresponding repulsive-gas value before multiplying by the available volume. Both corrections vanish relative to the ideal-gas terms at sufficiently low density.

Model air as a dilute ideal gas of molecules described by the hard sphere model. The number density and mean free path areThe factor $\sqrt{2}$ accounts for the relative motion of the colliding molecules. A molecular collision diameter has to be estimated because none is supplied. Taking $d\simeq 0.37\mathrm{nm}$, appropriate as an order-of-magnitude size for the main constituents of air, givesThus the mean free path is about $7\times 10^{-8}\mathrm{m}$, or $70\mathrm{nm}$. A slightly smaller assumed diameter $0.35\mathrm{nm}$ gives about $76\mathrm{nm}$, illustrating the intended precision of the estimate.