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Charge released per specified active or composite mass under complete stoichiometric utilization. For $n$ electrons per formula unit, active-material capacity is $nF/(3.6M)$ in mAh per gram.

For a degenerate isotropic parabolic electron band, electrical conductivity is $\sigma =n{e}^2{\tau }_{\sigma }/m$ and electronic thermal conductivity is ${\kappa }_e={\pi }^{2}n{k}_B^{2}T{\tau }_{\kappa }/(3m)$. Their ratio is ${\kappa }_e/(\sigma T)=L_0{\tau }_{\kappa }/{\tau }_{\sigma }$, where $L_0=({\pi }^2/3)(k_B/e{)}^2$ is the Lorenz number. Nearly elastic scattering that relaxes charge and heat currents similarly gives ${\tau }_{\kappa }\simeq {\tau }_{\sigma }$ and the Wiedemann-Franz law. Inelastic scattering can relax heat current and directed momentum differently, invalidating the simplest identification even in a good metal.

In a dilute electron gas, thermal kinetic energies become relativistic near $k_{B}T=m_e{c}^2$, corresponding to $T_{\ast }\simeq 5.93\times 10^9\mathrm{K}$. This is a crossover in the momentum distribution, not an abrupt temperature transition. Degenerate electrons can instead be relativistic at much lower temperature if their Fermi momentum exceeds $m_{e}c$.

At fixed surface density, a total-pressure alpha disk supported by radiation pressure has viscous heating $Q^{+}\propto T_c^8$ but radiative cooling $Q^{-}\propto T_c^4$ when electron scattering supplies a nearly constant opacity. Heating therefore wins after a positive temperature perturbation.

For symmetric touching parabolic dispersions with constant density of states $D$ per energy band, including spin degeneracy, charge neutrality fixes $\mu =0$. The Fermi-Dirac distribution gives $N_c=D{k}_{B}T\ln 2$ and the thermal internal energy, including electrons and holes, is $\Delta U=({\pi }^2/6)D(k_{B}T{)}^2$. Thus $C=({\pi }^2/3)D{k}_B^{2}T$. Its large ratio to $N_c$ is compatible with the equipartition theorem because this excitation count grows with temperature and the heat capacity includes both species.

For a three-dimensional free electron model, the spin-summed density of states at the Fermi energy is $g(E_F)=3N/(2E_F)$. A shell estimate treats $N_{\mathrm{act}}\sim g(E_F)k_{B}T$ electrons as thermally active and assigns each an excitation energy of order $k_{B}T$. For the illustrative choice $3k_{B}T/2$, $\Delta U\sim (9/4)N{k}_B^2{T}^2/E_F$ and differentiation gives $C_V\sim (9/2)N{k}_{B}T/T_F$. The numerical coefficient is heuristic: the controlled Sommerfeld expansion instead givesBoth estimates express Pauli exclusion principle suppression of the classical heat capacity when $T\ll T_F$. A shell argument establishes scaling rather than a universal prefactor.

For a Planck photon distribution, the hot-electron Kompaneets equation gives $\Delta n=yx{e}^x(e^x-1{)}^{-2}[x\coth (x/2)-4]$, where $x=h\nu /(k_B{T}_{\gamma })$. The equivalent thermodynamic temperature perturbation has the displayed form, while the Rayleigh-Jeans law gives $\Delta T/T=-2y$. The distortion changes sign at $x\simeq 3.830$. This first-order expression assumes a small Compton y parameter and nonrelativistic thermal electrons.

Release of electrons from a heated material; one source of electrons in an X-ray tube.

Heating a metal gives some electrons enough energy to escape its surface. A heated diode cathode can therefore supply an electric current across a vacuum gap to a collecting diode anode. This makes temperature a control for the supply of electrons. Conversely, measuring that supply at several temperatures lets us estimate the surface work function: the energy barrier measured from the Fermi level to the nearby vacuum level. A Richardson plot turns the strong exponential function dependence into a slope.

Three questions must be kept separate: how many electrons cross the surface, how many traverse the gap, and how many the collecting diode anode records. The Richardson-Dushman law answers the first under specified surface assumptions. Space charge can obstruct the second, while incomplete collection can alter the third.

The coefficient $A$ has dimensions of electric current density divided by squared temperature. For ideal nonrelativistic particles with the free electron mass, two spin states, a planar uniform surface, and unit escape transmission, the thermionic tail-flux derivation giveswhere $e$ is the positive elementary charge and $h$ is the Planck constant. The two area units differ by a factor of $10^4$. A fitted $A$ need not equal $A_0$: reflection at the surface, the electronic band structure, and uncertain emitting area affect the inferred coefficient. A constant escape transmission $\tau$ would give $A=\tau A_0$ in this simple model; an energy-dependent transmission generally requires an integral instead.

The name joins Owen Willans Richardson with Saul Dushman. Dushman's 1923 paper develops a thermodynamic treatment of electron emission analogous to evaporation and evaluates the prefactor using quantum ideas.^[ref]

Consider a planar surface with outward normal along $z$. For an auxiliary equilibrium electron gas just outside it, let $n_v$ be the total number density, counting both signs of $v_z$. Its Maxwell-Boltzmann velocity distribution has normal-component densityAn electron with $v_z>0$ crosses more area per unit time when it moves faster. The outgoing number flux therefore uses $v_z{f}_z$, rather than just the probability of $v_z>0$:This integral contributes only $T^{1/2}$. Holding a classical bulk number density fixed and imposing a normal kinetic energy threshold $W$ would instead give $\Gamma =n\sqrt{k_{B}T/(2\pi m_e)}{e}^{-W/(k_{B}T)}$; that argument alone does not produce $T^2$.

For the ideal surface model, measure external electron energies from the metal's chemical potential $\mu$. A freely propagating external state has $E-\mu =W+p^2/(2m_e)$. When $W\gg k_{B}T$, its Fermi-Dirac distribution reduces to the Maxwell–Boltzmann statistics tail. Counting two spin states per phase space cell of volume $h^3$ givesEach of the three Gaussian integrals supplies $T^{1/2}$. Combining this $T^{3/2}$ state count with the $T^{1/2}$ crossing factor givesThe auxiliary atmosphere supplies an equilibrium state-counting argument; an operating collector removes the outgoing electrons. It need not maintain a real equilibrium gas throughout the gap. Unit transmission and the ideal free electron state count are substantive assumptions, so this calculation does not determine an arbitrary real surface's measured $A$.

In particular, electrons inside a metal generally form a degenerate Fermi gas. A classical distribution for the escaping tail does not imply a classical distribution for the whole interior. Richardson himself emphasized this distinction in his lecture.^[ref]

The Thomson charge-to-mass experiment infers the magnitude of an electron's charge-to-mass ratio by combining electric and magnetic deflection of cathode rays. The central idea is to determine speed separately from magnetic curvature: a known speed turns the observed gyroradius into a measurement of $e/m$.

In 1897, J. J. Thomson used two grounded slits in an evacuated glass tube, then aluminium plates $5\mathrm{cm}$ long, $2\mathrm{cm}$ wide and $1.5\mathrm{cm}$ apart. A battery supplied their electric potential difference. Deflected cathode rays produced a phosphorescence patch on the glass; an external scale and movable luminous pointer recorded its position in darkness. Greater evacuation reduced gas electrical conductivity that screened the electric field; reversing plate polarity reversed deflection. External coils supplied the magnetic field, calibrated by induction against larger coils. He adjusted equal electric and magnetic deflections. ^[ref]

His comparison included air, hydrogen, carbon dioxide, and aluminium versus platinum cathodes, supporting a common carrier. He explicitly noted uncorrected magnetic bending outside the plates. His tables give $m/e$, the reciprocal of the magnitude used here. The inferred ratio alone could not separate electric charge from mass. ^[ref]

To connect this readout with the Lorentz force, derive a simple nonrelativistic model in modern SI notation. Let the uniform deflecting region have length $\ell$, followed by a field-free drift of length $L$. Assume a narrow speed distribution, small deflection angles, negligible collisions, and matching electric and magnetic field profiles. For an electron with speed $v$, the electric acceleration has magnitude $eE/m$, the transit time is approximately $\ell /v$, and the subsequent drift adds exit transverse velocity times $L/v$. Thus the displacement magnitudes measured separately with only electric field or only magnetic field areThe electric expression follows from constant transverse acceleration; the magnetic expression uses its small-deflection approximation. Equal displacement magnitudes then giveThis is an ideal derivation of the comparison principle, not a numerical reanalysis of the historical tube. Different exterior magnetic field profiles invalidate cancellation of the common geometrical factor. The comparison determines $e/m$ through deflection and the known apparatus dimensions; it does not require a downstream semicircle.

For a second, deliberately different teaching geometry, combine a velocity selector with a separate region containing only a uniform magnetic field $B_a$ perpendicular to the selected velocity. Here $B_s$ belongs to the velocity selector, while $B_a$ belongs to the analyzer: they need not be equal. The magnetic Lorentz force supplies the centripetal acceleration, soIf the analyzer turns the beam through a semicircle before it hits a screen in the entry plane, the separation $D$ between entrance and impact is the diameter: $r=D/2$. For a shorter observed arc, the screen displacement is not generally a radius; fit the appropriate circle geometry. This two-stage velocity selector and curvature apparatus is an original ideal teaching model, not a claim that Thomson's Fig. 2 used a separate selector and semicircular analyzer.

For a worked example, stipulate $E=2.00\times 10^3\mathrm{V}{\mathrm{m}}^{-1}$, $B_s=1.00\times 10^{-3}\mathrm{T}$, $B_a=2.00\times 10^{-4}\mathrm{T}$, and $D=0.1136\mathrm{m}$. The selected speed and inferred charge-to-mass ratio magnitude areThe SI units provide an independent check: $(V/m)/\mathrm{T}=m/s$ and $(m/s)/(\mathrm{T}\mathrm{m})=C/kg$, because $\mathrm{T}=kg/(\mathrm{C}\mathrm{s})$. The charge-to-mass ratio of the electron is negative; this calculation gives its magnitude. The bending direction gives the sign of the electric charge once the magnetic field direction is known. Scaling both $e$ and $m$ by the same positive factor preserves every displayed trajectory, so an independent electric charge measurement is required to determine mass separately.

The curvature formula assumes nonrelativistic momentum $p=mv$. At higher speeds, the relativistic cyclotron orbit gives $p=\gamma mv=e{B}_{a}r$, hence $e/m=\gamma v/(B_{a}r)$, where $\gamma$ is the Lorentz factor; the ideal straight-through selection $v=E/B_s$ still follows from vanishing Lorentz force. The stipulated speed is about $0.0067$ times the speed of light, making $\gamma -1$ about $2.2\times 10^{-5}$. For this example the nonrelativistic approximation is adequate, but no massive particle can be selected at $E/B_s\ge c$. Finite slit widths, collisions with residual gas, uncertain magnetic field calibration, and fields outside the ideal boundaries limit a real inference; repeat at several selected speeds and analyzer strengths to test consistency of $e/m$ rather than trusting one spot position.

A free nonrelativistic charge driven by a weak plane electromagnetic wave has peak acceleration $|q|E_0/m$. Its time-averaged Larmor formula power is $q^4{E}_0^2/(12\pi {ϵ}_0{m}^2{c}^3)$. Divide by incident light intensity ${ϵ}_{0}c{E}_0^2/2$ to obtain the displayed cross section, independent of frequency. For an electron, it is $8\pi r_e^2/3$. The approximation assumes negligible Compton recoil kinematics and small driven velocity compared with the speed of light.

Alongside the 1927 Davisson–Germer experiment, George Paget Thomson developed an independent transmission test. Alexander Reid first used celluloid; Thomson used thin metal films. A narrow electron beam crossed a film about $10^{-6}\mathrm{cm}$ thick, reaching a photographic plate perpendicular to the beam. Rings appeared. Accelerating voltages of $20$–$60\mathrm{kV}$ set the de Broglie wavelengths; metal structures known from X-ray diffraction allowed quantitative comparison. A downstream magnetic field deflected ring-forming electrons like those passing through holes, identifying the carriers as electrons rather than X-rays. Celluloid's uncertain structure limited its quantitative interpretation.^[ref]

The Davisson–Germer experiment uses crystal reflection and angular electric current scans; this transmission camera records a spatial pattern behind the specimen. Both probe interference in electron diffraction, with different specimen preparations and readouts. A developed photographic plate records accumulated detections, not a direct image of the wavefunction.

Why rings? In an ideal film containing many small crystals with uniformly distributed orientations, a family of lattice planes with spacing $d$ can satisfy Bragg's law at every azimuth about the incident beam. First-order elastic scattering then forms a cone of opening half-angle $2\theta$, where $2d\sin \theta =\lambda$. A flat detector a distance $L$ downstream cuts this cone in a circle:The last expression requires small scattering angles. It explains a useful control: hold the specimen and camera distance fixed, increase accelerating voltage, and the ring radii should shrink with the electron accelerating-voltage wavelength. It predicts positions, not intensities: preferred crystal orientations, film thickness and repeated quantum scattering can alter a real pattern.

For a chosen example, take $L=0.200\mathrm{m}$ and $d=0.200\mathrm{nm}$. The exact relativistic electron accelerating-voltage wavelength gives $\lambda =8.589\mathrm{pm}$ at $20\mathrm{kV}$ and $4.176\mathrm{pm}$ at $80\mathrm{kV}$. The corresponding exact ring radii are $8.594\mathrm{mm}$ and $4.176\mathrm{mm}$. These are calculated examples with chosen spacings and camera distance, not Thomson's measurements. The higher voltage lies outside the range quoted from his lecture, deliberately illustrating a prediction rather than reconstructing his data.

Three-body recombination captures an electron while a second free electron carries away energy, the inverse of direct electron-impact ionization. Its rate scales as the parent-ion density times the square of electron density. It is usually negligible compared with two-body radiative recombination and dielectronic recombination in the low-density solar corona, but matters in sufficiently dense plasmas.

For the uniform noninteracting electron model with two spin states and kinetic energy of nonrelativistic particles $E=p^2/(2m_e)$, count all available one-particle states up to an energy $E\ge 0$:This extensive density of states has units of inverse energy; $g(E)/V$ is the density of states per unit volume. Availability and occupation are different: $g(E)$ is positive above $E_F$, but at zero temperature those states are empty. The occupied energy distribution is $g(E)f(E)$, with the Fermi-Dirac distribution reduced to $f(E)=1$ below $E_F$ and $f(E)=0$ above it. The convention at the exact boundary does not affect a continuum integral.

With the dimensionless quantity $x=E/E_F$, the occupied count per $dx$, divided by the total particle number, is $3\sqrt{x}/2$ for $0<x<1$. Its area is one. Weighting this by $E$ gives the internal energy and its mean per particle:Thus the highest occupied energy and the mean energy differ. For the Fermi energy of a model electron gas, $U/N\simeq 1.0155\mathrm{eV}$. At fixed $N$, $E_F\propto V^{-2/3}$, so the derivative $P=-(\partial U/\partial V{)}_N$ gives the electron degeneracy pressure $P=2n{E}_F/5$. The same state-counting geometry therefore predicts a nonzero pressure at zero temperature.

For tree electron-Positron annihilation into three photons, all six orders along the fermion line must be summed. Write $D(r)=/r-m$ and $S(r)=D(r{)}^{-1}=(/r+m)/(r^2-m^2)$. Contracting an internal photon with its momentum gives $S(R)/qS(r)=S(R)-S(r)$ when $q=r-R$. Contracting an end photon removes its adjacent Dirac propagator by the external Dirac equation, with a minus sign at the electron end and a plus sign at the positron end. For either fixed order of the two uncontracted photons, the three insertion positions therefore telescope to zero. Summing their two orders proves the Ward identity. The mass terms in both remaining Dirac propagators are essential when $m\ne 0$. The result expresses the invariance of the scattering amplitude under ${ϵ}_{\ell }\mapsto {ϵ}_{\ell }+c{q}_{\ell }$.

For an original classical calculation of the return step, neglect the residual ion's Coulomb force after release. Take a spatially uniform driving electric field $E(t)=E_0\cos \omega t$, an electron of electric charge $-e$ and mass $m$, and zero initial position and velocity at $t_i$. With $\alpha =\omega t_i$ and $\beta =\omega t$, integrating Newton's second law givesA return requires $\beta >\alpha$ and $x(t)=0$. Define the cycle-averaged free-electron quiver kinetic energySearching the first-return solutions for release phases $0<\alpha <\pi /2$ gives a maximum near $\alpha =0.3134$ and $\beta =4.3990$, with $K_{\mathrm{return}}\simeq 3.1731U_p$. For a chosen ionization energy $I_p$, recombination into the initial state can therefore emit a photon with energy up to approximately $I_p+3.17U_p$ in this model. This is an upper return-energy scale, not a prediction of the emitted yield. It omits release probabilities, quantum interference, propagation, phase matching and depletion of the neutral gas.

The calculation explains why a long driving wavelength can increase the available return energy: at fixed $E_0$, $U_p$ grows as ${\omega }^{-2}$. It does not say that conversion efficiency improves. The electron must return with a probability amplitude for emission that survives the microscopic dynamics and adds coherently across the gas.

An effective low-energy lattice model obtained from a strongly repulsive Hubbard model near one electron per site. The retained Hilbert space excludes double occupancy. Its usual nearest-neighbor Quantum Hamiltonian has projected hopping $-t{\sum }_{⟨ij⟩,\sigma }({\tilde{c}}_{i\sigma }^{†}{\tilde{c}}_{j\sigma }+h.c.)$ and antiferromagnetic superexchange $J{\sum }_{⟨ij⟩}({\mathbf{S}}_i\cdot {\mathbf{S}}_j-n_i{n}_j/4)$. Virtual intermediate states with one double occupancy cost energy $U$, giving $J=4t^2/U$ to leading order. Three-site hopping terms of the same expansion order are omitted in this simplified version.

At half filling the projected hopping is blocked. Doped holes restore charge motion coupled to spin correlations. Nearest-neighbor singlet pairing can support a d-wave bond order ansatz on a square lattice, but the Quantum Hamiltonian alone is not a proof of a unique superconducting ground state for all parameters.

The Tolman electronic parameter compares ligand electronic influence using the symmetric CO stretching band of a matched metal-carbonyl probe, classically $\mathrm{Ni}(\mathrm{CO}{)}_3\mathrm{L}$. Lower frequency commonly indicates greater net electron release into the remaining CO ligands. It combines donation and acceptance and is not a direct measurement of octahedral splitting.

A researcher in semiconductor optics whose team observed Rydberg excitons in cuprous oxide in 2014. These bound electron–electron hole pairs reached principal quantum numbers as high as 25 and extended over micrometers, allowing a crystal to host excitations resembling highly excited hydrogen atoms.^[ref]