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The Fermi wavevector is the radius of the occupied sphere in wavevector coordinates for an isotropic Fermi gas. In the uniform two-component electron model, $k_F=(3{\pi }^{2}n{)}^{1/3}$ and $E_F={\hslash }^2{k}_F^2/(2m_e)$ for nonrelativistic particles. Its units are inverse length, whereas the Fermi momentum has units of momentum.

The sandwich compound $\mathrm{Fe}({\eta }^5-{\mathrm{C}}_5{\mathrm{H}}_5{)}_2$, conventionally Fe(II), $d^6$, with 18 valence electrons.

Field electron emission releases electrons from matter through electric-field-assisted quantum tunnelling at the surface. The emission mechanism and its rate are distinct from the subsequent acceleration and Larmor formula radiation of an electron already travelling through an approximately uniform electric field.

For a two-dimensional electron gas in a perpendicular magnetic field of magnitude $B>0$, the filling factor is the electron number per unit area $n_s$ divided by the orbital degeneracy per unit area $eB/h$ of one spin branch of a Landau level. Here $e$ is the positive elementary charge and $h$ the Planck constant. The nominal filling factor varies continuously with $n_s/B$; it need not be an integer.

For a dispersion relation $E(k)=E_0+t\cos (ka)$, $t<0$, occupied states at zero temperature form the interval $|k|<k_F$ in the first Brillouin zone. Its occupied fraction is $f=2k_F/(2\pi /a)=k_{F}a/\pi$. A spin-degenerate band therefore holds $2f$ electrons per site. If the band is independent of the other two wavevector coordinates, its Fermi surface consists of two planes. For $t>0$ the occupied region is instead near the zone edges, bounded by $|k|=\pi (1-f)/a$.

A battery's charge reservoir means a finite amount of electric charge can pass through the external electrical circuit before its usable chemical state is depleted. It does not mean a large net electric charge is sitting inside the casing. For one net cell chemical reaction transferring $n$ electrons per reaction event, an ideal reaction inventory obeyswhere $F$ is the Faraday constant and $\xi$ the extent of reaction in moles, assuming all the electric current serves that reaction. At fixed temperature and pressure, its reversible electromotive force obeys ${\Delta }_{r}G=-nF\mathcal{E}$: a decrease in Gibbs free energy provides the maximum reversible electrical work per mole. The electromotive force also depends on composition; the Daniell cell illustrates this with thermodynamic activity.^[ref]

As an original finite-reservoir exercise, stipulate usable $Q_{\max }=360\mathrm{C}$ and hold the preceding $3\mathrm{V}$, $1\Omega$, $4\Omega$ parameters constant until an abrupt cutoff. Then $t=Q_{\max }/I=600\mathrm{s}$, available electrical work is $\mathcal{E}{Q}_{\max }=1080\mathrm{J}$, load work is $V_L{Q}_{\max }=864\mathrm{J}$, and internal dissipation is $216\mathrm{J}$. This explicitly finite idealization predicts ten minutes, not an experimentally measured runtime. Real accessible charge depends on discharge conditions and the chosen terminal-voltage cutoff. Charge alone is also not an energy: converting it to delivered work requires the terminal electric potential difference throughout discharge.

Take $N$ equally spaced scattering sites at $x_j=jd$ on a line, a normally incident plane wave, equal site scattering amplitudes, and observation far enough away that outgoing directions are effectively parallel. Retain only one scattering event per amplitude, as in the Born approximation. For angle $\varphi$ from the outward normal, write the adjacent-site quantum phase difference as $\delta =kd\sin \varphi$, with $k=2\pi /\lambda$; reversing its overall sign leaves the intensity unchanged. Summing a geometric series givesAt $\delta =2\pi m$ use the continuous limit $I/I(0)=1$, rather than treating $0/0$ as a missing peak. The integer $m$ is the diffraction order: $d\sin \varphi =m\lambda$. This is a surface-row spacing, not the plane spacing in Bragg's law. The plotted intensity is normalized to its central value; it is not an integrated detection probability or an absolute collector electric current.

At $100\mathrm{V}$, $\lambda \simeq 0.12264\mathrm{nm}$ and the positive first-order angle is $\arcsin (\lambda /d)\simeq 29.4^{\circ }$. At $400\mathrm{V}$, the wavelength halves and this angle becomes about $14.2^{\circ }$. Increasing $N$ narrows each peak through more selective interference, while increasing momentum brings a fixed-order peak nearer the normal. Unequal site amplitudes, multiple scattering, a three-dimensional crystal lattice structure factor, and finite detector acceptance can change this simple picture; it is an explanatory model with hypothetical inputs.

Neglecting interactions and thermal pairs, the Fermi-Dirac distribution $f(p)=[\exp ((\epsilon (p)-{\mu }_{\mathrm{kin}})/(k_{B}T))+1{]}^{-1}$ with kinetic energy $\epsilon (p)=\sqrt{m_e^2{c}^4+p^2{c}^2}-m_e{c}^2$ determines the electron contribution to a stellar equation of state. The kinetic chemical potential is fixed by the electron number density, not an independently chosen pressure fraction:The electron kinetic internal energy per volume is $({\pi }^2{\hslash }^3{)}^{-1}{\int }_0^{\infty }{p}^2\epsilon (p)f(p)dp$. In the nondegenerate limit $P_e=n_e{k}_{B}T$; at zero temperature the integrals become the equation of state of a cold electron gas. These are limits of one component and must not be added together as independent electron pressures. When electron-positron thermal pair abundance matters, both charge species must be included, with charge neutrality fixing their net density.

In the envelope approximation, a quantum well confines electron and hole motion perpendicular to the plane. Averaging their screened attraction over the confinement envelopes gives the displayed effective in-plane potential. Its narrow-well limit approaches the ideal two-dimensional exciton; finite width softens the short-distance attraction. Dielectric mismatch, nonparabolic bands and strong electron–hole confinement mixing require more than this single-subband model.

Work in natural units with $c=1$, neglect the electron mass, and put the initial proton at rest with mass $M$. Let $k$ and $k^{\prime }$ be the initial and final electron four-momenta, $q=k-k^{\prime }$, and $Q^2=-q^2>0$. Define the laboratory energy transfer ${\nu }_{\mathrm{lab}}=E-E^{\prime }=P\cdot q/M$. This ${\nu }_{\mathrm{lab}}$ has units of energy; the invariant $P\cdot q$ has units of energy squared. The Bjorken scaling variable and the final hadron invariant mass obeyThese follow by squaring $k-k^{\prime }$ and $P+q$, using $k^2=k^{\prime 2}=0$ and $P=(M,0)$. Solving the Bjorken x definition gives the outgoing energy and then the scattering angle:Alternatively, substituting $Q^2=4E{E}^{\prime }{\sin }^2(\theta /2)$ before solving gives $E^{\prime }=E/[1+2E{\sin }^2(\theta /2)/(Mx)]$. A chosen $x,E,Q^2$ must satisfy $E^{\prime }>0$ and $0\le {\sin }^2(\theta /2)\le 1$. In addition, $W$ must exceed the lightest allowed inelastic hadron threshold; positive $W^2$ alone is insufficient.

For an original numerical example, take $M=0.938\mathrm{GeV}$, $x=0.3$, $E=15\mathrm{GeV}$ and $Q^2=3{\mathrm{GeV}}^2$. Then ${\nu }_{\mathrm{lab}}\simeq 5.330\mathrm{GeV}$, $E^{\prime }\simeq 9.670\mathrm{GeV}$, $\theta \simeq 8.25^{\circ }$, and $W\simeq 2.807\mathrm{GeV}$. Varying $2\le Q^2/{\mathrm{GeV}}^2\le 5$ and $10\le E/\mathrm{GeV}\le 20$ at this same Bjorken x keeps every setting physical and gives $W^2>5.54{\mathrm{GeV}}^2$, above the proton-pion threshold and the stricter chosen continuum selection $W>2\mathrm{GeV}$. These are calculated settings, not historical measurements.

A screen whose fluorescence makes the spatial distribution of incident electrons or other radiation visible.

The formal charge of a chemical atom in a Lewis structure assigns it all of its nonbonding electrons and half of the electrons in its connections:Here $v$ is the neutral chemical atom's valence electron count, $n$ is the number of nonbonding electrons assigned to it, and $b$ is the number of bonding electrons incident on it. Count electrons, not electron pairs: two lone pairs mean $n=4$, and two double connections mean $b=8$. The number $\mathrm{FC}$ gives electric charge in units of the positive elementary charge. Equal splitting is a bookkeeping convention, regardless of the actual electron density. See the IUPAC Gold Book, formal charge.

For $\mathrm{O}=\mathrm{C}=\mathrm{O}$, each oxygen has $v=6$, $n=4$, $b=4$, giving formal charge $6-4-4/2=0$. Carbon has $v=4$, $n=0$, $b=8$, giving formal charge $4-0-8/2=0$. For ${\mathrm{O}}^{-}-\mathrm{C}\equiv {\mathrm{O}}^{+}$, the left oxygen has formal charge $6-6-2/2=-1$, carbon still has formal charge zero, and the right oxygen has formal charge $6-2-6/2=+1$. The net electric charge is zero in either drawing.

The Lewis structure with no separated formal charges is the dominant contributor in this elementary comparison; the charge-separated drawing is a less important contributor, rather than an equally dominant alternative. This is a qualitative selection rule, not a calculation of energy or numerical chemical resonance weights. See Flowers et al., Chemistry 2e, 7.4: comparing carbon dioxide structures. A zero formal charge does not imply zero partial atomic charge: the latter depends on partitioning the actual electron density, which a Lewis structure does not supply.

As a short consistency derivation, summing $n+b/2$ over all chemical atoms counts every drawn electron once: a nonbonding electron is assigned once, and a bonding electron contributes half at each of its two endpoints. Thuswhere $N_{\mathrm{valence}}$ is the total drawn valence electron count. The right side is the net electric charge in units of elementary charge. This catches a wrong budget even when individual octet rule checks look plausible.

A quasihole adds one zero at $\eta$ per electron; its local deficit carries positive electric charge $e/m$.^[ref]

Here is a transport counting outline, conditional on an incompressible quantum Hall fluid. Inserting one magnetic flux quantum $\Delta \Phi =h/e$ slowly compared with $\hslash /\Delta$, where $\Delta$ is the excitation energy gap, produces a circulating electric field by Faraday's law. The transverse sheet conductivity sends a radial electric current whose time integral has magnitudeThe opposite electric charge moves to the boundary, so the total electric charge remains conserved. The counted deficit belongs to a collective quasiparticle; no individual electron has been cut into pieces. The argument assumes the gapped fluid and its Hall response, and is not a consequence of the finite two-electron normalization integral.

The fractional quantum Hall effect is quantized transverse sheet conductivity at fractional filling factors, produced by interacting electrons. The Laughlin wavefunction describes the sequence $\nu =1/m$, with odd $m$, including $\nu =1/3$.^[ref]

For an original ideal readout example, take a Hall bar on the $\nu =1/3$ plateau with negligible longitudinal sheet conductivity. The quantum Hall conductivity tensor inversion givesHere $R_K$ is the von Klitzing constant; $R_H$ denotes transverse electrical resistance, rather than the bulk Hall coefficient. Thus a fractional sheet conductivity corresponds to a larger transverse electrical resistance. The fraction labels a collective quantum state: it does not mean that each electron has acquired one third of its usual electric charge.

A useful Franck-Hertz energy-loss model separates three operations: accelerating an electron, losing a discrete energy $\Delta$ in an inelastic collision, and selecting the surviving electrons with a retarding collector. The collector measures an accepted fraction of an outgoing energy distribution. This makes the experiment a small inverse problem: the electric current depends on both the collision physics and the readout.

For a deliberately simple one-channel model, let $u\ge 0$ be the incoming kinetic energy, and let $p(u)$ be the probability that one encounter excites the chemical atom. Set $p(u)=0$ for $u<\Delta$ and $0\le p(u)\le 1$ otherwise. The conditional probability distribution of outgoing kinetic energy $E$ is the discrete loss kernelHere $\delta$ is the Dirac delta function. The kernel has unit integral, no negative outgoing energy, and mean outgoing kinetic energy $u-p(u)\Delta$. It neglects the small recoil loss in an elastic collision, other excitation channels and ionization. The threshold permits excitation; it does not require $p(u)=1$ above threshold. In particular, choosing a single $\Delta =4.9\mathrm{eV}$ here isolates the characteristic channel discussed above without claiming it is mercury's lowest possible excitation.

The Franck-Hertz experiment connects a reproducible loss of electron kinetic energy with excitation of a chemical atom. The original mercury-vapor experiments were performed during 1912–1914 while investigating ionization; they were not designed as a test of the Bohr model. Their later interpretation distinguished internal excitation from removal of an electron. ^[ref].

In a modern mercury tube, a heated cathode emits electrons and a positive perforated grid accelerates them through Hg vapor. A collector slightly below the grid's electric potential rejects slow electrons. The oven controls vapor density and hence the mean free path. Keep heating and retarding settings fixed, sweep the accelerating electric potential difference $V$, and record collector electric current $I(V)$. These measured quantities indirectly probe atomic excitation. ^[ref].

The curve shows repeated current minima, typically about $4.9\mathrm{V}$ apart. In the collision model, an electron gains approximately $eV$ of kinetic energy, with $e$ the positive elementary charge. An inelastic collision transfers about $4.9\mathrm{eV}$ to internal excitation. Near the grid, this can prevent the electron crossing the retarding region, lowering collected electric current. Further acceleration restores collection; repeated excitation losses produce further minima. These represent multiple collisions, rather than equally spaced atomic energy eigenvalues. Contact potentials shift the first minimum, so compare several successive spacings instead of identifying the first dip's voltage with the excitation cost. ^[ref].

The optical comparison supplies a second observable: mercury radiation at a wavelength near $253.6\mathrm{nm}$ was detected through a window in the tube. Its inferred photon energy $hc/\lambda \approx 4.89\mathrm{eV}$ agrees with the characteristic collision loss, with $c$ the speed of light. This agreement supports the interpretation as a transition between internal atomic energies. It does not make the collector trace a complete spectrum. In particular, the lowest excited mercury level, labeled $6{}^3{P}_0$, is metastable because an electric-dipole selection rule forbids its direct transition to the ground state; the nearby $6{}^3{P}_1$ level emits the quoted ultraviolet line. Collision excitation and optical detection therefore have different sensitivities to the available excited states. ^[ref].

Initial electron kinetic energy spread and varying collision positions broaden real minima. Repeat traces at different retarding settings and oven temperatures to distinguish excitation from collection effects. This probes a discrete energy-transfer channel, rather than every energy eigenspace, its dimension, or an absolute ground-state energy. ^[ref].

James Franck and Gustav Ludwig Hertz presented their paper on 24 April 1914. Its heated platinum wire sat inside a cylindrical platinum mesh, with a collector 1–2 mm beyond it. A heated bath and pumping prepared the vapor; guard rings suppressed leakage. They differentiated retarding-electric current scans to study outgoing energies.^[ref]

Their 1925 Nobel Prize preceded their lectures of 11 December 1926. James Franck described choosing pressure for many collisions before the mesh but few afterward.^[ref] Gustav Ludwig Hertz explained how a field-free collision region narrowed excitation thresholds.^[ref]

Define a signed offset $V_c$ so that the accelerating region adds kinetic energy $e(V-V_c)$. This convention concerns this gap; collector contact fields belong in the effective $V_r$. An electron emitted with initial kinetic energy $E_0$ and undergoing $n$ identical excitation losses reaches the grid withEach loss excites a fresh chemical atom initially in its ground state. These are repeated payments of the same excitation cost. The index $n$ counts events along an electron trajectory, not the ordinal number of an atomic excited state. The outgoing energy also does not locate the last collision: after an early collision an electron can regain enough kinetic energy to cross the barrier.

At fixed $V$, let $g(E_0)$ be the normalized initial-energy probability density function within the ensemble of electrons reaching the grid, and $w_n(E_0,V)$ the conditional probabilities for $n$ losses within that ensemble. Then the ideal collection law isTake ${\sum }_n{w}_n=1$ and exclude paths with insufficient kinetic energy for a loss at its actual collision position. The weights contain collision positions and rates, mean free paths and angular acceptance. This equation is a bookkeeping model, not a derivation of those weights. Multiplying all incident particles by one constant transmission is an additional approximation; real transmission and $I_0$ can vary with $V$.

One analytical extreme assumes $E_0=0$, one-dimensional motion and an immediate loss whenever the kinetic energy reaches $\Delta$. With $d=\Delta /e$, $x=V-V_c\ge 0$ and $0<V_r<d$, the number of losses and remaining kinetic energy areHere $\lfloor \cdot \rfloor$ is the floor function. Collection switches off for an interval of width $V_r$ after each reset, then recovers; the pattern repeats every $d$. For $d=4.9\mathrm{V}$, $V_c=1.2\mathrm{V}$ and $V_r=1.0\mathrm{V}$, an applied $V=11.4\mathrm{V}$ gives $n=2$, $r=0.4\mathrm{V}$ and rejection, whereas $V=12.2\mathrm{V}$ gives $n=2$, $r=1.2\mathrm{V}$ and collection. With $V_r=0$ the dips disappear in this idealization. With no collisions, there is only a single collection onset. Both limits expose the separate roles of loss and filtering.

To draw a smooth illustrative curve, extend this periodic rule mathematically to all $x$ and convolve it with a chosen normal distribution of voltage jitter, of standard deviation $\sigma$:Here $\Phi$ is the standard normal cumulative distribution function. Each summand is the smoothed indicator of the accepted interval $kd+V_r<x<(k+1)d$. The result satisfies $0\le C_{\sigma }\le 1$ and has period $d$. For the figure choose $\sigma =0.18\mathrm{V}$ and plot only well beyond the startup region. This is a chosen phenomenological curve, not a Boltzmann equation transport prediction or a fit to measured electric current. Its symmetric troughs center on $V_c+nd+V_r/2$, rather than the reset voltages $V_c+nd$. Finite excitation rates, several loss channels, space charge, initial angular and energy spreads, and real electrode geometry can change the trough shape and phase; a single convolution does not solve these effects.

For electrons directed normally toward an ideal planar collector, a retarding electric potential difference $V_r>0$ imposes the barrier $e{V}_r$. Write its transfer as $T(E)=H(E-e{V}_r)$, where $H$ is the Heaviside step function and $e$ the positive elementary charge. Thus a perfectly sharp loss need not produce a sharp drop in electric current: the drop also depends on where the outgoing energy lies relative to the barrier. For an oblique incident electron at angle $\theta$ to the normal, the same planar model instead tests $E{\cos }^2\theta >e{V}_r$; angular scattering can change collection even at unchanged total kinetic energy.

For example, choose a monoenergetic incoming beam with $u=5.6\mathrm{eV}$, $p=0.75$ and $\Delta =4.9\mathrm{eV}$. After one encounter, $25\%$ have $5.6\mathrm{eV}$ and $75\%$ have $0.7\mathrm{eV}$. A $1.0\mathrm{V}$ retarder collects only $0.25I_0$, whereas a $0.5\mathrm{V}$ retarder collects $I_0$; here $I_0$ is the incident electric current with fixed geometrical acceptance and no background. The mean outgoing kinetic energy is $1.925\mathrm{eV}$, independently checked by $5.6-0.75(4.9)$. These are chosen inputs and calculated fractions, not historical measurements.

More generally, let $f(E;V)$ be the normalized outgoing probability density function at a fixed accelerating electric potential difference $V$, for the normally incident idealization. A retarding scan givesThe second equation follows by differentiating the lower integration limit, with Dirac delta functions understood for sharp beams. It explains why an integrated collector readout can reveal an energy distribution. With angles present it probes the distribution of normal kinetic energy; finite instrumental resolution smooths it, and numerical differentiation amplifies noise. Keep the incoming beam fixed during this scan. A scan of $V$ changes $f$ itself and cannot be differentiated by the same inversion formula.

In a one-carrier Drude model, the Hall coefficient is $R_H=1/(nq)$. Combining its inferred electron density with atomic density ${\rho }_m{N}_A/M$ gives the displayed number of mobile electrons per atom. The molar mass and mass density must be supplied independently. Multiple bands, anisotropic carrier motion and Hall factors limit this simple interpretation; field reversal helps distinguish the transverse response from voltage offsets.

For a two-dimensional free electron gas with two spin states and $z$ electrons per square cell of side $a$, counting occupied wavevectors gives $z/a^2=2\pi k_F^2/(2\pi {)}^2$. The square first Brillouin zone has half-width $\pi /a$. A monovalent circle lies inside its faces; a divalent circle crosses its faces but not its corners. A real periodic potential can deform and reconnect the crossings, as in Fermi-surface reconstruction in a weak square potential; valence alone is insufficient to prove insulation.