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An absorption edge is an abrupt change in photon absorption as incident photon energy crosses an electronic excitation threshold. For inner-shell photoionization, a new vacancy-production channel opens near the shell's binding energy. This threshold concerns incident energy; a Characteristic X-ray concerns a subsequent difference of atomic energies. They are different observations. Barkla's lecture explicitly treats characteristic emission as requiring a critical excitation condition, while discussing characteristic radiation separately from scattering.^[ref]

In the stipulated X-ray fluorescence model, the K threshold is $10.0\mathrm{keV}$, whereas the K lines are $8.5$ and $9.7\mathrm{keV}$. An absorber made of a different material could have its own absorption edge between those line energies, preferentially removing the upper line. Actual near-edge structure and attenuation values require material data; the following deliberately chosen coefficients are not such data.

An independent site has one empty state and an occupied set with binding energy $-ϵ$ and vibrational partition function $z_s$. Its occupied probability is $\theta =z_s{e}^{\beta (\mu +ϵ)}/[1+z_s{e}^{\beta (\mu +ϵ)}]$. The denominator enforces the one-molecule-per-site constraint even though the vapour may obey classical statistics. Multiple occupancy or lateral interactions require a different model.

Binding energy is the energy required to separate a bound state into specified free constituents. For an electron bound in a chemical atom, neglecting this cost is an assumption of the free-electron Compton scattering model.

For $\rho =k/(2\pi G{r}^2)$ truncated at $r_0$, the enclosed mass is $M(r)=2kr/G$. The shell-assembly integral gives $W=-\int GM(r)dM/r=-G{M}_0^2/r_0$. The origin has divergent density but finite total mass and binding energy.

A Characteristic X-ray comes from an electronic transition filling an inner electron shell vacancy. The emitted photon carries the difference between atomic energies. Vacancy creation by direct electron impact requires kinetic energy above the shell's binding energy, exceeding the line energy. Higher tube voltage need not move the spectral line.^[ref]

An original stipulated shell model illustrates the distinction without using measured atomic data. Let positive binding energies be $B_K=9.0\mathrm{keV}$ and $B_L=1.2\mathrm{keV}$, ignoring relaxation corrections. Filling a K-shell vacancy from the L shell releasesAn $8.5\mathrm{kV}$ tube can energetically emit a $7.8\mathrm{keV}$ bremsstrahlung photon, yet cannot directly remove this K-shell electron. A $12\mathrm{kV}$ tube can create that vacancy by electron impact; if radiative filling occurs, the line is still $7.8\mathrm{keV}$, not $12\mathrm{keV}$. This argument concerns direct electron impact and an initially unexcited target; it does not describe every possible excitation route.

The occupied-to-vacant probability ratio of the adsorption-site grand partition function is $\theta /(1-\theta )=z_s{e}^{\beta ({\mu }_s+ϵ)}$. Taking its logarithm yields the displayed chemical potential. binding energy and vibrational energies must share a consistent reference so that zero-point energy is not counted twice.

A chemical shift in photoelectron spectroscopy is a difference in binding energy for the same kind of core electron in different chemical surroundings, using comparable energy references. Identifying a chemical element assigns its family of lines; resolving chemical shifts within that family asks how its chemical atoms are bonded. For example, distinguishing a metal from its chemically modified surface requires more than detecting that element in both places.

The removal energy compares the initial electronic state with the state left after emission. Redistribution of the remaining electrons changes that final state's energy; a measured shift therefore includes relaxation as well as initial bonding effects.^[ref]

This makes chemical shift in photoelectron spectroscopy useful but conditional. Compare the same electron shell, the same photoelectron binding energy reference, and compatible sample conditions. Agreement with a chemically appropriate reference supports an assignment; a line position by itself need not uniquely determine an oxidation state. An unresolved sum of two chemical components can even have a centroid where neither component lies. The following Gaussian mixture distribution shows exactly how that happens without claiming a material identification.

Consider a single electron of mass $m_e$ and a point atomic nucleus of mass $M$ and positive electric charge $Ze$, with $e>0$ the elementary charge. Use nonrelativistic classical mechanics, omit external fields and interactions with other electrons, and writeHere $\mu$ is the Coulomb two-body reduced mass, ${ϵ}_0$ the vacuum permittivity, and $h$ the Planck constant. The separation $r$ and relative speed $v$ describe the two-particle relative coordinate; neither body is silently pinned in place. The electron's own radius about the centre of mass is $Mr/(M+m_e)$.

For the assumed circular orbit, Coulomb's law provides the required centripetal acceleration, and the imposed orbital angular momentum condition isSubstitute $v=n\hslash /(\mu r)$ into the first equation and cancel $r^{-2}$ to obtainThe kinetic energy is $K=\mu v^2/2=\kappa /(2r)$ and the Coulomb potential energy is $U=-\kappa /r$, with zero energy at infinite separation. HenceThis derives the energy sequence from the stated circular model; the modern Schrödinger equation derivation is the distinct series-termination quantization of the Coulomb radial equation. The radius above scales as $n^2/Z$, whereas the binding energy scales as $\mu Z^2/n^2$. The Bohr radius article supplies the corresponding Coulomb potential length scale. The nonrelativistic assumption requires $v_n\ll c$, with $c$ the speed of light.

For an original numerical example choose $Z=1$, $M=1.67262192369\times 10^{-27}\mathrm{kg}$, $m_e=9.1093837015\times 10^{-31}\mathrm{kg}$, $e=1.602176634\times 10^{-19}\mathrm{C}$, $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, ${ϵ}_0=8.8541878128\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$, and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. Treat these as stipulated constants for the calculation. They give $\mu /m_e=0.99945568$, $r_1=0.0529465\mathrm{nm}$, $E_1=-13.598287\mathrm{eV}$ and $v_1/c\simeq 0.00730$, where $\mathrm{eV}$ denotes an electronvolt. At $n=2$, $r_2=4r_1$ and $E_2=E_1/4$; the ground-state ionization energy is $-E_1$.

Taking $M\to \infty$ gives $\mu \to m_e$. A finite nuclear mass reduces every binding energy by $\mu /m_e$ and increases each separation by $m_e/\mu$. This follows from relative motion, not a fitted correction to individual spectral lines. Replacing the atomic nucleus by a heavier isotope changes the same factor, providing another check of the model.

Compton scattering redistributes photon energy and momentum in a collision with a charged subatomic particle, here an electron. For a free electron initially at rest in the chosen inertial frame, an incident photon of energy $E=hc/\lambda$ leaves with energy $E^{\prime }=hc/{\lambda }^{\prime }$ after deflection through $\theta$. The angle $\theta \in [0,\pi ]$ is between the incident and outgoing photon directions; it is not the electron recoil angle. Four-momentum conservation giveswhere $m_e$ is the electron rest mass and $T_e$ its final kinetic energy. The photon energy loss pays for the electron recoil: neither energy nor momentum disappears.

For an original worked example, stipulate $E=70\mathrm{keV}$, the rounded electron rest energy $m_e{c}^2=511\mathrm{keV}$, and the rounded Compton wavelength ${\lambda }_C=2.426\mathrm{pm}$; $\mathrm{keV}$ means $10^3$ electronvolts. Treat these rounded inputs as the example's constants. They give the following calculated values, with energies rounded to three decimal places:

Since $0\le 1-\cos \theta \le 2$, every outgoing photon has $0<E^{\prime }\le E$ and $T_e\ge 0$. At zero deflection there is no recoil; backscattering transfers the most energy. At fixed $\theta$, the absolute wavelength increase is independent of incident energy, while the fractional energy loss grows with $E/(m_e{c}^2)$. This makes Compton scattering useful for inferring energy transfers from directions in photon detectors.

The free-electron-at-rest assumption matters. An electron in a bound state has binding energy and an initial momentum distribution. Neglecting these is useful when the transferred energy is large compared with the relevant binding energy and the initial momentum is small compared with the resolved momentum transfer. Otherwise the outgoing photon energies can broaden or depart from the stationary-target formula; an initially moving electron can even give energy to the photon.

A material may also scatter X-rays through a coherent elastic scattering channel with an effectively unshifted component. When an intact chemical atom of mass $M$ takes the recoil, the corresponding Compton wavelength $h/(Mc)$ is much smaller than the electron value. The single free-electron curve does not predict that component's weight. More generally, energy versus angle is a kinematic constraint; angular wave intensity requires a differential scattering cross-section, together with target and detector information.

Electron-ion recombination combines a free electron with a positive ion, reducing the number of free charge carriers. The released binding energy must be carried away, for example by a photon or another particle.

In an ionization chamber, recombination before charge collection reduces the measured signal. A simplified local model has loss rate $\alpha n_e{n}_i$, with coefficient $\alpha$ depending on the species and conditions. Increasing the collection field can reduce the time available for recombination, but complete collection must be established experimentally rather than assumed from the applied voltage alone.

The fixed-target production threshold concerns the incident accelerator beam, whereas antiproton mass from momentum and velocity concerns an outgoing secondary beam. To see the distinction, consider a free proton hitting a stationary free proton, producingThe final electric charge is $+2e$ and final baryon number is $3-1=2$, as initially. Approximate all four final rest masses by the same proton rest mass $m$. With metric signature $(+---)$, define the squared centre-of-mass energy by $W^2=c^2(P_1+P_2{)}^2$. The incident proton has total energy $E=m{c}^2+K$, and the target four-momentum is $(mc,0)$. Four-momentum conservation and the mass shell relation giveAt threshold the four final particles have zero relative momentum in their centre-of-momentum frame, so $W_{\mathrm{thr}}=4m{c}^2$. HenceOnly $2m{c}^2$ of new rest energy is created. The larger incident kinetic energy is required because the final system must also carry the initial laboratory momentum. The incident total energy at threshold is $7m{c}^2$; neither this nor $K_{\mathrm{thr}}$ is the selected secondary momentum.

The 1955 authors explicitly qualified the stationary-nucleon threshold: target nucleon motion inside an atomic nucleus can lower the required incident energy.^[ref]

The kinematic reason is visible for two on-shell incoming nucleons:A target momentum directed against the beam increases $W^2$. Applying this illustration to a bound nucleon also requires binding energy and residual-nucleus recoil; the free stationary-target $6m{c}^2$ result is therefore not a universal sharp threshold for a copper target. A low observed secondary momentum is entirely compatible with a much larger accelerator kinetic energy.

In a hydrogen–helium gas with negligible doubly ionized helium, define the nuclear number densities $n_H=n_{H^0}+n_{H^{+}}$ and $n_{He}=n_{H{e}^0}+n_{H{e}^{+}}$. Neglecting nuclear binding energy in the abundance conversion, the baryon number density and helium mass fraction satisfyThe ionization fractions $X_{H^{+}}=n_{H^{+}}/n_H$ and $X_{H{e}^{+}}=n_{H{e}^{+}}/n_{He}$ count ionized nuclei within each species. Charge neutrality gives $n_e=n_H{X}_{H^{+}}+n_{He}{X}_{H{e}^{+}}$, henceThus complete single ionization gives $\mathcal{F}=1-3Y_p/4$, rather than one electron per baryon. This fraction uses a different denominator from $n_e/n_H$. At sufficiently high temperature to doubly ionize helium, the helium contribution must instead count two electrons per such nucleus.

In the Newtonian virial theorem convention, the gravitational radius of a self-gravitating system of total mass $M$ and gravitational potential energy $W<0$ isIt converts global binding energy into a characteristic length without selecting a physical outer boundary. For example, a uniform-density sphere of radius $R$ has $W=-3G{M}^2/(5R)$, so $R_g=5R/3$, larger than its surface radius. For an isolated system in virial equilibrium, $2K+W=0$ gives the mass-weighted mean-square speed $⟨v^2⟩=GM/R_g$ in the centre-of-mass frame.

The same phrase is also used for relativistic lengths proportional to $GM/c^2$, including the Schwarzschild radius. The definition here uses the system's Newtonian binding energy and depends on its mass distribution; specifying the convention avoids confusing these different scales.

Before local expulsion, a linear star-formation law $\psi =\nu {\rho }_g$ without return gives ${\rho }_g={\rho }_0{e}^{-\nu t}$. If $q=\gamma E_{\mathrm{SN}}$ is the energy supplied per unit stellar mass, accumulated energy is $q{\rho }_0(1-e^{-\nu t})$. A fixed binding energy per unit remaining gas mass $B(r)$ gives the threshold $q(e^{\nu t}-1)=B(r)$. This assumes a specified retained gravitational profile and neglects exchange between volumes.

John Cockcroft and Ernest Walton used accelerated protons to produce the disintegration of lithium in 1932. A high-voltage accelerator supplied the projectile beam, while the energetic alpha particles leaving the target provided the reaction signal. The reaction $p+{}^7\mathrm{Li}\to 2\alpha$ releases nuclear binding energy, so the outgoing particles can carry far more kinetic energy than the incident proton. This connected a controlled electrical accelerator with a quantitative nuclear reaction.^[ref]

A physicist.^[ref][ref]

Kai Siegbahn developed electron spectroscopy as a quantitative probe of chemical and electronic structure. In X-ray photoelectron spectroscopy, incident X-rays eject electrons whose measured kinetic energies reveal their binding energies after the instrument's reference offset is accounted for. Small changes of a core-level binding energy reflect the atom's chemical environment. Resolving and calibrating these shifts makes the spectrum sensitive to electronic surroundings as well as elemental identity, linking the spectrometer's design with a physical inference.^[ref]

Physics.^[ref]

The nuclear reactionconserves electric charge and total nucleon count: $3+1=2+2$ and $7+1=4+4$. Its nuclear reaction Q-value is the initial minus final rest energy. It is additional to the incoming beam kinetic energy; the accelerator triggers a rearrangement whose products can carry much more kinetic energy than the incident proton.

Use one consistent set of neutral-atom isotope masses, in unified atomic mass units, from the National Institute of Standards and Technology:These are isotope masses, not element-averaged atomic weights.^[ref] With $\mathrm{u}{c}^2=931.49410372\mathrm{MeV}$,^[ref]The four electron rest masses cancel between sides. Differences of total electronic binding energies do not cancel exactly: this atomic-mass result approximates the bare-nucleus nuclear reaction Q-value at the displayed $0.001\mathrm{MeV}$ scale, not at arbitrary precision. No mixture of rounded nuclear binding energies is needed. At chosen incident kinetic energy $T=0.500\mathrm{MeV}$ on a stationary target, the two outgoing alpha particles together carry approximately $Q+T=17.846\mathrm{MeV}$ before losses in the target and detector window.

Equal product masses imply equal kinetic energies and opposite momenta in the center of mass frame. The laboratory frame has nonzero initial momentum because the proton is moving. Conservation of momentum therefore forbids the two laboratory momenta from always being equal and opposite. The products are equal in laboratory kinetic energy only for particular emission directions, not for every event.

An exact two-body calculation makes this distinction quantitative. Write $a=m_p{c}^2$, $b=m_{\mathrm{Li}}{c}^2$, $h=m_{\alpha }{c}^2$, and $T$ for incident kinetic energy. Express every quantity below in the same energy units; $\mathcal{P}=pc$ is momentum times the speed of light. For a stationary target,Here $W$ is the center-of-mass energy, and $\beta$ is the center-of-mass speed divided by the speed of light. Each outgoing alpha particle has center-of-mass total energy $E_{\ast }=W/2$ and momentum-times-$c$ magnitude ${\mathcal{P}}_{\ast }=\sqrt{E_{\ast }^2-h^2}$. A Lorentz transformation gives the paired laboratory kinetic energies, with ${\theta }_{\ast }$ measured from the incident beam in the center of mass frame:Their sum is independent of ${\theta }_{\ast }$ and equals $Q+T$. Their difference is $2\gamma \beta {\mathcal{P}}_{\ast }\cos {\theta }_{\ast }$; the symmetry belongs to the product interchange and the center of mass frame.

For the chosen $T=0.500\mathrm{MeV}$ model, subtract $Z$ electron rest masses from each neutral-atom mass, neglect electronic binding energies, and use $m_e{c}^2=0.51099895069\mathrm{MeV}$.^[ref] At ${\theta }_{\ast }=0$, the paired laboratory kinetic energies are approximately $9.979$ and $7.867\mathrm{MeV}$; at ${\theta }_{\ast }=90^{\circ }$ both are approximately $8.923\mathrm{MeV}$. In the latter case both products have a forward laboratory momentum component, so even equal laboratory kinetic energies do not imply back-to-back laboratory trajectories. A real thick target and finite detector acceptance would broaden or select these ideal kinematic possibilities.

Formation of a deuteron from a neutron and a proton with photon emission. For an initial pair at rest, two-body energy and momentum conservation gives $E_{\gamma }=B-B^2/[2(m_p+m_n)]$ in natural units, where $B=m_p+m_n-m_d$. The missing binding energy becomes nuclear recoil.

Take the hydrogen ground-state binding-energy magnitude as $E_B\simeq 13.6\mathrm{eV}=2.18\times 10^{-18}\mathrm{J}$, and use $a_0\simeq 5.29\times 10^{-11}\mathrm{m}$ as its characteristic radius. A definite volume convention isIt givesA quantum atom has no sharp outer surface. If instead its characteristic volume is taken as $a_0^3$, the estimate is $1.47\times 10^{13}\mathrm{J}{\mathrm{m}}^{-3}$ and the ratio is about $4.2\times 10^2$. Both conventions place the two densities roughly three orders of magnitude apart at order-of-magnitude accuracy. The comparison uses the magnitude of the binding energy, not its negative sign relative to ionization.