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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.

For emission take integers $n_i>n_f\ge 1$. Applying conservation of energy to the model's internal energy difference, while neglecting the much smaller whole-atom photon recoil, givesThus Johannes Robert Rydberg's Rydberg formula is a difference of two energy levels expressed as a spectroscopic wavenumber. The same constants determine every line; there is no new parameter for each electronic transition. With the chosen inputs, $R_H=1.096775834\times 10^7{\mathrm{m}}^{-1}$. All wavelengths below are calculated in vacuum, not air, and are rounded model predictions rather than historical measurements.

The conventional names commemorate Theodore Lyman and Johann Jakob Balmer. Fixing $n_f=1$ produces the ultraviolet Lyman series, beginning with the Lyman-alpha line; fixing $n_f=2$ produces the Balmer series. As $n_i\to \infty$, the wavelength approaches $n_f^2/R_H$: $91.1763\mathrm{nm}$ at the Lyman limit and $364.705\mathrm{nm}$ for the Balmer series. Crowding follows directly from $E_{n+1}-E_n\sim 2B_H/n^3$. Discreteness therefore coexists with a series limit.

Correct line positions do not give wave intensities, lifetimes or electric-dipole selection rules for hydrogen. Those require a theory of the wavefunctions and their coupling to electromagnetic radiation. The circular model also omits spin, relativistic corrections and the Lamb shift. Its successful gross energy levels do not make its assumed orbits modern atomic orbitals.

The ground electron configuration of caesium has a single $6s$ electron, with $J=1/2$. Addition of angular momentum giving $F=3,4$ fixes the nuclear spin to $I=7/2$. With dimensionless angular momentum operators, write the Fermi contact interaction asThe two hyperfine levels therefore differ by $4A$. In the convention ${\mathbf{\mu }}_I=g_I{\mu }_B\mathbf{I}$ and $g_s\simeq 2$,This exposes a missing ingredient in a prediction from the supplied data: $Z$ and $g_I$ alone do not specify the valence electron contact density. Screening effect and penetration of the closed core strongly affect it. If $g_I$ instead denotes the total nuclear moment in units of ${\mu }_B$, it must first be divided by $I$ in the displayed convention.

An explicitly crude unscreened hydrogen-like ion model uses $|{\psi }_{6s}(0){|}^2=Z^3/(\pi 6^3{a}_0^3)$ and gives ${\nu }_0\simeq 5.76\times 10^{11}\mathrm{Hz}$, or ${\omega }_0\simeq 3.62\times 10^{12}{\mathrm{s}}^{-1}$. It describes neither the shielding nor the radial function of neutral caesium. Using a pure unit-charge $6s$ hydrogen orbital instead gives only $3.46\mathrm{MHz}$, illustrating why a unique useful estimate cannot follow from an unspecified effective charge.

For physical neutral ${}^{133}\mathrm{Cs}$, the independently known zero-magnetic field clock transition isThis measured frequency, used to define the second, is supplementary physical information, not a consequence of $Z=55$ and the rounded $g_I$ alone.^[ref] The direct electronic transition is in the microwave range and is a magnetic dipole transition, rather than a direct optical electric dipole transition. An optical implementation needs a two-field Raman transition.

Using the corrected weak-pulse Ramsey spectrum of part (b), the peak at $\delta =0$ has height $(\Omega {\tau }_p{)}^2$. The quantum interference fringes have spacing $2\pi /T$ in angular frequency, with zeros at $(2k+1)\pi /T$, and the sinc function envelope has its first zeros at $\delta =\pm 2\pi /{\tau }_p$. For $T\gg {\tau }_p$, the central peak has approximately full width at half maximum $\pi /T$ in angular frequency. The clock transition between the two $M_F=0$ states has no first-order Zeeman energy, although a quadratic one remains.

Measure each energy level by the wavenumber $\tilde{E}=(E-E_g)/(hc)$ in ${\mathrm{m}}^{-1}$. An upward electronic transition of vacuum wavelength $\lambda$ adds $1/\lambda$. Treat the quoted wavelengths as vacuum values to their limited accuracy; air corrections would be smaller than the precision justified here.

The strong ground-state absorption must be $5s^2{}^1{S}_0\to 5s5p{}^1{P}_1$ at $461\mathrm{nm}$. The weak ground-state electronic transition at $689\mathrm{nm}$ must reach $5s5p{}^3{P}_1$: the electric-dipole selection rule on $J$ excludes the other two triplet state energy levels, while weak spin-orbit coupling admixes singlet state character and relaxes the approximate spin selection rule.

Assume the usual ordering $E({}^3{P}_0)<E({}^3{P}_1)<E({}^3{P}_2)$, as expected from a positive effective spin-orbit coupling constant. The three nearby strong spectral lines then join these energy levels to a common $5s6s{}^3{S}_1$ energy level. Larger lower-state energy means a smaller gap and longer wavelength, so $679,688,707\mathrm{nm}$ correspond to $J=0,1,2$ respectively. The remaining $1.12\mu \mathrm{m}$ spectral line joins $5s5p{}^1{P}_1$ and $5s6s{}^1{S}_0$. Assume the latter is the upper energy level. These assignments and orderings agree with independently tabulated strontium energy levels.^[ref]

Starting with the two ground-state gaps, the construction isThe resulting energy levels, rounded to avoid suggesting spectroscopic precision, are

Both $461\mathrm{nm}$ and $1.12\mu \mathrm{m}$ exhibit the normal Zeeman effect. Each connects two singlet state energy levels, with Landé g-factor $g_J=1$ for the $J=1$ energy level and no first-order shift for $J=0$. Their frequency shifts are $0,\pm {\mu }_{B}B/h$. The weak $689\mathrm{nm}$ spectral line also has three components, but $g({}^3{P}_1)=3/2$ gives shifts $0,\pm 3{\mu }_{B}B/(2h)$; a three-component pattern alone does not establish the normal Zeeman effect. The $679\mathrm{nm}$ spectral line likewise has three components with $g({}^3{S}_1)=2$.

The electronic transition starts from $A_{1g}$, so the electric-dipole test reduces to whether the final electronic irreducible representation occurs among $A_{2u}+E_{1u}$. The excited $E_{1u}$ state matches the $(x,y)$ dipole components and is therefore electric-dipole allowed.

The $B_{2u}$ state matches neither $z$ nor $(x,y)$ and is electric-dipole forbidden at the equilibrium geometry. Both excited states are singlets, so the difference is spatial symmetry rather than the spin selection rule. Inversion parity alone does not decide the result: both final states are ungerade, but only the matching vector representation is allowed.

In a simple metal, conduction electrons can move collectively in response to an incident electromagnetic wave. The Drude model gives, with time dependence $e^{-i\omega t}$,In the low-loss range $\gamma \ll \omega <{\omega }_p$, the relative permittivity is approximately $1-{\omega }_p^2/{\omega }^2<0$. The wavevector inside is predominantly imaginary, so the field is evanescent and penetrates only a short distance. Surface currents reradiate the field; little energy enters a good conductor. The normal-incidence Fresnel equations giveMany ordinary metals have their plasma frequency in the ultraviolet, placing visible light below this reflection edge. For $\omega >{\omega }_p$ the free-carrier contribution permits propagation; at $\omega \gg {\omega }_p$, ${ϵ}_r\to 1$ and $R\sim {\omega }_p^4/(16{\omega }^4)\to 0$. The electrons' inertial response becomes too small to screen the rapidly oscillating field.

Ultraviolet transparency is a simple-metal model prediction, not a universal property of real metals. Electronic transitions between bands and bound-electron absorption can remain strong, shift the plasma edge and prevent transparency; transmission also depends on thickness and attenuation. The robust explanation is the change from a reflective negative-permittivity range to a possible propagating range above the plasma edge.

Take $q=-e$ as the signed electron charge. The Lorentz oscillator model is a driven damped bound charge:The first term is inertia, the second is a phenomenological frictional loss of coherent motion to other degrees of freedom, and the third is the linear restoring force from binding to the atom. The driving term is the electric force. The natural angular frequency ${\omega }_0$ represents the electronic transition; this classical oscillator describes its near-resonant linear response. The friction convention is $m\gamma \dot{u}$, so no factor two is hidden in $\gamma$.

For independent electronic transitions, add the electric dipole moment responses:The positive constants $S_j$ include electron density and transition atomic oscillator strength. Each term adds an absorption peak to ${ϵ}_r^{\prime \prime }$ and a dispersive positive-to-negative contribution to ${ϵ}_r^{\prime }-1$, as in the middle row of the figure.

Between well-separated absorption lines, where the medium is transparent, the refractive index generally increases with frequency: $d{n}_r/d\omega >0$, or $d{n}_r/d\lambda <0$. This is normal dispersion. Near a resonance the real response turns down with increasing frequency, producing an interval of anomalous dispersion, $d{n}_r/d\omega <0$, accompanied by appreciable absorption. The absorptive and dispersive parts are linked by the Kramers-Kronig relations; the steep variation is not independent of the loss spectrum.

The ruby R-line shift compares the same emission feature under two conditions. It is distinct from the Stokes shift, which compares absorption and emission features. In ruby, chromium ions in the corundum crystal provide the electronic transitions; compression changes their local environment and transition energies.^[ref]

For consistently defined vacuum wavelengths, the reciprocal relation givesThus a positive wavelength shift means a negative reciprocal shift, and $E=hc/\lambda$ gives lower photon energy; $h$ is the Planck constant and $c$ the speed of light. For spectroscopic wavenumber, use vacuum wavelength and keep that convention consistent. A calibration specified in air wavelength must retain its convention or be converted consistently. The reciprocal relation is nonlinear: a line exactly linear in $\Delta \lambda$ is not exactly linear in $\Delta (1/\lambda )$.

For a purely mathematical check with chosen vacuum wavelengths ${\lambda }_0=700.0\mathrm{nm}$ and $\Delta \lambda =2.0\mathrm{nm}$, the reciprocal shift is $-40.7000\mathrm{c}{\mathrm{m}}^{-1}$. The small-shift approximation $-\Delta \lambda /{\lambda }_0^2$ gives $-40.8163\mathrm{c}{\mathrm{m}}^{-1}$, about $0.286\%$ too large in magnitude. These chosen values illustrate the conversion; they are not measured ruby lines or an additional pressure calibration.

An otherwise forbidden electronic transition gains intensity when accompanying molecular vibration supplies a symmetry factor that permits a nonzero electric dipole transition moment. A symmetry-allowed total vibronic state specifies a possible transition, while the coupling determines its actual strength.

X-ray fluorescence is Characteristic X-ray emission following excitation by incident photons. In a simple initially unexcited chemical atom, photoionization can eject an inner electron if the incident photon supplies its binding energy. An outer electron can then fill the vacancy, releasing an emitted photon with the difference of shell binding energies. Nonradiative relaxation competes with this route, so one absorbed incident photon need not yield one escaping characteristic photon.

Choose a separate stipulated electron shell model with values $B_K=10.0$, $B_L=1.5$ and $B_M=0.3$, all in $\mathrm{keV}$. Ignore subshell splitting, recoil and changes of binding energy during relaxation. The possible shell differences areThese labels denote simplified shell groups, not resolved subshell electronic transitions. With $hc=1.23984\mathrm{keV}\mathrm{nm}$, their calculated wavelengths are respectively $0.145864$, $0.127819$ and $1.03320\mathrm{nm}$. An incident $9.8\mathrm{keV}$ photon exceeds both K-line energies but cannot directly create the K vacancy in this initially unexcited model. A $12.0\mathrm{keV}$ photon can, leaving approximately $2.0\mathrm{keV}$ of photoelectron kinetic energy. This comparison fixes neither the excitation probability nor the radiative branching probability.

An X-ray tube accelerates electrons from a negative cathode to a positive target anode. A heated cathode supplies electrons by thermionic emission; the evacuated envelope limits collisions before the target. This differs from the gas-discharge tube in the Röntgen X-ray discovery experiment.^[ref]

In bremsstrahlung, an electron changes velocity in the target's electric fields, emitting a photon while losing kinetic energy. Different encounters allow a continuous range of transferred energies. This is distinct from the discrete electronic transitions producing Characteristic X-rays.

Changing the accelerating voltage changes the available kinetic energy per electron and the X-ray tube cutoff. Changing the number of arriving electrons changes the number of opportunities for photon emission. These are distinct controls: a larger electric current does not by itself give each electron more energy. For an original idealized budget, $V=50\mathrm{kV}$ and tube electric current $i=0.20\mathrm{mA}$ give electrical power $Vi=10\mathrm{W}$ and arrival rate $i/e\simeq 1.25\times 10^{15}{\mathrm{s}}^{-1}$, where $e$ is the elementary charge. This is input power, not emitted X-ray power; obtaining the latter requires a conversion efficiency.