University Physics IV · Atomic Physics · 11.9
Inner-Shell X-rays & Moseley's Law
Knock an electron out of the innermost shell and the hole closes again in about a femtosecond, sometimes by emitting a photon. Those photons carry the sharpest fingerprint an element has, and reading them is how the periodic table got its ordering and how an electron microscope names what it is looking at.
Build the model
Connect the measurement to the mechanism.
Strip a 1s electron out of a heavy atom and the thirty or forty electrons outside barely matter. The hole sits inside every other shell, so an electron falling into it from n = 2 travels in an almost pure Coulomb field of charge Z − 1: the nucleus, minus the one 1s partner still sitting there. That is the entire content of Moseley's law.
A hydrogenic formula written for one electron, with Z replaced by Z − σ and a single number σ standing in for the whole rest of the atom, reproduces K-series frequencies across forty elements to about one percent. The payoff is that the square root of frequency plotted against atomic number is a straight line, so one line in a spectrum reads off Z with no chemistry at all, and Z rather than atomic weight is what orders the periodic table. The cost is that σ is a fitted stand-in.
It carries no information about configuration, it conceals the spin-orbit doublet that splits every K line in two, and it ignores relativity, so the straight line bends upward past Z of about 50. And the photon is never guaranteed: the same vacancy can be filled by ejecting an outer electron instead, and in light atoms that Auger channel wins more than nine times out of ten.
- Simple definition
- Moseley's law states that the square root of a characteristic X-ray frequency rises linearly with atomic number, √ν = A(Z − σ), where σ is one screening constant standing in for every other electron in the atom.
- Example
- For the Kα line σ = 1, so E = 10.204 eV × (Z − 1)². Copper, Z = 29, gives 10.204 × 784 = 8.00 keV against a measured 8.05 keV — a 0.6% error from a one-electron formula applied to a 29-electron atom.
One measured wavelength returns an integer: the line names the element with no chemistry involved.
ν in Hz, Z the atomic number; σ = 1 for the K series, 7.4 for the L series
Molybdenum, Z = 42: 10.204 × 41² = 17.15 keV, against 17.48 keV measured.
R∞hc = 13.606 eV; the 3/4 is 1/1² − 1/2², the n = 2 → 1 jump
Tungsten, Z = 74: 1.890 × 66.6² = 8.38 keV, within 0.2% of the measured 8.40 keV.
5/36 = 1/2² − 1/3²; σ = 7.4 stands for the K and L electrons already in place
The continuum stops where one electron gives a single photon everything it has: 60 kV → 20.7 pm.
V₀ the tube voltage in kV, λₘᵢₙ in nm; hc = 1239.84 eV⋅nm. No Z appears.
Zinc, Z = 30: 0.474 predicted, 0.474 measured. Aluminium radiates only 3.6% of its K vacancies.
Dimensionless, 0 to 1; Γ are level widths in eV. The fit holds above Z ≈ 20.
The background a characteristic line has to stand above, and why heavy anodes at high kV are bright.
Intensity per unit photon energy from the anode; integrating gives total ∝ Z V₀²
Two spectra come off one anode
Point 70 keV electrons at a metal anode and two different things come back. Most electrons decelerate in the nuclear fields and radiate a broad continuum — bremsstrahlung — whose intensity per unit photon energy Kramers wrote as I(E) ∝ Z(Eₘₐₓ − E). It stops dead at Eₘₐₓ = eV₀, the Duane-Hunt limit, because no photon can carry more than the whole kinetic energy of one electron; at 60 kV that is 60 keV, or λₘᵢₙ = 20.7 pm. Standing on that continuum are a few narrow lines whose energies do not move when you change the voltage. Raise V₀ and the cut-off slides to higher energy while the whole continuum brightens as V₀²; the lines only grow taller. Change the anode metal and the lines jump somewhere else entirely. That single observation — position set by the target, not by the beam — is what tells you the lines come from the atom's own level structure and not from the collision.
Why a one-electron formula works on a fifty-electron atom
A K vacancy is a hole in the innermost shell of an atom that may have fifty electrons outside it. An electron dropping from n = 2 into that hole spends its whole journey inside every other shell, where the enclosed charge is the nucleus plus the one 1s electron still present: Z − 1. So the transition is hydrogenic, and E(Kα) = (3/4)(13.606 eV)(Z − 1)². Copper: 10.204 × 28² = 8.00 keV, measured 8.05. Molybdenum: 10.204 × 41² = 17.15 keV, measured 17.48. Tungsten: 10.204 × 73² = 54.4 keV, measured 59.3. The error runs from 0.6% at Z = 29 to 8.3% at Z = 74, and at the heavy end it is not a screening problem — it is relativity. The 1s electron of tungsten moves at roughly Zα = 74/137 = 0.54 c, so it is bound more deeply than a non-relativistic formula allows, and Moseley's straight line curves upward.
Moseley's straight line, and what it settled
Moseley photographed the K lines of about forty elements in 1913-14 and plotted √ν against atomic number. The points fell on a straight line crossing the axis at Z = 1, with slope √((3/4)cR∞) = 4.97 × 10⁷ Hz^½. Two consequences followed at once. First, the abscissa is an integer you can count, so the ordering principle of the periodic table is nuclear charge and not atomic weight — which puts cobalt before nickel even though cobalt is the heavier of the two, 58.93 against 58.69. Second, a gap in the line is a missing element whose line position you can predict in advance; the holes at Z = 43, 61, 72 and 75 were later filled by technetium, promethium, hafnium and rhenium. Chemistry had ordered the elements by similarity and mass. One photographic plate replaced that with counting.
You must make the vacancy first: the K edge is the threshold
No line appears until something creates the hole. The tube must supply at least the K-shell binding energy EK, which is the same energy as the K absorption edge — the step in the attenuation curve where photons suddenly become able to eject a 1s electron. For molybdenum EK = 20.0 keV, so a 15 kV tube produces bremsstrahlung and no K lines at all, however long you run it. Above threshold the characteristic intensity grows roughly as (V₀ − VK)1.6, which is why analytical work uses an overvoltage of two to three rather than sitting just above the edge. The edge always lies above every line in its own series: Kβ is the 3p → 1s jump, and the series limit, ∞ → 1s, is the edge itself. Molybdenum: Kα 17.5 keV, Kβ 19.6 keV, edge 20.0 keV.
Selection rules decide which lines exist, and the doublet
Inner-shell transitions obey the same electric-dipole rules as optical ones: Δl = ±1 and Δj = 0, ±1. That is why the strong K lines come from 2p and 3p, and why there is no 2s → 1s line at all, even though the 2s electron is right there and the energy would be within a few tens of eV of Kα. The same rules predict the doublet. Spin-orbit coupling splits the 2p level into 2p₃/₂ and 2p₁/₂, giving Kα₁ and Kα₂ with intensities in the ratio of their degeneracies 2j + 1, that is 4:2, or 2:1. The splitting climbs steeply with Z — 20 eV in copper, 105 eV in molybdenum, 1.34 keV in tungsten — so a spectrometer that shows one copper Kα line will cleanly resolve two in tungsten.
In light atoms the photon is the minority channel
Filling the hole does not have to make a photon. The atom can hand the released energy to another electron and eject it instead: an Auger electron, leaving a doubly ionised atom. The two channels compete, and the fluorescence yield ωK is the photon's share. Auger rates are almost independent of Z while the radiative rate climbs roughly as Z⁴, so ωK ≈ Z⁴/(Z⁴ + 9.0 × 10⁵) to within a few percent above Z ≈ 20. Aluminium manages 0.036, thirty-six photons per thousand vacancies; copper 0.44; tungsten 0.96. This is the practical ceiling on X-ray fluorescence analysis of light elements, and the reason Auger spectroscopy is the technique of choice for them: what does not leave as a photon leaves as an electron, and an electron cannot escape from more than about a nanometre of solid, so it reports on the surface.
Change one variable at a time
Make the relationship visible.
Set Z to 74 and step the voltage up from 65 kV: only continuum until the tube passes the 69 keV K edge, then a Kα spike that climbs with the overvoltage. Sweep Z instead and both lines slide right as (Z − 1)² while the cut-off holds still and the continuum brightens.
Kα ENERGY17.15 keV
K-EDGE THRESHOLD20.69 keV
CUT-OFF WAVELENGTH0.0177 nm
OVERVOLTAGE V/EK3.38
Live interpretationKα ENERGY: 17.15 keV. K-EDGE THRESHOLD: 20.69 keV. CUT-OFF WAVELENGTH: 0.0177 nm. OVERVOLTAGE V/EK: 3.38
Catch the common trap
Explain before calculating.
A molybdenum-anode tube (Z = 42, Kα 17.5 keV, Kβ 19.6 keV, K edge 20.0 keV) is run at 18 kV. What comes out of it?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn unknown metal foil bombarded by electrons emits a strong characteristic line at wavelength 154.2 pm. Using E(Kα) = 10.204 eV × (Z − 1)², identify the element.
- Photon energy from wavelength: E = hc/λ = 1239.84 eV⋅nm ÷ 0.1542 nm = 8040 eV = 8.04 keV.
- Invert the Moseley form: (Z − 1)² = E / 10.204 eV = 8040 / 10.204 = 788.0.
- Take the square root: Z − 1 = √788.0 = 28.07, so Z = 29.07.
- Atomic number is an integer, so Z = 29 — copper. The stray 0.07 is the residual error left by describing 28 other electrons with the single screening constant σ = 1, not a fractional charge.
AnswerZ = 29, copper. 154.2 pm is the Cu Kα centroid, the standard laboratory diffraction wavelength.
MediumA tungsten-anode tube (Z = 74, K edge 69.5 keV) runs at 100 kV. Find the shortest wavelength it emits, estimate the Kα energy from Moseley's law and compare it with the measured 59.3 keV, and estimate what fraction of the K vacancies produce a photon rather than an Auger electron.
- Duane-Hunt limit: λₘᵢₙ = hc/(eV₀) = 1239.84 eV⋅nm ÷ 100 000 eV = 0.01240 nm = 12.40 pm. This depends on the tube voltage alone — a copper anode at 100 kV would give exactly the same cut-off.
- Moseley: E(Kα) = 10.204 eV × (74 − 1)² = 10.204 × 5329 = 54 400 eV = 54.4 keV.
- That is 8.3% below the measured 59.3 keV, and the gap is not screening but relativity: the 1s electron of tungsten moves at about Zα = 74/137 = 0.54 c, so it is bound more deeply than any non-relativistic hydrogenic formula allows.
- Threshold check: 100 kV exceeds the 69.5 keV K edge, so the K lines are excited, at an overvoltage of 100/69.5 = 1.44 — modest, so the lines will not stand far above the continuum.
- Fluorescence yield: 74⁴ = 3.00 × 10⁷, so ωK = 3.00 × 10⁷ / (3.00 × 10⁷ + 9.0 × 10⁵) = 0.97.
Answerλₘᵢₙ = 12.4 pm; Moseley gives 54.4 keV against 59.3 keV measured, 8.3% low because it ignores relativity; ωK ≈ 0.97, so nearly every K vacancy in tungsten radiates.
HardA spectrometer records Kα centroids of 6.400 keV from iron (Z = 26) and 17.44 keV from molybdenum (Z = 42). A third sample gives 22.10 keV. Fit √E = a(Z − σ) to the two standards, identify the unknown, and say how far the fit can be trusted.
- Take square roots of the standards: √6.400 = 2.5298 keV^½ at Z = 26, and √17.44 = 4.1761 keV^½ at Z = 42. Moseley's law says these and every other Kα point lie on one straight line.
- Slope: a = (4.1761 − 2.5298) / (42 − 26) = 1.6463/16 = 0.10289 keV^½ per unit Z.
- Screening constant, from either standard: σ = 26 − 2.5298/0.10289 = 26 − 24.587 = 1.41.
- Unknown: √22.10 = 4.7011 keV^½, so Z = σ + √E/a = 1.41 + 4.7011/0.10289 = 1.41 + 45.69 = 47.10.
- Z must be an integer, so Z = 47 — silver, whose measured Kα centroid is 22.105 keV. The 0.10 is fit error, not a fractional charge.
- σ came out 1.41 rather than the textbook 1 because one parameter is absorbing the Kα doublet, relativistic corrections and the limited Z range at once, so extrapolating this fit to Z = 80 would be unsafe. Interpolating is fine: the same fit puts palladium at 21.05 keV and cadmium at 22.98 keV, about 1 keV either side of silver.
AnswerZ = 47, silver. The two-point fit gives a = 0.1029 keV^½ per unit Z and σ = 1.41, and places the unknown at Z = 47.10, which must be read as the integer 47.